List comprehension to evaluate to true if list contains a value

Question

I have a long list of values and want a list comprehension to evaluate to True (and print "True" only once if any value in the list is the integer 1).

I can print "True" for each instance a 1 is found but cannot see how to just have it return a single True.

Code

a = [0,0,1,1,0,1]

b = [print("True") for i in a if i == 1]
print('\n')
#c = [print("True") if any i in a is True] # doesn't work, syntax error



d = [print("TRUE") if any(i == 1)]

btw, You almost never want to have `print()` inside a list comprehension. — quamrana, Jun 16 '21 at 07:54
Because a list comprehension is a way of transforming one list into another. No printing involved. — quamrana, Jun 16 '21 at 07:55
`print('{}'.format(any(a)))` would work, although it would print False if no 1 found. — Sherlock Bourne, Jun 16 '21 at 07:57
I didn't know that, the first comprehension line works but prints 'True' for each instance. — Windy71, Jun 16 '21 at 07:57
@Sayse is right, we sometimes do not think of the easiest idea — Sherlock Bourne, Jun 16 '21 at 08:03

score 1 · Accepted Answer · answered Jun 16 '21 at 07:57

1

Did you mean to convert the resulting list to bool()?

a = [0,0,1,1,0,1]
b = bool([i for i in a if i == 1])
print(b)

answered Jun 16 '21 at 07:57

quamrana

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Thats what I meant quamrana, I will rewrite the question to reflect that in case it helps someone else. Much appreciated. – Windy71 Jun 16 '21 at 08:00

score 1 · Answer 2 · answered Jun 16 '21 at 08:01

1

if your list only contains zeros and ones you could just print(any(a)) otherwise you could do this

a = [0,0,1,0,2,0]
b =[x==1 for x in a]
print(any(b))

returns True

answered Jun 16 '21 at 08:01

Valentin C

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List comprehension to evaluate to true if list contains a value

2 Answers2