如何从数字元组列表中形成数字列表？

对数字列表执行简单的统计计算 数字列表的最大值、 最小值和总和 digits = [1, 2, 3, 4, 5, 6, 7, 8, 9, 0] min(digits) 0 max(digits) 9 sum(digits) 45 列表解析 squares = [value2 for value in range(1,11)] print(squares) for循环为for value in range(1,11)，将值1~10提供给表达式value2。这里的for 语句末尾没有冒号。 结果 [1, 4, 9, 16, 25, 36, 49, 64, 81, 100] 使用列表的一部分 切片 players = [‘charles’, ‘martina’, ‘michael’, ‘florence’, ‘eli’]  print(players[0:3]) 输出也是一个列表 [‘charles’, ‘martina’, ‘michael’] 没有指定第一个索引，从列表头开始： players = [‘charles’, ‘martina’, ‘michael’, ‘florence’, ‘eli’] print(players[:4]) [‘charles’, ‘martina’, ‘michael’, ‘florence’] 输出最后三个可使用players[-3:]： 遍历切片 遍历列表的部分元素，在for循环中使用切片。 players = [‘charles’, ‘martina’, ‘michael’, ‘florence’, ‘eli’]

print(“Here are the first three players on my team:”)  for player in players[:3]: print(player.title()) 输出 Here are the first three players on my team: Charles Martina Michael 复制列表 复制列表，创建一个包含整个列表的切片，同时省略起始索引和终止索引（[:]）。 始于第一个元素，终止于最后一个元素的切片，即复制整个列表。 my_foods = [‘pizza’, ‘falafel’, ‘carrot cake’]  friend_foods = my_foods[:] print(“My favorite foods are:”) print(my_foods) print("\nMy friend’s favorite foods are:") print(friend_foods) 输出 My favorite foods are: [‘pizza’, ‘falafel’, ‘carrot cake’] My friend’s favorite foods are: [‘pizza’, ‘falafel’, ‘carrot cake’] 元组 不可变的列表为元组 元组用圆括号而不是方括号 dimensions = (200, 50)  print(dimensions[0]) print(dimensions[1]) 输出 200 50 遍历元组中的所有值 dimensions = (200, 50) for dimension in dimensions: print(dimension) 输出 200 50 修改元组变量 不能修改元组的元素，可以给存储元组的变量赋值，可重新定义整个元组 dimensions = (200, 50) print(“Original dimensions:”) for dimension in dimensions: print(dimension) #修改元素 dimensions = (400, 100） print("\nModified dimensions:") for dimension in dimensions: print(dimension) #两次输出分别为 Original dimensions: 200 50 Modified dimensions: 400 100

你可以试试看

def process_data(data):
    return int(''.join(map(str,data)))

out=[process_data(i) for i in a]
# [3141, 3114, 3411, 3411, 3114, 3141, 1341, 1314, 1431, 1413, 1134,...]

要么

def process_data(data):
    num=data[0]
    for i in data[1:]:
        num=num\*0+i
    return num
out=[process_data(i) for i in a]
# [3141, 3114, 3411, 3411, 3114, 3141, 1341, 1314, 1431, 1413, 1134,...]

回答来源：stackoverflow