string sieving with python

Question

Assume you get a string. Example:

s = "FishCatRainComp_CatWater_JamDog"

And you define a sieve. A sieve - is a list of words you would like to catch up(once if multiple occurrence in s), for example:

sieve = ["Dog","Cat"]

Passing a string through a sieve should produce a string, in our case that would be:

out = "CatDog"

What would be the most elegant way to achieve the result?

Your simple way suggests that the input should in fact be `"DogCat"` (and not `"CatDog"` as you've mentioned). — barak manos, Nov 10 '16 at 17:12
In any case, you could use `''.join([word for word in sieve if word in s])`. — barak manos, Nov 10 '16 at 17:14

barak manos · Accepted Answer · 2016-11-10T17:28:05.590

4

Here is the most elegant way that comes to mind:

''.join([word for word in sieve if word in s])

Given that the order of the words in the input string should be reflected in the output string:

def SieveString(s,sieve):
    zipped = zip(sieve,[s.index(word) for word in sieve if word in s])
    zipped.sort(key=lambda x:x[1])
    return ''.join(word for word,index in zipped)

edited Nov 10 '16 at 17:28

answered Nov 10 '16 at 17:15

barak manos

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That will produce `"DogCat"` for the sample op posted above. Not `"CatDog"` which is the correct answer. – Omada Nov 10 '16 at 17:16
@Omada: See my comment to the question, which originally provided a coding example giving the same output, and asked for a more elegant manner of doing it. – barak manos Nov 10 '16 at 17:18
@Omada nice catch. Perhaps barak should iterate through sieve backwards with sieve[::-1] – Mark Hannel Nov 10 '16 at 17:18
@MarkHannel No. That would reverse the whole string. – Christian Dean Nov 10 '16 at 17:18
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@barakmanos I think you are misunderstanding how a sieve is supposed to work. The output should have the words in the same order as they appeared in `s` (at least as far as my understanding is) – Omada Nov 10 '16 at 17:19
@Omada: I think you are missing my point. The question **originally** gave a coding example which resulted in **this** output (`"DogCat"`), asking for a more elegant manner. I notified OP about that (since he/she also mentioned the output `"CatDog"`), and answered it as is. The OP has later removed the original coding example. – barak manos Nov 10 '16 at 17:21
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@Omada: In any case, I've added a solution referring to this ("new") restriction. – barak manos Nov 10 '16 at 17:28

Francisco · Answer 2 · 2016-11-10T17:31:13.637

If you want to maintain the order of the strings as they appear in s, you could do something like this:

found = re.findall('({})'.format('|'.join(re.escape(w) for w in sieve)), s)

You would then have to remove the repeated strings:

def remove_repeated(seq):
    seen = set()
    seen_add = seen.add
    return [x for x in seq if not (x in seen or seen_add(x))]

print(''.join(remove_repeated(found)))

This solution might be longer, but it has better asymptotic behaviour

You could otherwise sort the string by their index:

>>> sorted([word for word in sieve if word in s], key=lambda word: s.index(word))

string sieving with python

2 Answers2