Python 在问答频道中刷题积累到的小技巧（二）

from this import s,d
txt = ''
for t in s:
    txt += d.get(t,t)
with open('zen.txt','w') as f:
    f.write(txt)
print('\nThe Zen of Python is written to a file named "zen.txt".')

'''
The Zen of Python, by Tim Peters
Beautiful is better than ugly.
Explicit is better than implicit.
Simple is better than complex.
Complex is better than complicated.
Flat is better than nested.
Sparse is better than dense.
Readability counts.
Special cases aren't special enough to break the rules.
Although practicality beats purity.
Errors should never pass silently.
Unless explicitly silenced.
In the face of ambiguity, refuse the temptation to guess.
There should be one-- and preferably only one --obvious way to do it.
Although that way may not be obvious at first unless you're Dutch.
Now is better than never.
Although never is often better than *right* now.
If the implementation is hard to explain, it's a bad idea.
If the implementation is easy to explain, it may be a good idea.
Namespaces are one honking great idea -- let's do more of those!
The Zen of Python is written to a file named "zen.txt".
'''

>>> # sum(iterable, start=0) --> 返回值：start + sum(iterable)
>>> sum([1,2,3])
6
>>> sum([1,2,3], start=5)
11
>>> # = start + sum([1,2,3]) # start在前是原因的
>>> sum([1,2,3], -5)
1
>>> sum([[1,2,3],[4,5]],[6]) # 换成列表的加法
[6, 1, 2, 3, 4, 5]
>>> [6]+[1,2,3]+[4,5]
[6, 1, 2, 3, 4, 5]
>>> sum([[1,2,3],[4,5]],[]) # 这就是第2参数用个空列表就能降维列表的原因
[1, 2, 3, 4, 5]

>>> listn = lambda ls,n=2:sum(map(lambda x:[x]*n,ls), [])
>>> a = [*range(1,5)]
>>> a
[1, 2, 3, 4]
>>> listn(a,3)
[1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4]
>>> [i//3 for i in range(3*5+1)]
[0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5]
>>> [i//3 for i in range(3*5)]
[0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4]
>>> [i//3+1 for i in range(4*3)]
[1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4]
>>> [1+i//3 for i in range(4*3)]
[1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4]
>>> a = [*range(1,5)]
>>> a
[1, 2, 3, 4]
>>> listn = lambda ls,n:sum(map(lambda x:[x]*n,ls),[])
>>> listn(a,3)
[1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4]
>>> [1+i//3 for i in range(4*3)] # 整数列表用整除运算推导也容易
[1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4]
>>> listn('abcde',3)
['a', 'a', 'a', 'b', 'b', 'b', 'c', 'c', 'c', 'd', 'd', 'd', 'e', 'e', 'e']

with open('sentence.txt', 'r') as f:
    data = f.read()
scores = sum([s.split() for s in data.split('\n')], [])
for s in scores:
    print(s)

import pandas as pd
df = pd.read_csv('某瓣电影排行榜.csv')
lis = [(director,film) for director,film in zip(df['导演'],df['影名'])]
dic = {}
for i,n in enumerate(lis):
    dic[n[0]] = dic.get(n[0], []) + [lis[i][1]] # 技巧点 ，当然pandas会有更好直接筛选方法
for k,v in sorted(dic.items(),key=lambda x:len(x[1]), reverse=True):
    if len(v)>=3:
        print(f"【导演】：{k}\n【片数】：{len(v)}\n【作品】：{' '.join(dic[k])}\n")

lst = df["导演"].tolist()
import numpy as np
dict(sorted(dict(zip(*np.unique(lst, return_counts=1))).items(),key=lambda x:x[1],reverse=True))
from collections import Counter
dict(Counter(lst))  # 当然Counter()更简单，返回的就是降序字典

class bTree:
    def __init__(self, rooot=None, lchild=None, rchild=None):
        self.root = root
        self.left = lchild
        self.right = rchild
    def __repr__(self):
        if not (self.left or self.right): return f'{self.root}'
        return f'[{self.left if self.left else "-"}<{self.root}>{self.right if self.right else "-"}]'
    def preOrder(self):
        '''前序遍历'''
        if not self: return []
        return [self.root]+bTree.preOrder(self.left)+bTree.preOrder(self.right)
    def inOrder(self):
        '''中序遍历'''
        if not self: return []
        return bTree.inOrder(self.left)+[self.root]+bTree.inOrder(self.right)
    def postOrder(self):
        '''后序遍历'''
        if not self: return []
        return bTree.postOrder(self.left)+bTree.postOrder(self.right)+[self.root]
    def Height(self):
        if not self: return 0
        lH = bTree.Height(self.left)
        rH = bTree.Height(self.right)
        return max(lH,rH)+1
exp = bTree('+')
exp.left  = bTree('*')
exp.right = bTree('/')
exp.left.left   = bTree(1)
exp.left.right  = bTree(2)
exp.right.left  = bTree(3)
exp.right.right = bTree(4)
order = exp.inOrder()
print(order)
tmp = ' '.join(map(str,order))
print(tmp,'=',eval(tmp))
print(exp)
print(repr(exp))
print(exp.preOrder())
print(exp.postOrder())