How to check if a list has an empty string

Question

Suppose I have a list ['', 'Tom', 'Jane', 'John', '0'], and I wrote the following to check if it has empty string '':

if any('' in s for s in row[1:]):
        print "contains empty string, continue now"
        continue

I expect this to only pick the empty string, but this also picks a list that contains '0'. The '0' string is valid and I do not want to filter it out. So how can I check for an empty string in a python list?

Answer 1

You can just check if there is the element '' in the list:

if '' in lst:
     # ...

if '' in lst[:1]:  # for your example
     # ...

Here is an example:

>>> lst = ['John', '', 'Foo', '0']
>>> lst2 = ['John', 'Bar', '0']
>>> '' in lst
True
>>> '' in lst2
False

Answer 2

any('' in s for s in row[1:])

checks, for each string s in the list row[1:], whether the empty string is a substring of it. That's always true, trivially. What you mean is

any(s == '' for s in row[1:])

But this can be abbreviated to

'' in row[1:]

Answer 3

Instead of using:

if any('' in s for s in row[1:]):

change your condition to:

if '' in your_list:

where your_list is name of list you are examining.

Answer 4

Let's take your example and give the list a name:

list = ['', 'Tom', 'Jane', 'John', '0']
if '' in list:
    do something

That is all you need!

Answer 5

How about this:

if any(len(x) == 0 for x in row):
    print("contains empty string, continue now")
    continue