Preserving the order in difference between two lists

Question

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?

`set([])` is the same as `set()` and anything minus the empty set is itself. What are you trying to do? — Two-Bit Alchemist, Sep 03 '15 at 19:56

Padraic Cunningham · Answer 1 · 2015-09-03T20:03:55.333

2

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

edited Sep 03 '15 at 20:03

answered Sep 03 '15 at 19:57

Padraic Cunningham

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But... but... nothing is in `l_match` by definition. What is the point? – Two-Bit Alchemist Sep 03 '15 at 19:57
@Two-BitAlchemist, I presumed the OP actually means to use a list with values, I think the order is the point of the question or else it makes zero sense bar removing dupes from l. – Padraic Cunningham Sep 03 '15 at 19:58
What's gained from making `l_match` a set? What's different here from doing `ele not in l_match`? – Robin James Kerrison Sep 03 '15 at 20:03
2

@RobinJamesKerrison, in a real world use case where the list actually has values, it is the difference between quadratic and linear – Padraic Cunningham Sep 03 '15 at 20:04
@PadraicCunningham I hope so, too; I just don't understand then why it is empty in the example code. – Two-Bit Alchemist Sep 03 '15 at 20:06

score 1 · Answer 2 · answered Jun 22 '22 at 18:05

1

This is old af but, in case someone is still wondering about it, a little googling gave me this really simple solution.

x = [1, 2, 6, 8, 2, 3]
y = [2, 6]
sorted(set(x) - set(y), key=x.index)

output -> [1, 8, 3]

answered Jun 22 '22 at 18:05

Gabriel Pena

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score 0 · Answer 3 · answered Sep 03 '15 at 20:00

0

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]

answered Sep 03 '15 at 20:00

Robin James Kerrison

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score 0 · Answer 4 · edited May 23 '17 at 11:44

0

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))])

edited May 23 '17 at 11:44

Community

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answered Sep 03 '15 at 20:03

Edwin Torres

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Preserving the order in difference between two lists

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