Preserving the order in difference between two lists

问题

I have two lists l and l_match. l_match is an empty list.

l = ['gtttaattgagttgtcatatgttaataacg',
     'tttaattgagttgtcatatgttaataacgg',
     'ttaattgagttgtcatatgttaataacggt',
     'taattgagttgtcatatgttaataacggta',
     'aattgagttgtcatatgttaataacggtat']

l_match = []

print list(set(l) - set(l_match))

gives the output

['aattgagttgtcatatgttaataacggtat',
 'tttaattgagttgtcatatgttaataacgg',
 'ttaattgagttgtcatatgttaataacggt',
 'taattgagttgtcatatgttaataacggta',
 'gtttaattgagttgtcatatgttaataacg']

I want the output the same order as the input. i.e. in the above case the output should be

['gtttaattgagttgtcatatgttaataacg',
 'tttaattgagttgtcatatgttaataacgg',
 'ttaattgagttgtcatatgttaataacggt',
 'taattgagttgtcatatgttaataacggta',
 'aattgagttgtcatatgttaataacggtat']

Can you suggest edits?

回答1:

Just make l_match a set:

l_match = []

st =  set(l_match)

print([ele for ele in l if ele not in st])

If l can have dupes use an OrderedDict to get unique values from l:

from collections import OrderedDict
print([ele for ele in OrderedDict.fromkeys(l) if ele not in st])

Obviously l_match would contain values in the real world or a simple l[:] = OrderedDict.fromkeys(l) would suffice to remove dupes from l and keep the order

回答2:

You should look through l and include each element therein in your result array only if it's not in l_match. This will preserve the order. In python, the statement is a single line:

print [entry for entry in l if entry not in l_match]

回答3:

What about this: How do you remove duplicates from a list in whilst preserving order?

l = ['gtttaattgagttgtcatatgttaataacg', 'tttaattgagttgtcatatgttaataacgg', 'ttaattgagttgtcatatgttaataacggt', 'taattgagttgtcatatgttaataacggta', 'aattgagttgtcatatgttaataacggtat']
seen = set()
seen_add = seen.add
print([ x for x in l if not (x in seen or seen_add(x))])

来源：https://stackoverflow.com/questions/32384312/preserving-the-order-in-difference-between-two-lists

标签

python

list

set

set-difference