Question

我需要合并二维列表和单维列表而不会丢失元素。

我使用循环来实现结果，但我想知道是否有更好的方法。

list1 = ["a","b","c","d"]
list2 = [["1","2","3"],["4","5","6"],["7","8"]]
max_column_count = len(list1)
expected_result = [list1]
for row in list2:
    if max_column_count > len(row):
        columns = max_column_count - len(row)
        row += [''] * columns
    expected_result.append(row)
print(expected_result)

输出

[['a', 'b', 'c', 'd'], ['1', '2', '3', ''], ['4', '5', '6', ''], ['7', '8', '', '']]

Answer 1

如果您作为输出发布的是预期输出，那么使用itertools中的chain将是一种方法：

>>> mx_len = len(max([list1,*list2]))
>>> 
>>> mx_len
4
>>> [x+['']*(mx_len-len(x)) for x in itertools.chain([list1], list2)]
[['a', 'b', 'c', 'd'], ['1', '2', '3', ''], ['4', '5', '6', ''], ['7', '8', '', '']]
>>>
>>> #another way by unpacking list2 in a list with list1
>>>
>>> [x+['']*(mx_len-len(x)) for x in itertools.chain([list1, *list2])]
[['a', 'b', 'c', 'd'], ['1', '2', '3', ''], ['4', '5', '6', ''], ['7', '8', '', '']]

另一种方式是双压缩效果，例如使用zip_longest转置两个列表并用''填充缺失值然后再次压缩列表以返回到原始形状，这样：

>>> l1 = itertools.zip_longest(list1, *list2, fillvalue='')
>>> 
>>> l2 = list(zip(*l1))
>>>
>>> l2
[('a', 'b', 'c', 'd'), ('1', '2', '3', ''), ('4', '5', '6', ''), ('7', '8', '', '')]

Answer 2

list1 = ["a","b","c","d"]
list2 = [["1","2","3"],["4","5","6"],["7","8"]]
list3 = []
list3.append(list1)
list3.append(list2)
>>>
list3 =[['a', 'b', 'c', 'd'], [['1', '2', '3'], ['4', '5', '6'], ['7', '8']]]

Answer 3

如果希望结果列表包含相同大小的列表，则填充空字符串：

list1 = ["a","b","c","d"]
list2 = [["1","2","3"],["4","5","6"],["7","8"]]
req_len = len(list1)
result = [list1] + [org + [''*(req_len - len(org))] for org in list2]
print result

Answer 4

>>> list1 = ["a","b","c","d"]
... list2 = [["1","2","3"],["4","5","6"],["7","8"]]
... list3 = [list1] + map(lambda x: x + ['']*(len(list1)-len(x)),list2)
>>> list3
6: [['a', 'b', 'c', 'd'],
 ['1', '2', '3', ''],
 ['4', '5', '6', ''],
 ['7', '8', '', '']]
>>>

这与您正在做的事情基本相同，但更简洁。如果你不了解地图功能，这是一个学习的好时机（https://docs.python.org/3/library/functions.html#map）

在Python中合并2d列表和1d列表的最佳方法？

4 个答案: