regex: find all words with certain letters but not other

Question

can anyone is help me with that:

I need to find all words from list containing letters [t OR d] AND [k OR c] but not any of [s,z,n,m]

I figured out first part, but don't know how to include stop list:

\w*[t|d]\w*[k|c]\w*

in Python notation

Thank you in advance

`if (re.search('[td]', input) or re.search('[kc]', input)) and not re.search('[sznm]', input)` — nhahtdh, Feb 09 '15 at 12:16
Note that your current regex will only find words where `t` or `d` appears before `k` or `c` - is that what you want? — jonrsharpe, Feb 09 '15 at 12:17

score 2 · Answer 1 · answered Feb 09 '15 at 12:25

2

You can use 2 steps. First find t|d AND k|c, then filter out matches with unwanted letters.

Since you said you figured out first part, here is the second:

matches = [i for i in matches if not re.search(r'[sznm]', i)]    
print(matches)

answered Feb 09 '15 at 12:25

user

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Math · Answer 2 · 2015-02-09T12:42:43.717

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If you need the t or d appearing before k or c, use : [^sznm\s\d]*[td][^sznm\s\d]*[kc][^sznm\s\d]*.

[^sznm\s\d] means any character except z, n, m, s, whitespace characters (\s) or numbers (\d).

edited Feb 09 '15 at 12:42

answered Feb 09 '15 at 12:26

Math

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[^sznm] is not equal to **all letters except s, z, n, m** this is mean that word can containing signs spaces and ... – M.javid Feb 09 '15 at 12:37
yep, thus the precision, but I added white space and numbers – Math Feb 09 '15 at 12:40
ok, new suggested pattern is more acceptable but not excepted signs. – M.javid Feb 09 '15 at 12:47
Your solution is working, but it made whole a bit complicated (it's just my issue), so I decided to avoid regex - thank you – AKarpun Feb 09 '15 at 15:17

Padraic Cunningham · Answer 3 · 2015-02-09T14:35:06.867

1

s = "foobar foo".split()

allowed = ({"k", "c"}, {"r", "d"})
forbid = {"s","c","z","m"}

for word in s:
    if all(any(k in st for k in word) for st in allowed) and all(k not in forbid for k in word):
        print(word)

Or using a list comp with set.intersection:

words = [word for word in s if all(st.intersection(word) for st in allowed) and not denied.intersection(word)]

edited Feb 09 '15 at 14:35

answered Feb 09 '15 at 12:27

Padraic Cunningham

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score 1 · Accepted Answer · edited May 23 '17 at 11:56

1

Based on answer of Padraic

EDIT We both missed this condition

[t OR d] AND [k OR c]

So - fixed accordingly

s = "detected dot knight track"

allowed = ({"t","d"},{"k","c"})
forbidden = {"s","z","n", "m"}

for word in s.split():
    letter_set = set(word)
    if all(letter_set & a for a in allowed) and letter_set - forbidden == letter_set:
        print(word)

And the result is

detected
track

edited May 23 '17 at 11:56

Community

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answered Feb 09 '15 at 12:36

volcano

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Thank you, regex was not feasible idea in long run – AKarpun Feb 09 '15 at 15:13

score 0 · Answer 5 · answered Feb 09 '15 at 12:32

0

Use this code:

import re
re.findall('[abcdefghijklopqrtuvwxy]*[td][abcdefghijklopqrtuvwxy]*[kc][abcdefghijklopqrtuvwxy]*', text)

answered Feb 09 '15 at 12:32

M.javid

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tommy.carstensen · Answer 6 · 2015-02-09T12:44:45.980

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I really like the answer by @padraic-cunningham that does not make use of re, but here is a pattern, which will work:

pattern = r'(?!\w*[sznm])(?=\w*[td])(?=\w*[kc])\w*'

Positive (?=...) and negative (?!...) lookahead assertions are well documented on python.org.

edited Feb 09 '15 at 12:44

answered Feb 09 '15 at 12:37

tommy.carstensen

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score 0 · Answer 7 · answered Feb 09 '15 at 12:43

0

You need to use lookarounds.

^(?=.*[td])(?!.*[sznm])\w*[kc]\w*$

ie,

>>> l = ['fooktz', 'foocdm', 'foobar', 'kbard']
>>> [i for i in l if re.match(r'^(?=.*[td])(?!.*[sznm])\w*[kc]\w*$', i)]
['kbard']

answered Feb 09 '15 at 12:43

Avinash Raj

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regex: find all words with certain letters but not other

7 Answers7