Removing an element from a list based on a predicate

Question

I want to remove an element from list, such that the element contains 'X' or 'N'. I have to apply for a large genome. Here is an example:

input:

codon=['AAT','XAC','ANT','TTA']

expected output:

codon=['AAT','TTA']  

Answer 1

For basis purpose

>>> [x for x in ['AAT','XAC','ANT','TTA'] if "X" not in x and "N" not in x]
['AAT', 'TTA']

But if you have huge amount of data, I suggest you to use dict or set

And If you have many characters other than X and N, you may do like this

>>> [x for x in ['AAT','XAC','ANT','TTA'] if not any(ch for ch in list(x) if ch in ["X","N","Y","Z","K","J"])]
['AAT', 'TTA']

NOTE: list(x) can be just x, and ["X","N","Y","Z","K","J"] can be just "XNYZKJ", and refer gnibbler answer, He did the best one.

Answer 2

Another not fastest way but I think it reads nicely

>>> [x for x in ['AAT','XAC','ANT','TTA'] if not any(y in x for y in "XN")]
['AAT', 'TTA']

>>> [x for x in ['AAT','XAC','ANT','TTA'] if not set("XN")&set(x)]
['AAT', 'TTA']

This way will be faster for long codons (assuming there is some repetition)

codon = ['AAT','XAC','ANT','TTA']
def pred(s,memo={}):
    if s not in memo:
        memo[s]=not any(y in s for y in "XN")
    return memo[s]

print filter(pred,codon)

Here is the method suggested by James Brooks, you'd have to test to see which is faster for your data

codon = ['AAT','XAC','ANT','TTA']
def pred(s,memo={}):
    if s not in memo:
        memo[s]= not set("XN")&set(s)
    return memo[s]

print filter(pred,codon)

For this sample codon, the version using sets is about 10% slower

Answer 3

There is also the method of doing it using filter

    lst = filter(lambda x: 'X' not in x and 'N' not in x, list)