我有两个列表,['A', 'B', 'C', 'D']
和[1, 2, 3, 4]
。这两个列表的项目数始终相同。我需要将每个字符串乘以其数字,因此我要寻找的最终产品是:
['A', 'B', 'B', 'C', 'C', 'C', 'D', 'D', 'D', 'D']
嵌套列表推导式也可以使用:
>>> 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']
zip
返回一个由 (char, count)
元组组成的列表:>>> 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
循环会执行count
次,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']
对于一个好的、高效的实现,我会使用 itertools.repeat
:
>>> 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']
>>>
letters = ['A', 'B', 'C', 'D']
times = [1, 2, 3, 4]
np.repeat(letters, times).tolist()
#output
['A', 'B', 'B', 'C', 'C', 'C', 'D', 'D', 'D', 'D']
代码非常直观,查看内联注释即可
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']
您可以使用zip()
来实现:
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']
或者你也可以使用 zip()
和 列表推导式
来完成:
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']