Question

我有一个[＆＃39; A＆＃39;，＆＃39; B＆＃39;，＆＃39; C＆＃39;，＆＃39; D＆＃39;]和[1,2]的列表，3,4，两个列表将始终具有相同数量的项目。我需要将每个字符串乘以它的数字，所以我要找的最终产品是：

[＆＃39; A＆＃39;，＆＃39; B＆＃39;，＆＃39; B＆＃39;，＆＃39; C＆＃39;，＆＃39; C＆＃39;，＆＃39; C＆＃39;，＆＃39; D＆＃39;，＆＃39; D＆＃39;，＆＃39; D＆＃39;，＆＃39; D＆＃39;]。

任何帮助将不胜感激，谢谢。

Answer 1

嵌套列表理解也有效：

>>> l1 = ['A', 'B', 'C', 'D'] 
>>> l2 = [1, 2, 3, 4]
>>> [c for c, i in zip(l1, l2) for _ in range(i)]
['A', 'B', 'B', 'C', 'C', 'C', 'D', 'D', 'D', 'D']

以上zip返回(char, count)元组：

>>> t = list(zip(l1, l2))
>>> t
[('A', 1), ('B', 2), ('C', 3), ('D', 4)]

然后对于每个元组，第二个for循环执行count次以将该字符添加到结果中：

>>> [char for char, count in t for _ in range(count)]
['A', 'B', 'B', 'C', 'C', 'C', 'D', 'D', 'D', 'D']

Answer 2

我会使用itertools.repeat来实现一个非常有效的实现：

>>> letters = ['A', 'B', 'C', 'D']
>>> numbers = [1, 2, 3, 4]
>>> import itertools
>>> result = []
>>> for letter, number in zip(letters, numbers):
...     result.extend(itertools.repeat(letter, number))
...
>>> result
['A', 'B', 'B', 'C', 'C', 'C', 'D', 'D', 'D', 'D']
>>>

我也认为它非常易读。

Answer 3

代码非常简单，请参阅内联评论

l1 = ['A', 'B', 'C', 'D'] 
l2 = [1, 2, 3, 4]
res = []
for i, x in enumerate(l1): # by enumerating you get both the item and its index
    res += x * l2[i] # add the next item to the result list
print res

输出

['A', 'B', 'B', 'C', 'C', 'C', 'D', 'D', 'D', 'D']

Answer 4

您可以使用zip()这样做：

a = ['A', 'B', 'C', 'D']
b = [1, 2, 3, 4]

final = []
for k,v in zip(a,b):
    final += [k for _ in range(v)]

print(final)

输出：

>>> ['A', 'B', 'B', 'C', 'C', 'C', 'D', 'D', 'D', 'D']

或者您也可以使用zip()和list comprehension：

来实现

a = ['A', 'B', 'C', 'D']
b = [1, 2, 3, 4]
final = [k for k,v in zip(a,b) for _ in range(v)]
print(final)

输出：

>>> ['A', 'B', 'B', 'C', 'C', 'C', 'D', 'D', 'D', 'D']

将一个字符串列表乘以int，逐个元素列表

4 个答案: