Python : list comprehension + filter

Question

I have a string from which I wish to extract vowels and consonants along with their indices.

I am thinking about :

a = "HELLO"
list_ = list(a)
vow = [(x, i) for i,x in enumerate(list_) if x in ['A', 'E', 'I', 'O', 'U']]
cons = [(x, i) for i, x in enumerate(list_) if x not in j for j in list_ if j in ['A', 'E', 'I', 'O', 'U']]

However, I am getting NameError (saying j is not defined). Why am I not able to nest my list comprehension.

My desired output :

vows : [('E', 1), ('O', 4)]
cons : [('H', 0), ('L', 2), ('L', 3)]

List comprehensions can be nested, there is something wrong about the way I am doing it — Darshan Chaudhary
– Darshan Chaudhary, Commented Oct 2, 2015 at 6:33

Anand S Kumar · Accepted Answer · 2015-10-02 06:39:53Z

The issue with your second list comprehension is that ideally the if condition should go at the end (after the second for loop , only then j would we accessible). But you really do not need that, simply check if x is not in the list of vowels. Example -

cons = [(x, i) for i,x in enumerate(list_) if x not in ['A', 'E', 'I', 'O', 'U']]

Demo -

>>> a = "HELLO"
>>> list_ = list(a)
>>> vow = [(x, i) for i,x in enumerate(list_) if x in ['A', 'E', 'I', 'O', 'U']]
>>> cons = [(x, i) for i,x in enumerate(list_) if x not in ['A', 'E', 'I', 'O', 'U']]
>>> vow
[('E', 1), ('O', 4)]
>>> cons
[('H', 0), ('L', 2), ('L', 3)]

You can make this a bit faster by using set for vowels and you do not really need list_ , you can enumerate over a itself, and get exactly same result. Example -

vowel_set = {'A', 'E', 'I', 'O', 'U'}
vow = [(x, i) for i,x in enumerate(a) if x in vowel_set]
cons = [(x, i) for i,x in enumerate(a) if x not in vowel_set]

Demo -

>>> a = "HELLO"
>>> vowel_set = {'A', 'E', 'I', 'O', 'U'}
>>> vow = [(x, i) for i,x in enumerate(a) if x in vowel_set]
>>> cons = [(x, i) for i,x in enumerate(a) if x not in vowel_set]
>>> vow
[('E', 1), ('O', 4)]
>>> cons
[('H', 0), ('L', 2), ('L', 3)]

Serge Ballesta · Accepted Answer · 2015-10-02 06:40:35Z

2

The correct syntax would be to explicitely make the inner loop a list comprehension:

cons = [(x, i) for i, x in enumerate(list_) if x not in [ j for
            j in list_ if j in ['A', 'E', 'I', 'O', 'U']]]

But here the correct way to get consonants would be simply (as already said by others):

cons = [(x, i) for i,x in enumerate(list_) if x not in ['A', 'E', 'I', 'O', 'U']]

answered Oct 2, 2015 at 6:40

Serge Ballesta

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Comments

Ignacio Vazquez-Abrams · Accepted Answer · 2015-10-02 06:34:34Z

2

Rule #1 about nested LCs in Python: The outer loop goes first.

cons = [(x, i) for j in list_ if j in ['A', 'E', 'I', 'O', 'U'] for i, x in enumerate(list_) if x not in j]

But this will give the wrong result since you should just be using not in in the first place instead.

answered Oct 2, 2015 at 6:34

Ignacio Vazquez-Abrams

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Comments

Hayley Guillou · Accepted Answer · 2015-10-02 06:37:15Z

2

You're over-complicating it

a = "HELLO"
list_ = list(a)
vow = [(x, i) for i,x in enumerate(list_) if x in ['A', 'E', 'I', 'O', 'U']]
cons = [(x, i) for i, x in enumerate(list_) if x not in ['A', 'E', 'I', 'O', 'U']]

See Anand S Kumar's answer for details :)

The mistake in your nesting is that you need to nest it again

cons = [(x, i) for i, x in enumerate(list_) if x not in [j for j in list_ if j in ['A', 'E', 'I', 'O', 'U']]]

edited Oct 2, 2015 at 6:37

answered Oct 2, 2015 at 6:33

Hayley Guillou

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Open AI - Opting Out · Accepted Answer · 2015-10-02 10:54:10Z

2

You can check for membership in a string:

>>> 'I' in 'AEIOU'
True

You could convert each character in a string to a member of a set, for slightly faster searching:

>>> 'Z' in set('AEIOU')
False

You don't need the second inner loop:

>>> vowels = set('AEIOU')
>>> vows = [(char, index) for index, char in enumerate(word) if char in vowels]
>>> cons = [(char, index) for index, char in enumerate(word) if char not in vowels]

edited Oct 2, 2015 at 10:54

answered Oct 2, 2015 at 6:52

Open AI - Opting Out

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2 Comments

Darshan Chaudhary Over a year ago

But how can you also get indices this way ? if 'L' in 'HELLO': print 'HELLO'.find('L') This will only give the index of the first occurrence of L (here 2)

Open AI - Opting Out Over a year ago

My distinction is to check against a string of vowels rather than a list of vowels.

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