用另一个列表替换所有子列表的第三个元素

Question

我有一个列表的列表，我想 replace/update 使用新列表的所有子列表的第三个元素。

lst1 = [['a','b','c', 3],['d','e','f', 9],['g','h','i', 'j']]            
lst2 = [2, 3, 4]

期望的输出：

lst_new = [['a','b', 2, 3],['d','e', 3, 9],['g','h', 4, 'j']]

Answer 1

可以使用numpy吗？就像：

arr1 = np.array(lst1)                                                                 
arr1[:, 2] = lst2

输出：

array([['a', 'b', '2', '3'],
       ['d', 'e', '3', '9'],
       ['g', 'h', '4', 'j']], dtype='<U1')

Answer 2

lst1 = [['a','b','c', 3],['d','e','f', 9],['g','h','i', 'j']] 
lst2 = [2,3,4]

#We make a copy so that this 'lst_new' variable does not point to lst1
lst_new = lst1.copy()


for num, lst in zip(lst2, lst_new):
    lst[2] = num
#At this point, your lst_new will be changed

Answer 3

试试这个：

lst1 = [['a','b','c', 3],['d','e','f', 9],['g','h','i', 'j']]            
lst2 = [2, 3, 4]

for x,y in zip(lst1,lst2): #loops over both lst1 and lst2
    x[2] = y

输出：

[['a', 'b', 2, 3], ['d', 'e', 3, 9], ['g', 'h', 4, 'j']]

Answer 4

lst1 = [['a', 'b', 'c', 3], ['d', 'e', 'f', 9], ['g', 'h', 'i', 'j']]
lst2 = [2, 3, 4]

for x in range(0, len(lst2)):
    lst1[x][2] = lst2[x]

print(lst1)

Answer 5

只是为了有另一个选择的乐趣：

map(lambda e, subls: [*subls[:2], e, *subls[3:]], lst2, lst1))

它利用了map, lambda and the unpacking operator *。

用另一个列表替换所有子列表的第三个元素

Replace the third element of all sublists with another list

python

list

nested-lists