如何将索引1的元素转换为嵌套列表中的upper（）？

Question

How can I convert each element at index 1 in the inner list to upper() using only list comprehension.

ed_sheeran = [ [j.upper() for j[1] in i ] for i in ed_sheeran ] # My code but not working.


ed_sheeran =  [[' ', 'w', 'h', 'i', 't', 'e', ' ', 'l', 'i', 'p', 's', ',', ' ', 'p', 'a', 'l', 'e', ' ', 'f', 'a', 'c', 'e', ' '], [' ', 'b', 'r', 'e', 'a', 't', 'h', 'i', 'n', 'g', ' ', 'i', 'n', ' ', 's', 'n', 'o', 'w', 'f', 'l', 'a', 'k', 'e', 's', ' '], [' ', 'b', 'u', 'r', 'n', 't', ' ', 'l', 'u', 'n', 'g', 's', ' ', 's', 'o', 'u', 'r', ' ', 't', 'a', 's', 't', 'e', ' ']]

for i in ed_sheeran:
    i[1] = i[1].upper()    
print(ed_sheeran)

Output: 
[[' ', 'W', 'h', 'i', 't', 'e', ' ', 'l', 'i', 'p', 's', ',', ' ', 'p', 'a', 'l', 'e', ' ', 'f', 'a', 'c', 'e', ' '], [' ', 'B', 'r', 'e', 'a', 't', 'h', 'i', 'n', 'g', ' ', 'i', 'n', ' ', 's', 'n', 'o', 'w', 'f', 'l', 'a', 'k', 'e', 's', ' '], [' ', 'B', 'u', 'r', 'n', 't', ' ', 'l', 'u', 'n', 'g', 's', ' ', 's', 'o', 'u', 'r', ' ', 't', 'a', 's', 't', 'e', ' ']]

Answer 1

只需将列表的其他部分缝合在一起，确保所有元素都在列表中。

[[x[0], x[1].upper()] +  x[2:] for x in ed_sheeran]

Answer 2

只是枚举子列表，如果索引值为1，则将单个元素更改为上

[[y.upper() if i==1 else y for i,y in enumerate(x)]  for x in ed_sheeran]

Answer 3

.upper() will return a new string, not update it inplace - and, on the other hand, the attribution statement = is not meant to be used inside expressions such as list comprehensions.

Replacing a list element is possible in an expression, though, if one call the __setitem__ method on the list, which is ordinarily done by the = . If one wants to "look clean" by not calling the dunder __setitem__ methos directly, it is possible to use operator.setitem to do that as an expression:

from operator import setitem
ed_sheeran = [setitem(part, 1, part[1].upper()) for part in ed_sheeran ]

Or, if you prefer without importing setitem :

ed_sheeran = [part.__setitem__(1, part[1].upper()) for part in ed_sheeran]

Actually, this is done inplace, so you don't even need to re-assign the outter-list - although if you don't want to create a new-list in memory, and use a generator expression you will need to "consume" the generator. The "any" built-in is a nice one in this case:

any(part.__setitem__(1, part[1].upper()) for part in ed_sheeran)

(this works only because __setitem__ returns None)