嵌套 ifs 中的列表理解

Question

I am a newbie trying to understand list comprehensions in python. My question is different from another posts.

I was asked to write list comprehension code to get the following output:

All odd numbers from 1 to 30 (both inclusive). Those that are multiples of 5 will be marked with an 'x'.

[1, 3, '5x', 7, 9, 11, 13, '15x', 17, 19, 21, 23, '25x', 27, 29]

For this, I tried to get it with normal for and if ways. This is my solution and it worked:

odds = []

for i in list(range(1,30+1)):
  if i%2 !=0:
    odds.append(i)
    if i%5 == 0:
      odds.append(f'{i}x')
      odds.remove(i)

print(odds)

In the image you can find my failed list comprehension attempt. I need some light to place the rest of the stuff correctly.

Thank you!

Answer 1

You cannot solve this problem in one line using list comprehension alone. You need the ternary operator (enclosed in the parentheses).

[(n if n%5 else f'{n}x') for n in range(1,31) if n%2]

Answer 2

If you don't care about the order of the items, you can avoid needing the ternary operator ( conditional expression ) by concatenating the lists resulting from two list comprehensions, eg

[n for n in range(1,31,2) if n%5 != 0] + [f'{n}x' for n in range(1,31,2) if n%5 == 0]

... which produces:

[1, 3, 7, 9, 11, 13, 17, 19, 21, 23, 27, 29, '5x', '15x', '25x']

Alternatively, at the risk of making this look like code golf:

[[f'{n}x',n][min(n%5,1)] for n in range(1,31,2)]

... produces:

[1, 3, '5x', 7, 9, 11, 13, '15x', 17, 19, 21, 23, '25x', 27, 29]

I've used the expression min(n%5,1) to index the list [f'{n}x',n] and thus select either item 0 or item 1 in the list, depending on whether n is divisible by 5 or not -- also without using the conditional expression .

Answer 3

Alternate workaround:

numlist = [(i,f'{i}x')[not i%5] for i in range(31) if i%2]
print(numlist)
# [1, 3, '5x', 7, 9, 11, 13, '15x', 17, 19, 21, 23, '25x', 27, 29]