前言
网传的七天学Python的路线如下，我觉得可以在学过此表中前几天的内容后，就可以回头来学习一下列表推导式：它综合了列表、for循环和条件语句。
第一天：基本概念(4小时) ： print,变量，输入，条件语句。
第二天：基本概念(5小时) ：列表，for循环，while循环，函数，导入模块。
第三天：简单编程问题(5小时) ：交换两个变量值，将摄氏度转换为华氏温度，求数字中各位数之和， 判断某数是否为素数， 生成随机数，删除列表中的重复项等等。
第四天：中级编程问题(6小时) ：反转-个字符串(回文检测)，计算最大公约数，合并两个有序数组，猜数字游戏，计算年龄等等。
第五天：数据结构(6小时) ：栈，队列，字典，元组，树，链表。
第六天：面向对象编程(OOP) (6小时) ：对象，类，方法和构造函数，面向对象编程之继承。
第七天：算法(6小时) ：搜索(线性和二分查找)、 排序(冒泡排序、 选择排序)、递归函数(阶乘、斐波那契数列)时间复杂度(线性、二次和常量)。
列表推导式
list comprehension或译为列表解析式，是一种创建列表的简洁语法；也可认为它是一个简版的for循环，但执行效率高于for循环。python 2.7+ 开始又引入了集合推导式、字典推导式，原理与列表推导式相近。
语法规范：
out_list = [out_express for out_express in input_list if out_express_condition]
其中，if 条件可有可无；for 循环可以嵌套多层，内外层循环的变量不可以同名；
推导式中也可以嵌套推导式，内外层推导式的变量互不影响，可以同名；
推导表达式out_express尽可能用内置函数，省得import或def function()。
入门实例
>>> [i for i in range(20)] [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19] >>> [i for i in range(40) if i%2==0] [0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38] >>> [i*2 for i in range(20)] [0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38] >>> [i+j for i in range(5) for j in range(5)] [0, 1, 2, 3, 4, 1, 2, 3, 4, 5, 2, 3, 4, 5, 6, 3, 4, 5, 6, 7, 4, 5, 6, 7, 8] >>> [i+j for i in range(10) for j in range(10)] [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18] >>> {i+j for i in range(10) for j in range(10)} {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18} >>> [(i,j,k) for i in 'abc' for j in range(2) for k in range(2)] [('a', 0, 0), ('a', 0, 1), ('a', 1, 0), ('a', 1, 1), ('b', 0, 0), ('b', 0, 1), ('b', 1, 0), ('b', 1, 1), ('c', 0, 0), ('c', 0, 1), ('c', 1, 0), ('c', 1, 1)] >>> [chr(i) for i in range(97,123)] ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z'] >>> {i:ord(i) for i in [chr(i) for i in range(97,123)]} {'a': 97, 'b': 98, 'c': 99, 'd': 100, 'e': 101, 'f': 102, 'g': 103, 'h': 104, 'i': 105, 'j': 106, 'k': 107, 'l': 108, 'm': 109, 'n': 110, 'o': 111, 'p': 112, 'q': 113, 'r': 114, 's': 115, 't': 116, 'u': 117, 'v': 118, 'w': 119, 'x': 120, 'y': 121, 'z': 122} >>> dic={i:ord(i) for i in [chr(i) for i in range(97,123)]} >>> dic['x'] 120 >>>
注：
列表推导式外用‘[...]’，换成‘{...}’就是集合推导式；有键值对就是字典推导式。
推导式外用‘(...)’会得到一个“生成器”，如果需要“元组推导式”另要用tuple()函数转换；
生成器还有一个特性，只能被遍历一次，遍历过后就会被清空。
>>> (i for i in range(20)) <generator object <genexpr> at 0x0000000002CF3890> >>> type(i for i in range(20)) <class 'generator'> >>> tuple(i for i in range(20)) (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19) >>> >>> g = (i for i in range(10)) >>> g <generator object <genexpr> at 0x03CE71E8> >>> a = [i for i in g] >>> a [0, 1, 2, 3, 4, 5, 6, 7, 8, 9] >>> b = [i for i in g] >>> b [] >>>
生成过程：
>>> [i for i in range(10)]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
#这个推导式用for循环赋值的代码：
1. my_list = [] 2. for i in range(10): 3. my_list.append(i) 4. print(my_list)
注：当然这么简单的列表，可以不用推导式更不需要写代码来生成。
python有更加省事的方法来直接赋值：
方法一：
1. >>> a = []; a[:] = range(10) 2. >>> a 3. [0, 1, 2, 3, 4, 5, 6, 7, 8, 9] 4. >>>
方法二：此方法是我在学习用星号“*”给序列解包时偶尔试出来的
>>> *a,=range(10) # 此处变量a后的逗号“,”必不可少 >>> a [0, 1, 2, 3, 4, 5, 6, 7, 8, 9] >>> >>> # 或者： >>> a = [*range(10)] >>> a [0, 1, 2, 3, 4, 5, 6, 7, 8, 9] >>> >>> # 若要取部分，把不要的元素“赋值”给下划线 _ >>> _,*a=range(10) >>> a [1, 2, 3, 4, 5, 6, 7, 8, 9] >>> >>> # 元组的赋值，结尾用逗号： >>> tp = *range(1,11), >>> tp (1, 2, 3, 4, 5, 6, 7, 8, 9, 10) >>>
序列解包 * ：
>>> Cards = [*range(2,10),*list('TJQKA')] >>> Cards [2, 3, 4, 5, 6, 7, 8, 9, 'T', 'J', 'Q', 'K', 'A'] >>> Cards = *range(2,10),*tuple('TJQKA') >>> Cards (2, 3, 4, 5, 6, 7, 8, 9, 'T', 'J', 'Q', 'K', 'A') >>> # tuple 赋值时连最外的()都可以省掉
map()函数解包 *
1. >>> n = 12345678 2. >>> [int(i) for i in str(n)] 3. [1, 2, 3, 4, 5, 6, 7, 8] 4. >>> [*map(int,str(n))] 5. [1, 2, 3, 4, 5, 6, 7, 8] 6. >>>
还有一个特别的：单循环的变量可用 _ 代替：
>>> [_ for _ in range(10)] [0, 1, 2, 3, 4, 5, 6, 7, 8, 9] >>> [_*_ for _ in range(10)] [0, 1, 4, 9, 16, 25, 36, 49, 64, 81] >>> [str(_) for _ in range(2,10)] ['2', '3', '4', '5', '6', '7', '8', '9'] >>>
列表元素间的运算：
# 加法 >>> listAdd = lambda a,b:[i+j for m,i in enumerate(a) for n,j in enumerate(b) if m==n] >>> a = [1,2,3,4] >>> b = [2,4,7,11,16] >>> listAdd(a,b) [3, 6, 10, 15] >>> listAdd([0]+a,b) [2, 5, 9, 14, 20] >>> listAdd(a,b[1:]) [5, 9, 14, 20] >>> listAdd(a[1:],b) [4, 7, 11] >>>
# 其它运算 >>> listOP = lambda a,b,c=0:[i-j if c==2 else i*j if c==3 else i/j if c==4 else i//j if c==5 else i%j if c==6 else i+j for m,i in enumerate(a) for n,j in enumerate(b) if m==n] >>> a = [1,2,3,4] >>> b = [2,4,7,11,16] >>> listOP(a,b) [3, 6, 10, 15] >>> listOP(a,b,1) [3, 6, 10, 15] >>> listOP(a,b,2) [-1, -2, -4, -7] >>> listOP(b,a,2) [1, 2, 4, 7] >>> listOP(a,b,2) [-1, -2, -4, -7] >>> listOP(a,b,3) [2, 8, 21, 44] >>> listOP(a,b,4) [0.5, 0.5, 0.42857142857142855, 0.36363636363636365] >>> listOP(b,a,4) [2.0, 2.0, 2.3333333333333335, 2.75] >>> listOP(a,b,5) [0, 0, 0, 0] >>> listOP(b,a,5) [2, 2, 2, 2] >>> listOP(b[1:],a,5) [4, 3, 3, 4] >>> listOP(a,b,6) [1, 2, 3, 4] >>> listOP(b,a,6) [0, 0, 1, 3] >>> listOP(b[2:],a,6) [0, 1, 1] >>>
初阶实例
1. 1000~2021中包含7的数字有多少
>>> sum([1 for i in range(1000,2022) if '7' in str(i)]) 273 >>> [i for i in range(1000,2022) if '7' in str(i)] [1007, 1017, 1027, 1037, 1047, 1057, 1067, 1070, 1071, 1072, 1073, 1074, 1075, 1076, 1077, 1078, 1079, 1087, 1097, 1107, 1117, 1127, 1137, 1147, 1157, 1167, 1170, 1171, 1172, 1173, 1174, 1175, 1176, 1177, 1178, 1179, 1187, 1197, 1207, 1217, 1227, 1237, 1247, 1257, 1267, 1270, 1271, 1272, 1273, 1274, 1275, 1276, 1277, 1278, 1279, 1287, 1297, 1307, 1317, 1327, 1337, 1347, 1357, 1367, 1370, 1371, 1372, 1373, 1374, 1375, 1376, 1377, 1378, 1379, 1387, 1397, 1407, 1417, 1427, 1437, 1447, 1457, 1467, 1470, 1471, 1472, 1473, 1474, 1475, 1476, 1477, 1478, 1479, 1487, 1497, 1507, 1517, 1527, 1537, 1547, 1557, 1567, 1570, 1571, 1572, 1573, 1574, 1575, 1576, 1577, 1578, 1579, 1587, 1597, 1607, 1617, 1627, 1637, 1647, 1657, 1667, 1670, 1671, 1672, 1673, 1674, 1675, 1676, 1677, 1678, 1679, 1687, 1697, 1700, 1701, 1702, 1703, 1704, 1705, 1706, 1707, 1708, 1709, 1710, 1711, 1712, 1713, 1714, 1715, 1716, 1717, 1718, 1719, 1720, 1721, 1722, 1723, 1724, 1725, 1726, 1727, 1728, 1729, 1730, 1731, 1732, 1733, 1734, 1735, 1736, 1737, 1738, 1739, 1740, 1741, 1742, 1743, 1744, 1745, 1746, 1747, 1748, 1749, 1750, 1751, 1752, 1753, 1754, 1755, 1756, 1757, 1758, 1759, 1760, 1761, 1762, 1763, 1764, 1765, 1766, 1767, 1768, 1769, 1770, 1771, 1772, 1773, 1774, 1775, 1776, 1777, 1778, 1779, 1780, 1781, 1782, 1783, 1784, 1785, 1786, 1787, 1788, 1789, 1790, 1791, 1792, 1793, 1794, 1795, 1796, 1797, 1798, 1799, 1807, 1817, 1827, 1837, 1847, 1857, 1867, 1870, 1871, 1872, 1873, 1874, 1875, 1876, 1877, 1878, 1879, 1887, 1897, 1907, 1917, 1927, 1937, 1947, 1957, 1967, 1970, 1971, 1972, 1973, 1974, 1975, 1976, 1977, 1978, 1979, 1987, 1997, 2007, 2017] >>>
1000~2021中“包含7且能被7整除”的数字有多少
>>> sum([1 for i in range(1000,2022) if '7' in str(i) and i%7==0]) 39 >>> [i for i in range(1000,2022) if '7' in str(i) and i%7==0] [1057, 1071, 1078, 1127, 1176, 1197, 1267, 1274, 1337, 1372, 1379, 1407, 1470, 1477, 1547, 1575, 1617, 1673, 1687, 1701, 1708, 1715, 1722, 1729, 1736, 1743, 1750, 1757, 1764, 1771, 1778, 1785, 1792, 1799, 1827, 1876, 1897, 1967, 1974] >>>
小于1000000的所有正整数一共包含有多少个数字“7”
>>> num=lambda n:sum([str(i).count('7') for i in [i for i in range(1,n+1)] if '7' in str(i)]) >>> num(999999) 600000 >>> # if '7' in str(i) 可省掉，即0也合计结果一样
2. 求所有在100到1000之间的水仙花数
水仙花数定义：指一个正整数的各位数字的立方和等于其本身。
(1). 通常的解法，条件表达式比较麻，如果是10位数呢
>>> for i in range(100,1000): if i==(i //100)**3 + (i//10%10)**3 + (i%10)**3: print(i, end=' ') 153 370 371 407 >>> >>> # 改写成列表推导式： >>> [i for i in range(100,1000) if i==(i //100)**3 + (i//10%10)**3 + (i%10)**3] [153, 370, 371, 407] >>>
(2). 把数字转成字符串，然后遍历计算立方和
>>> >>> for i in range(100,1000): k=0 for j in str(i): k+=int(j)**3 if k==i: print(i,end=' ') 153 370 371 407 >>> >>> # 转成列表推导式： >>> [n for i,n in enumerate([sum([int(i)**3 for i in str(j)]) for j in range(100,1000)]) if i+100==n] [153, 370, 371, 407] >>>
3. 一维与二维列表间的互转
>>> *a,=range(1,10) >>> a [1, 2, 3, 4, 5, 6, 7, 8, 9] >>> b=[a[i:i+3] for i in range(0,len(a),3)] >>> b [[1, 2, 3], [4, 5, 6], [7, 8, 9]] >>> >>> c=[j for i in b for j in i] >>> c [1, 2, 3, 4, 5, 6, 7, 8, 9] >>> >>> sum(b,[]) [1, 2, 3, 4, 5, 6, 7, 8, 9] >>> # 很高级的方法 # pythonic
4. 实现二维列表的转置
行列互换，首行变首列，尾行变尾列，如下所示：
''''''''''''''''''' [ [1, 2, 3], [4, 5, 6], [7, 8, 9] ] ↓↓↓ [ [1, 4, 7], [2, 5, 8], [3, 6, 9] ] 推导式如下： ''''''''''''''''''' >>> arr=[[1,2,3], [4,5,6], [7,8,9]] >>> [[arr[i][j] for i in range(len(arr))] for j in range(len(arr[0]))] [[1, 4, 7], [2, 5, 8], [3, 6, 9]] >>>
使用zip()函数：优点不用考虑数组的行数和列数，但直接结果是元组的列表，需转换下
>>> arr=[[1,2,3], [4,5,6], [7,8,9]] >>> list(zip(*arr)) [(1, 4, 7), (2, 5, 8), (3, 6, 9)] >>> >>> arr=[[1,2,3], [4,5,6], [7,8,9]] >>> arr=[list(i) for i in zip(*arr)] >>> arr [[1, 4, 7], [2, 5, 8], [3, 6, 9]] >>> arr=[list(i) for i in zip(*arr)] >>> arr [[1, 2, 3], [4, 5, 6], [7, 8, 9]] >>> # 换一个3行4列的： >>> arr=[[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]] >>> arr=[list(i) for i in zip(*arr)] >>> arr [[1, 5, 9], [2, 6, 10], [3, 7, 11], [4, 8, 12]] >>> arr=[list(i) for i in zip(*arr)] >>> arr [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]] >>>
使用numpy库：特有numpy.array()可与list()相互转换
>>> import numpy as np >>> np.arange(1,10) array([1, 2, 3, 4, 5, 6, 7, 8, 9]) >>> # 或者： >>> np.array([*range(1,10)]) array([1, 2, 3, 4, 5, 6, 7, 8, 9]) >>> >>> list(np.arange(1,10)) [1, 2, 3, 4, 5, 6, 7, 8, 9] >>> >>> np.arange(1,10).reshape((3, 3)) array([[1, 2, 3], [4, 5, 6], [7, 8, 9]]) >>> arr=np.arange(1,10).reshape((3, 3)) >>> [list(i) for i in arr] [[1, 2, 3], [4, 5, 6], [7, 8, 9]] >>> >>> [j for i in arr for j in i] [1, 2, 3, 4, 5, 6, 7, 8, 9] >>> # 附：用numpy生成一个3行5列的随机数矩阵，数值范围[1,10)： >>> [list(i) for i in __import__('numpy').random.randint(1,10,(3,5))] [[6, 3, 6, 7, 8], [4, 9, 5, 8, 7], [8, 4, 1, 2, 1]] >>>
5. 求列表嵌套的最大深度
思路：遍历列表，只要还有一个元素是列表，就删除非列表元素然后进行降维；循环到所有元素都为非列表为止。代码如下，自定义函数就用了两个推导式一个循环语句
>>> def func(L): if not isinstance(L,list): # or: if type(L) is not list: return 0 k=1 while any([isinstance(i,list) for i in L]): k+=1 L=[j for i in [i for i in L if isinstance(i,list)] for j in i] return k >>> l = [1, 2, [3, [4, [5, 6], [7, [8], [[9, 10], [11, [12, 13, 14], 15]]]]]] >>> func(l) 7 >>>
注：内置函数 isinstance(i,list) 判断一个对象i是否为某个指定类型，等价于type(i) is list。例如： isinstance(123,int) 和 type(123) is int 都返回True。
6. 求斜边长小于100的勾股数组
代码如下，其中 A 有直角边互换的重复，B用条件a<b约束，C把约束条件放进 range() 函数中。
>>> A=[(a,b,c) for a in range(1,100) for b in range(1,100) for c in range(1,100) if a**2+b**2==c**2] >>> B=[(a,b,c) for a in range(1,100) for b in range(1,100) for c in range(1,100) if a<b and a**2+b**2==c**2] >>> C=[(a,b,c) for a in range(1,100) for b in range(a,100) for c in range(1,100) if a**2+b**2==c**2] >>> A==B False >>> len(A)==len(B)*2 True >>> B==C True >>> C [(3, 4, 5), (5, 12, 13), (6, 8, 10), (7, 24, 25), (8, 15, 17), (9, 12, 15), (9, 40, 41), (10, 24, 26), (11, 60, 61), (12, 16, 20), (12, 35, 37), (13, 84, 85), (14, 48, 50), (15, 20, 25), (15, 36, 39), (16, 30, 34), (16, 63, 65), (18, 24, 30), (18, 80, 82), (20, 21, 29), (20, 48, 52), (21, 28, 35), (21, 72, 75), (24, 32, 40), (24, 45, 51), (24, 70, 74), (25, 60, 65), (27, 36, 45), (28, 45, 53), (30, 40, 50), (30, 72, 78), (32, 60, 68), (33, 44, 55), (33, 56, 65), (35, 84, 91), (36, 48, 60), (36, 77, 85), (39, 52, 65), (39, 80, 89), (40, 42, 58), (40, 75, 85), (42, 56, 70), (45, 60, 75), (48, 55, 73), (48, 64, 80), (51, 68, 85), (54, 72, 90), (57, 76, 95), (60, 63, 87), (65, 72, 97)] >>>
7. 实现随机字符串(可作随机密码)
>>> import string,random >>> [''.join(random.sample(string.printable[:-6],10)) for _ in range(30)] ['\\T[~(]J#"+', '):0~He7Dam', 'zw=?>7a(&^', 'v<c@W:!&VP', 'y\\~W6{u:P1', 'R)il~3p+;y', "PGQ_'{.k15", 'Z"^w=P&3{R', 'yMGR[g_65$', "4.)Q7$COd'", 'WTptgYS$Nj', 'Ra$4Lrvu2)', ',V$z.C8>L(', '/YwfR#ZuM@', '>~){Q7ayUo', 'Ol]54z|a;\\', 'Dp80fV,\\-@', '[kB{he98&r', "E]$'Q@R-`0", 'm{qMBRD.p2', '=.Is;r>%/x', 'o7zS{DQ~Tx', 'hH:E{s?#Gt', 'WB]`%\\f.FT', 'Mbxu&8YEN_', '5Et+3dGAf%', 'k5#o_]2Y?T', '$K3(yD7wvJ', '^5kJ*Nn:jz', '8,q7/Oyb*3'] >>> >>> [''.join(random.sample(string.ascii_letters+string.digits*5,10)) for _ in range(30)] ['ugON2AoS10', '3E62mQ2sP8', 'sL76c4Ppyj', 'hS967O15bX', '7n8580rq01', 'B75C178051', '8Mvc0g52wd', 'Zv08H3GED8', '158F1Kd36o', '914FM222TK', 'n5I5aqY66h', '91Tz8P5yMf', '22K9tPLoHn', 'gR5862BZj9', '319pO53389', 'z31R67r811', 'E1duG7mzPS', 't6kx344cCU', '3b66u5yOc3', '387s3bj031', 'J665322viO', 'N4Y76QmfO9', '9d4038O7fD', '2lQ8D41z3G', 'l03P7146G4', 'n716wj2b9c', '4av2g6dDb7', '6q65ro2z43', 'LJC77i56xq', 'hHBGA547a5'] >>>
不使用string库，只用字母、数字和下划线：
>>> pwd = lambda x=10:[''.join(__import__('random').sample((*map(chr,[*range(48,58),*range(65,91),*range(97,123)]),'_'),8)) for _ in range(x)] >>> >>> print(*pwd()) NVKfO5Du HxT2qSJF _GzK1kD3 KVw1OWjB Ob8fRswa MFqvpEWK fPDzuj8e ZndGFAs7 VHMp3FtX Hc7o642q >>> print(*pwd()) c2HEwvkn I9wH1Vjm yOCaqzNR pXMqRuDg nUfTKXuV Co5Ebq7g mBCkDco1 ieUzSTpu y1s8zVct OiK3GFTw >>> print(*pwd(5)) xTd69oJW Ob6pFsaq 4XW5lw_Y aHJxiZgr Z0VAGhNB >>> print(*pwd(8)) BP1bpzlv VBvnFQE5 kmFZSLid WyCpqvK_ vOyQlB4c VSY8q67y 8lkGBRbt _I4MfqJk
8. 一个四层嵌套的推导式：求k等差数
“k等差数”定义：任意相邻两位之间的差的绝对值都为 k 的正十进制整数。
给定整数的位数 n 和 等差值 k，求所有 k等差数：
>>> iscdnum = lambda n,k:[num for num in range(10**(n-1),10**n) if all([i==k for i in [abs(int(j[0])-int(j[1])) for j in [str(num)[i:i+2] for i in range(len(str(num))-1)]]])] >>> iscdnum(2,5) [16, 27, 38, 49, 50, 61, 72, 83, 94] >>> iscdnum(3,2) [131, 135, 202, 242, 246, 313, 353, 357, 420, 424, 464, 468, 531, 535, 575, 579, 642, 646, 686, 753, 757, 797, 864, 868, 975, 979] >>> iscdnum(3,3) [141, 147, 252, 258, 303, 363, 369, 414, 474, 525, 585, 630, 636, 696, 741, 747, 852, 858, 963, 969] >>> iscdnum(3,4) [151, 159, 262, 373, 404, 484, 515, 595, 626, 737, 840, 848, 951, 959] >>> iscdnum(3,9) [909] >>> iscdnum(5,2) [13131, 13135, 13531, 13535, 13575, 13579, 20202, 20242, 20246, 24202, 24242, 24246, 24642, 24646, 24686, 31313, 31353, 31357, 35313, 35353, 35357, 35753, 35757, 35797, 42020, 42024, 42420, 42424, 42464, 42468, 46420, 46424, 46464, 46468, 46864, 46868, 53131, 53135, 53531, 53535, 53575, 53579, 57531, 57535, 57575, 57579, 57975, 57979, 64202, 64242, 64246, 64642, 64646, 64686, 68642, 68646, 68686, 75313, 75353, 75357, 75753, 75757, 75797, 79753, 79757, 79797, 86420, 86424, 86464, 86468, 86864, 86868, 97531, 97535, 97575, 97579, 97975, 97979] >>>
进阶实例
1. 乘法口诀表
>>> lst=['%sx%s=%s'%(j,i,i*j) for i in range(1,10) for j in range(1,i+1)] >>> lst ['1x1=1', '1x2=2', '2x2=4', '1x3=3', '2x3=6', '3x3=9', '1x4=4', '2x4=8', '3x4=12', '4x4=16', '1x5=5', '2x5=10', '3x5=15', '4x5=20', '5x5=25', '1x6=6', '2x6=12', '3x6=18', '4x6=24', '5x6=30', '6x6=36', '1x7=7', '2x7=14', '3x7=21', '4x7=28', '5x7=35', '6x7=42', '7x7=49', '1x8=8', '2x8=16', '3x8=24', '4x8=32', '5x8=40', '6x8=48', '7x8=56', '8x8=64', '1x9=9', '2x9=18', '3x9=27', '4x9=36', '5x9=45', '6x9=54', '7x9=63', '8x9=72', '9x9=81']
列印时，要注意它的项数通项公式是： An=n(n+1)/2+1
>>> for i in range(9): for j in range(i+1): print(lst[i*(i+1)//2+j],end='\t' if i!=j else '\n') 1x1=1 1x2=2 2x2=4 1x3=3 2x3=6 3x3=9 1x4=4 2x4=8 3x4=12 4x4=16 1x5=5 2x5=10 3x5=15 4x5=20 5x5=25 1x6=6 2x6=12 3x6=18 4x6=24 5x6=30 6x6=36 1x7=7 2x7=14 3x7=21 4x7=28 5x7=35 6x7=42 7x7=49 1x8=8 2x8=16 3x8=24 4x8=32 5x8=40 6x8=48 7x8=56 8x8=64 1x9=9 2x9=18 3x9=27 4x9=36 5x9=45 6x9=54 7x9=63 8x9=72 9x9=81 >>>
或者用join()直接把 \t \n 插入列表拼接成字符串，然后输出：
>>> print('\n'.join(['\t'.join([f'{j}x{i}={i*j}' for j in range(1,i+1)]) for i in range(1,10)])) 1x1=1 1x2=2 2x2=4 1x3=3 2x3=6 3x3=9 1x4=4 2x4=8 3x4=12 4x4=16 1x5=5 2x5=10 3x5=15 4x5=20 5x5=25 1x6=6 2x6=12 3x6=18 4x6=24 5x6=30 6x6=36 1x7=7 2x7=14 3x7=21 4x7=28 5x7=35 6x7=42 7x7=49 1x8=8 2x8=16 3x8=24 4x8=32 5x8=40 6x8=48 7x8=56 8x8=64 1x9=9 2x9=18 3x9=27 4x9=36 5x9=45 6x9=54 7x9=63 8x9=72 9x9=81 >>> >>> # f'{j}x{i}={i*j}' 等价于 '%sx%s=%s'%(j,i,i*j)
2. 求100以内的质数（或称素数）
>>> [k[0] for k in [(j,sum([j%i==0 for i in range(2,j)])) for j in range(2,100)] if k[1]==0] [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97] >>> >>> # 等价于： >>> [k[0] for k in [(j,sum([0 if j%i else 1 for i in range(2,j)])) for j in range(2,100)] if k[1]==0] [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97] >>>
上面两种方法都是累加sum布尔值bool的个数来计算的，可以用any() all()函数代替：
>>> [k[0] for k in [(j,[j%i==0 for i in range(2,j)]) for j in range(2,100)] if not any(k[1])] [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97] >>> >>> # 等价于： >>> [k[0] for k in [(j,[j%i!=0 for i in range(2,j)]) for j in range(2,100)] if all(k[1])] [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97] >>>
使用filter()和map()函数：
[i for i in filter(lambda x:all(map(lambda p:x%p,range(2,x))), range(2,100))] [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97] >>>
条件反过来就是100以内的合数：
>>> [k[0] for k in [(j,[j%i==0 for i in range(2,j)]) for j in range(2,100)] if any(k[1])] [4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81, 82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96, 98, 99] >>> [k[0] for k in [(j,[j%i!=0 for i in range(2,j)]) for j in range(2,100)] if not all(k[1])] [4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81, 82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96, 98, 99] >>> [i for i in filter(lambda x:not all(map(lambda p:x%p,range(2,x))), range(2,100))] [4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81, 82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96, 98, 99] >>>
求1000以内的质回文数（即是质数又是回文数）
>>> Pr=[str(k[0]) for k in [(j,[j%i!=0 for i in range(2,j)]) for j in range(2,1000)] if all(k[1])] >>> [int(p) for p in Pr if p[::-1] in Pr] [2, 3, 5, 7, 11, 101, 131, 151, 181, 191, 313, 353, 373, 383, 727, 757, 787, 797, 919, 929]
求1000以内的数，满足本身和它的回文数同是质数
>>> pstr=[str(k[0]) for k in [(j,[j%i!=0 for i in range(2,j)]) for j in range(2,1000)] if all(k[1])] >>> [int(p) for p in pstr if p[::-1] in pstr] [2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, 97, 101, 107, 113, 131, 149, 151, 157, 167, 179, 181, 191, 199, 311, 313, 337, 347, 353, 359, 373, 383, 389, 701, 709, 727, 733, 739, 743, 751, 757, 761, 769, 787, 797, 907, 919, 929, 937, 941, 953, 967, 971, 983, 991] >>>
分解质因数(N<2022)
>>> P = [k[0] for k in [(j,sum([j%i==0 for i in range(2,j)])) for j in range(2,2022)] if k[1]==0] >>> def func(n): res=[] t=P[::-1] m=t[-1] while n>=m: if n%m==0: n//=m res.append(m) else: t.pop() m=t[-1] return res >>> i = 2021 >>> print(i,'=','x'.join([str(j) for j in func(i)])) 2021 = 43x47 >>> >>> func(2000) [2, 2, 2, 2, 5, 5, 5] >>> func(3999) [3, 31, 43] >>>
3. 求出字符串的所有字串（可推广到所有可切片数据类型）
>>> L='abcd' >>> [L[i:j] for i in range(len(L)) for j in range(i+1,len(L)+1)] ['a', 'ab', 'abc', 'abcd', 'b', 'bc', 'bcd', 'c', 'cd', 'd'] >>>
注：凡是可以用[i:j]来切片的“容器类”数据类型都可用此推导式。如：
>>> L=[1,2,3,4] >>> [L[i:j] for i in range(len(L)) for j in range(i+1,len(L)+1)] [[1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [2], [2, 3], [2, 3, 4], [3], [3, 4], [4]] >>>
(1). 找出字符串s='aaabcddcbddba'中最长的回文字串
>>> s='aaabcddcbddba' >>> {s[i:j] for i in range(len(s)) for j in range(i+1,len(s)+1)} {'bcddcbdd', 'ddcbddb', 'abcddcbdd', 'aabcddc', 'aabcddcbdd', 'cbddba', 'aaa', 'cddcbd', 'cbddb', 'a', 'cdd', 'bcdd', 'aabcddcbddba', 'aabcd', 'abcddcbddb', 'aaabc', 'ab', 'cbdd', 'cddcbdd', 'aaabcddcb', 'aabcddcbd', 'aaabcdd', 'ba', 'ddcbddba', 'dba', 'db', 'cddcbddba', 'cbd', 'aaabcddcbddb', 'aaabcd', 'ddb', 'dcbdd', 'abcddc', 'abcd', 'abcdd', 'bcddcb', 'aaabcddcbdd', 'abcddcbddba', 'aabc', 'bcddc', 'bdd', 'cb', 'bcddcbddba', 'c', 'dcbddb', 'ddba', 'dcbd', 'b', 'aaab', 'dd', 'd', 'ddcbd', 'bcd', 'aa', 'abcddcbd', 'bcddcbddb', 'aaabcddcbd', 'cddc', 'ddcb', 'dc', 'abc', 'bddb', 'ddc', 'bcddcbd', 'bc', 'aabcdd', 'aab', 'aaabcddcbddba', 'cddcb', 'abcddcb', 'cd', 'bddba', 'aabcddcbddb', 'bd', 'ddcbdd', 'aaabcddc', 'dcb', 'dcbddba', 'aabcddcb', 'cddcbddb'} >>> # 使用字典推导式可去掉相同子串 >>> >>> [i for i in {s[i:j] for i in range(len(s)) for j in range(i+1,len(s)+1)} if i==i[::-1]] ['aaa', 'a', 'bcddcb', 'c', 'b', 'dd', 'd', 'aa', 'cddc', 'bddb'] >>> >>> [(len(i),i) for i in {i for i in [s[i:j] for i in range(len(s)) for j in range(i+1,len(s)+1)} if i==i[::-1]]] [(3, 'aaa'), (1, 'a'), (6, 'bcddcb'), (1, 'c'), (1, 'b'), (2, 'dd'), (1, 'd'), (2, 'aa'), (4, 'cddc'), (4, 'bddb')] >>> >>> max([(len(i),i) for i in [i for i in {s[i:j] for i in range(len(s)) for j in range(i+1,len(s)+1)} if i==i[::-1]]])[1] 'bcddcb' >>>
(2). 给定 L=[2, -3, 3, 50, 5, 0, -1]，输出其子序列中各元素合计数最大的子序列
>>> L = [2, -3, 3, 50, 5, 0, -1] >>> [L[i:j] for i in range(len(L)) for j in range(i+1,len(L)+1)] [[2], [2, -3], [2, -3, 3], [2, -3, 3, 50], [2, -3, 3, 50, 5], [2, -3, 3, 50, 5, 0], [2, -3, 3, 50, 5, 0, -1], [-3], [-3, 3], [-3, 3, 50], [-3, 3, 50, 5], [-3, 3, 50, 5, 0], [-3, 3, 50, 5, 0, -1], [3], [3, 50], [3, 50, 5], [3, 50, 5, 0], [3, 50, 5, 0, -1], [50], [50, 5], [50, 5, 0], [50, 5, 0, -1], [5], [5, 0], [5, 0, -1], [0], [0, -1], [-1]] >>> [sum(j) for j in [L[j:i] for i in range(len(L),0,-1) for j in range(len(L))]] [56, 54, 57, 54, 4, -1, -1, 57, 55, 58, 55, 5, 0, 0, 57, 55, 58, 55, 5, 0, 0, 52, 50, 53, 50, 0, 0, 0, 2, 0, 3, 0, 0, 0, 0, -1, -3, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0] >>> # 合并为一行代码： >>> max(sum(j) for j in [L[j:i] for i in range(len(L),0,-1) for j in range(len(L))]) 58 >>> >>> # 和最大的子序列为： >>> l = [L[i:j] for i in range(len(L)) for j in range(i+1,len(L)+1)] >>> m = max(sum(j) for j in [L[j:i] for i in range(len(L),0,-1) for j in range(len(L))]) >>> [i for i in l if sum(i)==m] [[3, 50, 5], [3, 50, 5, 0]] >>>
4. 根据方程式画出字符图
略：见相关文章《探究“一行代码画爱心”的秘密，去向心爱的人表白吧》
5. EXCEL表格列号字串转整数
>>> ExcelCol2Int = lambda s:sum([(ord(n)-64)*26**i for i,n in enumerate(list(s)[::-1])]) >>> ExcelCol2Int('A') 1 >>> ExcelCol2Int('AA') 27 >>> ExcelCol2Int('AX') 50 >>> ExcelCol2Int('CV') 100 >>> ExcelCol2Int('AAA') 703 >>> ExcelCol2Int('XFD') 16384 >>>
6. 打印Gray格雷码序列
什么是格雷码，什么是卡诺图？不懂的问度娘吧
Gray = lambda n:[(i>>1)^i for i in range(2**n)] GrayB = lambda n:[bin((i>>1)^i)[2:].rjust(n,'0') for i in range(2**n)] ''' Gray(0) [0] Gray(1) [0, 1] Gray(2) [0, 1, 3, 2] Gray(3) [0, 1, 3, 2, 6, 7, 5, 4] Gray(4) [0, 1, 3, 2, 6, 7, 5, 4, 12, 13, 15, 14, 10, 11, 9, 8] >>> GrayB(3) # 把输出排成矩阵，即三变量卡诺图 ['000', '001', '011', '010', '110', '111', '101', '100'] >>> GrayB(4) # 把输出排成方阵，即四变量卡诺图 ['0000', '0001', '0011', '0010', '0110', '0111', '0101', '0100', '1100', '1101', '1111', '1110', '1010', '1011', '1001', '1000'] '''
注：上面代码中的 (i>>1)^i 可以写成： i^i>>1 或 i^i//2，因为右移或整除的运算级别都比异或要高。验证代码如下：
1. >>> any([(i>>1)^i == i^i>>1 == i^i//2 for i in range(323)]) 2. True 3. >>>
另外，观察到Gray()输出的整数序列的规律，就想到用迭代法也能实现，并且只要2行代码：
# 用迭代法实现：
# 用迭代法实现： def iGray(n): if n==0: return [0] return iGray(n-1)+[i+2**(n-1) for i in iGray(n-1)[::-1]]
高阶实例
1. 杨辉三角形
方法一：公式递推
>>> def func(i): t=L=[1] while(i>1): i-=1 t=L+[t[n]+t[n+1] for n in range(len(t)-1)]+L return t >>> func(1) [1] >>> func(3) [1, 2, 1] >>> func(8) [1, 7, 21, 35, 35, 21, 7, 1] >>> >>> for i in range(1,10):print(func(i)) [1] [1, 1] [1, 2, 1] [1, 3, 3, 1] [1, 4, 6, 4, 1] [1, 5, 10, 10, 5, 1] [1, 6, 15, 20, 15, 6, 1] [1, 7, 21, 35, 35, 21, 7, 1] [1, 8, 28, 56, 70, 56, 28, 8, 1] >>>
方法二：组合公式的自定义函数
>>> def Combin(n,i): m,t=min(i,n-i),1 for j in range(0,m): t*=(n-j)/(m-j) return t >>> [Combin(8,i) for i in range(9)] [1, 8.0, 28.0, 55.99999999999999, 70.0, 55.99999999999999, 28.0, 8.0, 1] >>> [round(Combin(8,i)) for i in range(9)] [1, 8, 28, 56, 70, 56, 28, 8, 1] >>> [[round(Combin(j,i)) for i in range(j+1)] for j in range(10)] [[1], [1, 1], [1, 2, 1], [1, 3, 3, 1], [1, 4, 6, 4, 1], [1, 5, 10, 10, 5, 1], [1, 6, 15, 20, 15, 6, 1], [1, 7, 21, 35, 35, 21, 7, 1], [1, 8, 28, 56, 70, 56, 28, 8, 1], [1, 9, 36, 84, 126, 126, 84, 36, 9, 1]] >>>
根据组合公式用阶乘来计算：
C(m,n)=math.factorial(n)//(math.factorial(m)*math.factorial(n-m))
递归法，虽然没有小数精度的问题，但也有递归次数不能太大即n值有限制的缺点：
>>> def Comb(n,i): if i in [0,n]: return 1 elif i==1: return n else: return Comb(n-1,i-1)+Comb(n-1,i) >>> [Comb(8,i) for i in range(9)] [1, 8, 28, 56, 70, 56, 28, 8, 1] >>> print('\n'.join(['\t'.join([str(Comb(j,i)) for i in range(j+1)]) for j in range(10)])) 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1 >>>
方法三：使用现成的库函数
(1). itertools库combinations函数
>>> from itertools import combinations as comb >>> [[len(list(comb(range(j),i))) for i in range(j+1)] for j in range(10)] [[1], [1, 1], [1, 2, 1], [1, 3, 3, 1], [1, 4, 6, 4, 1], [1, 5, 10, 10, 5, 1], [1, 6, 15, 20, 15, 6, 1], [1, 7, 21, 35, 35, 21, 7, 1], [1, 8, 28, 56, 70, 56, 28, 8, 1], [1, 9, 36, 84, 126, 126, 84, 36, 9, 1]] >>> print('\n'.join(['\t'.join([str(len(list(comb(range(j),i)))) for i in range(j+1)]) for j in range(10)])) 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1 >>>
(2). scipy库comb函数
>>> from scipy.special import comb >>> [[round(comb(j,i)) for i in range(j+1)] for j in range(10)] [[1], [1, 1], [1, 2, 1], [1, 3, 3, 1], [1, 4, 6, 4, 1], [1, 5, 10, 10, 5, 1], [1, 6, 15, 20, 15, 6, 1], [1, 7, 21, 35, 35, 21, 7, 1], [1, 8, 28, 56, 70, 56, 28, 8, 1], [1, 9, 36, 84, 126, 126, 84, 36, 9, 1]] >>> print('\n'.join(['\t'.join([str(round(comb(j,i))) for i in range(j+1)]) for j in range(10)])) 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1 >>>
2. 斐波那契数列
(1). 引入自定义函数或lambda表达式：
>>> f=lambda n:n<3 and 1 or f(n-1)+f(n-2) >>> [f(i) for i in range(1,30)] [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229] >>>
注：这个lambda函数用了递归法，在项数大于40后速度超慢。
另：类似【n<3 and 1 or 2】这种表达式中，逻辑与、或有这样的特性：
and 两边都对取后面一个表达式， or 两边都对取前面一个表达式。
>>> n=5 >>> 1 and n 5 >>> n and 1 1 >>> 0 and n 0 >>> n and 0 0 >>> 1 or n 1 >>> n or 1 5 >>> 0 or n 5 >>> n or 0 5 >>>
(2). 引入临时推导表达式：
>>> N=50 # 项数=50 >>> f=[1,1] >>> [f.append(f[n-2]+f[n-1]) for n in range(2,N)] [None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None] >>> f [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155, 165580141, 267914296, 433494437, 701408733, 1134903170, 1836311903, 2971215073, 4807526976, 7778742049, 12586269025] >>> ###合成一行，临时变量接收[None]*n列表### >>> f=[1,1];t=[f.append(f[n-2]+f[n-1]) for n in range(2,N)] >>> f [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155, 165580141, 267914296, 433494437, 701408733, 1134903170, 1836311903, 2971215073, 4807526976, 7778742049, 12586269025] >>>
真的一行代码：
>>> N=50 # 项数=50 >>> t=[(f[n][0], f.append((f[n][1],f[n][0]+f[n][1]))) for f in ([[1,1]],) for n in range(N)] >>> t [(1, None), (1, None), (2, None), (3, None), (5, None), (8, None), (13, None), (21, None), (34, None), (55, None), (89, None), (144, None), (233, None), (377, None), (610, None), (987, None), (1597, None), (2584, None), (4181, None), (6765, None), (10946, None), (17711, None), (28657, None), (46368, None), (75025, None), (121393, None), (196418, None), (317811, None), (514229, None), (832040, None), (1346269, None), (2178309, None), (3524578, None), (5702887, None), (9227465, None), (14930352, None), (24157817, None), (39088169, None), (63245986, None), (102334155, None), (165580141, None), (267914296, None), (433494437, None), (701408733, None), (1134903170, None), (1836311903, None), (2971215073, None), (4807526976, None), (7778742049, None), (12586269025, None)] >>> [f[0] for f in t] [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155, 165580141, 267914296, 433494437, 701408733, 1134903170, 1836311903, 2971215073, 4807526976, 7778742049, 12586269025] >>> ###合并成一行### >>> [f[0] for f in [(f[n][0], f.append((f[n][1],f[n][0]+f[n][1]))) for f in ([[1,1]],) for n in range(N)]] [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155, 165580141, 267914296, 433494437, 701408733, 1134903170, 1836311903, 2971215073, 4807526976, 7778742049, 12586269025] >>>
3. 曼德勃罗集(Mandelbrot Set)分形
>>> print('\n'.join([''.join(['*'if abs((lambda a:lambda z,c,n:a(a,z,c,n))(lambda s,z,c,n:s(s,z*z+c,c,n-1) if n else z)(0,0.02*x+0.05j*y,40))<2 else ' ' for x in range(-78,20)]) for y in range(-20,21)])) * ** *********** ************ ********* * * ************ * * ****** * *************************** * *************************************** ***** ******************************************* *** ******************************************** * ************************************************** ******************************************************* * * ***************************************************** **** ******* * ******************************************************* ***************** ******************************************************* *********************** ********************************************************* *********************** ******************************************************** **** ******************************************************************************** ******************************************************************************************* **** ******************************************************************************** *********************** ******************************************************** *********************** ********************************************************* ***************** ******************************************************* **** ******* * ******************************************************* * * ***************************************************** ******************************************************* ************************************************** *** ******************************************** * ******************************************* *************************************** ***** ****** * *************************** * * * ************ * * ********* ************ *********** ** * >>>
附录
