Python using "IN" in a "IF ELSE" loop

Question

i have a list of tuples that i loop through in a simple for loop to identify tuples that contain some conditions.

    mytuplist = 
    [(1, 'ABC', 'Today is a great day'), (2, 'ABC', 'The sky is blue'), 
     (3, 'DEF', 'The sea is green'), (4, 'ABC', 'There are clouds in the sky')]

I want it to be efficient and readable like this:

    for tup in mytuplist:
        if tup[1] =='ABC' and tup[2] in ('Today is','The sky'):
            print tup

The above code does not work and nothing gets printed.

This code below works but is very wordy. How do i make it like the above?

for tup in mytuplist:
    if tup[1] =='ABC' and 'Today is' in tup[2] or 'The sky' in tup[2]:
        print tup

alecxe · Accepted Answer · 2016-06-21 18:56:39Z

7

You should use the built-in any() function:

mytuplist = [
    (1, 'ABC', 'Today is a great day'),
    (2, 'ABC', 'The sky is blue'),
    (3, 'DEF', 'The sea is green'),
    (4, 'ABC', 'There are clouds in the sky')
]

keywords = ['Today is', 'The sky']
for item in mytuplist:
    if item[1] == 'ABC' and any(keyword in item[2] for keyword in keywords):
        print(item)

Prints:

(1, 'ABC', 'Today is a great day')
(2, 'ABC', 'The sky is blue')

answered Jun 21, 2016 at 18:56

alecxe

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3 Comments

Andrew Over a year ago

imo this is the best way

jxn Over a year ago

How do we do the opposite, example, if item[1] != 'ABC' and any(keyword in item[2] for keyword in keywords): to get tuple 3 and 4?

alecxe Over a year ago

@jxn well, this would give you the tuple 3: if item[1] != 'ABC' and all(keyword not in item[2] for keyword in keywords):. Basically, we are asking for all keywords not to match.

Ignacio Vazquez-Abrams · Accepted Answer · 2016-06-21 18:56:38Z

3

You don't, because in with a sequence only matches the whole element.

if tup[1] =='ABC' and any(el in tup[2] for el in ('Today is', 'The sky')):

answered Jun 21, 2016 at 18:56

Ignacio Vazquez-Abrams

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Comments

chepner · Accepted Answer · 2016-06-21 18:58:39Z

1

Your second approach (which, however, needs to be parenthesized as x and (y or z) to be correct) is necessary tup[2] contains one of your key phrases, rather than being an element of your set of key phrases. You could use regular-expression matching at the cost of some performance:

if tup[1] == 'ABC' and re.match('Today is|The sky', tup[2])

answered Jun 21, 2016 at 18:58

chepner

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