CODEWITHSUNDEEP
pythonlist
Ippy
Ippy

Reputation: 139

Print list while excluding specific named values

I am looking for a basic way to exclude printing an item from a list (not remove/delete) if that named value = something specified, e.g. = 0

I thought it would be something basic like this, but I get an error that I is not defined.

a = 2
b = 0
c = 4
d = 0
e = 6

lis = [a,b,c,d,e]

while i != 0 in range(lis):
  print (lis[i])

Wanted result:
2
4
6

Seems like I am missing something simple? Or is this not as simple as I imagined?

Upvotes: 0

Views: 4043

Answers (3)

Praful Bagai
Praful Bagai

Reputation: 17392

Here are the options which you can use.

1) You can use either a normal loop

for i in lis:
    if i != 0:
        print i

OR

2) List Comprehensions

[i for i in lis if i!=0]

I would still say when coming to Stackoverflow, please consider going through the tutorials, because its the most basic thing in python.

Upvotes: 0

Alex
Alex

Reputation: 5646

for l in lis:
    if l != 0:
        print l

Iterate through the items in the list one by one. Every time, check the current item l to see if it equals 0. Print the item if it is non-zero.

Upvotes: 0

tzaman
tzaman

Reputation: 47820

Python syntax doesn't support a conditional where you're trying to apply it. There are several alternatives, though:

You could put the condition inside the loop:

for item in lis:
    if item != 0:
        print item

You could filter the list:

for item in filter(lambda i: i!=0, lis):  # filter(None, lis) works too
    print item

Or using a list comprehension (technically a generator expression here):

for item in (i for i in lis if i != 0):
    print item

Or skip the loop entirely:

print '\n'.join(str(i) for i in lis if i != 0)

Upvotes: 4

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