将列表中的字符串乘以另一个列表中的数字，逐个元素

Question

我有两个列表，['A', 'B', 'C', 'D'] 和 [1, 2, 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，所以我要寻找的最终产品是：

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

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']

在上面的zip 返回(char, count) 元组：

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

然后对每个元组执行第二个for 循环count 次以将字符添加到结果中：

>>> [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 来实现一个不错、高效的实现：

>>> 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']
>>>

我也觉得它的可读性很好。

Answer 3

代码非常简单，见内联 cmets

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']

或者你也可以这样做，使用zip()和list comprehension：

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']