How to remove all integer values from a list in python

Question

I am just a beginner in python and I want to know is it possible to remove all the integer values from a list? For example the document goes like

['1','introduction','to','molecular','8','the','learning','module','5']

After the removal I want the document to look like:

['introduction','to','molecular','the','learning','module']

Answer 1

To remove all integers, do this:

no_integers = [x for x in mylist if not isinstance(x, int)]

However, your example list does not actually contain integers. It contains only strings, some of which are composed only of digits. To filter those out, do the following:

no_integers = [x for x in mylist if not (x.isdigit() 
                                         or x[0] == '-' and x[1:].isdigit())]

Alternately:

is_integer = lambda s: s.isdigit() or (x[0] == '-' and x[1:].isdigit())
no_integers = filter(is_integer, mylist)

Answer 2

None of the items in your list are integers. They are strings which contain only digits. So you can use the isdigit string method to filter out these items.

items = ['1','introduction','to','molecular','8','the','learning','module','5']

new_items = [item for item in items if not item.isdigit()]

print new_items

Link to documentation: http://docs.python.org/library/stdtypes.html#str.isdigit

Answer 3

You can do this, too:

def int_filter( someList ):
    for v in someList:
        try:
            int(v)
            continue # Skip these
        except ValueError:
            yield v # Keep these

list( int_filter( items ))

Why? Because int is better than trying to write rules or regular expressions to recognize string values that encode an integer.

Answer 4

Please do not use this way to remove items from a list: (edited after comment by THC4k)

>>> li = ['1','introduction','to','molecular','8','the','learning','module','5']
>>> for item in li:
        if item.isdigit():
            li.remove(item)

>>> print li
['introduction', 'to', 'molecular', 'the', 'learning', 'module']

This will not work since changing a list while iterating over it will confuse the for-loop. Also, item.isdigit() will not work if the item is a string containing a negative integer, as noted by razpeitia.

Answer 5

You can also use lambdas (and, obviously, recursion), to achieve that (Python 3 needed):

 isNumber = lambda s: False if ( not( s[0].isdigit() ) and s[0]!='+' and s[0]!='-' ) else isNumberBody( s[ 1:] )

 isNumberBody = lambda s: True if len( s ) == 0 else ( False if ( not( s[0].isdigit() ) and s[0]!='.' ) else isNumberBody( s[ 1:] ) )

 removeNumbers = lambda s: [] if len( s ) == 0 else ( ( [s[0]] + removeNumbers(s[1:]) ) if ( not( isInteger( s[0] ) ) ) else [] + removeNumbers( s[ 1:] ) )

 l = removeNumbers(["hello", "-1", "2", "world", "+23.45"])
 print( l )

Result (displayed from 'l') will be: ['hello', 'world']

Answer 6

I personally like filter. I think it can help keep code readable and conceptually simple if used in a judicious way:

x = ['1','introduction','to','molecular','8','the','learning','module','5'] 
x = filter(lambda i: not str.isdigit(i), x)

or

from itertools import ifilterfalse
x = ifilterfalse(str.isdigit, x)

Note the second returns an iterator.