Finding the Differentiation Condition between 2 list in python

Question

I am a newbie in python and I was trying to figure out, how do I differentiate between 2 lists which are as shown below

['11-10-2017', '12:15 PM', 'B.ARTS', 'Linguistics', '', '', 'German', 'Name']

And

['', '', '', '', '', '', '', '']

The Problem is, that both the lists have '' element, and I want a solid Condition which satisfies that if a list has an item which is a string and not ''. It is also possible that list has 7 '' and just one item is a string.

`filter(None, your_list)` might be what you are looking for. — Unni, Sep 15 '17 at 05:58

score 2 · Accepted Answer · answered Sep 15 '17 at 06:41

2

You can just use any with the list as argument :

>>> any(['', '', '', '', '', '', '', ''])
False
>>> any(['', '', '', '', '', '', '', 'Test', ''])
True

If there's any element which is truthy (i.e. non empty), it will return True.

answered Sep 15 '17 at 06:41

Eric Duminil

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This wasn't exactly what I wanted, but it was the closest to my expected output. – Panda Sep 15 '17 at 17:06

pylang · Answer 2 · 2017-09-15T06:57:27.690

1

It seems you want to filter empty strings from a list:

lst = ['11-10-2017', '12:15 PM', 'B.ARTS', 'Linguistics', '', '', 'German', 'Name']
[item for item in lst if item]
# ['11-10-2017', '12:15 PM', 'B.ARTS', 'Linguistics', 'German', 'Name']

I want a solid Condition which satisfies that if a list has an item which is a string and not ''

The condition is if item. To clarify, '' is an empty string. During iteration, if the item is '', the condition is False and thus the item is excluded from the result. Otherwise, the condition is True and the result is added to the list. See also this post.

This behavior is because all objects in Python have "truthiness" - all objects are assumed True except for a few, such as False, 0, "", None and empty collections.

edited Sep 15 '17 at 06:57

answered Sep 15 '17 at 06:26

pylang

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Shouldn't it be same as `filter(None, lst)`? Please see my earlier comment. – Unni Sep 15 '17 at 06:29
1

@Unni, yes `list(filter(None, lst))` gives the same output in a different way. `filter` is the functional approach that uses an inherent default predicate, which may not be obvious to the OP. Here I am using a list comprehension to demonstrate that `''` is an empty, Falsey string, which gets filtered because it negates the condition. Still, I feel the OP may be looking for something else, which requires clarification in the question. – pylang Sep 15 '17 at 06:34

Unni · Answer 3 · 2017-09-15T07:17:20.770

0

I need a condition which is satisfied if there are no '', in the string.

You can use all for checking this.

In [1]: s1 = ['11-10-2017', '12:15 PM', 'B.ARTS', 'Linguistics', '', '', 'German', 'Name']
In [2]: s2 = ['11-10-2017', '12:15 PM']

In [4]: all(x for x in s1)
Out[4]: False

In [5]: all(x for x in s2)
Out[5]: True

edited Sep 15 '17 at 07:17

answered Sep 15 '17 at 06:14

Unni

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I need a condition which is satisfied if there are no `''`, in the string. – Panda Sep 15 '17 at 06:51

score 0 · Answer 4 · answered Sep 15 '17 at 07:28

The easiest way to see if there is not a '' in your list is to use not in:

tests = [
    ['11-10-2017', '12:15 PM', 'B.ARTS', 'Linguistics', '', '', 'German', 'Name'], 
    ['', '', '', '', '', '', '', ''], 
    ['a', 'b', 'c'],
    ['', '', '', '', '', 'x', '', '']]

for t in tests:
    print '' not in t, t

Which would display:

False ['11-10-2017', '12:15 PM', 'B.ARTS', 'Linguistics', '', '', 'German', 'Name']
False ['', '', '', '', '', '', '', '']
True ['a', 'b', 'c']
False ['', '', '', '', '', 'x', '', '']

Finding the Differentiation Condition between 2 list in python

4 Answers4