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?
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
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:]
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.
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!
How about this:
if any(len(x) == 0 for x in row):
print("contains empty string, continue now")
continue
'' in s->s == ''(or more triviallyif '' in row)