so given "needle" and "there is a needle in this but not thisneedle haystack"
I wrote
def find_needle(n,h):
count = 0
words = h.split(" ")
for word in words:
if word == n:
count += 1
return count
This is O(n) but wondering if there is a better approach? maybe not by using split at all?
How would you write tests for this case to check that it handles all edge cases?
I don't think it's possible to get bellow O(n) with this (because you need to iterate trough the string at least once). You can do some optimizations.
I assume you want to match "whole words", for example looking up foo should match like this:
foo and foo, or foobar and not foo.
^^^ ^^^ ^^^
So splinting just based on space wouldn't do the job, because:
>>> 'foo and foo, or foobar and not foo.'.split(' ')
['foo', 'and', 'foo,', 'or', 'foobar', 'and', 'not', 'foo.']
# ^ ^
This is where re module comes in handy, which will allows you to build fascinating conditions. For example \b inside the regexp means:
Matches the empty string, but only at the beginning or end of a word. A word is defined as a sequence of Unicode alphanumeric or underscore characters, so the end of a word is indicated by whitespace or a non-alphanumeric, non-underscore Unicode character. Note that formally, \b is defined as the boundary between a \w and a \W character (or vice versa), or between \w and the beginning/end of the string. This means that r'\bfoo\b' matches 'foo', 'foo.', '(foo)', 'bar foo baz' but not 'foobar' or 'foo3'.
So r'\bfoo\b' will match only whole word foo. Also don't forget to use re.escape():
>>> re.escape('foo.bar+')
'foo\\.bar\\+'
>>> r'\b{}\b'.format(re.escape('foo.bar+'))
'\\bfoo\\.bar\\+\\b'
All you have to do now is use re.finditer() to scan the string. Based on documentation:
Return an iterator yielding match objects over all non-overlapping matches for the RE pattern in string. The string is scanned left-to-right, and matches are returned in the order found. Empty matches are included in the result unless they touch the beginning of another match.
I assume that matches are generated on the fly, so they never have to be in memory at once (which may come in handy with large strings, with many matched items). And in the end just count them:
>>> r = re.compile(r'\bfoo\b')
>>> it = r.finditer('foo and foo, or foobar and not foo.')
>>> sum(1 for _ in it)
3
This does not address the complexity issue but simplifies the code:
def find_needle(n,h):
return h.split().count(n)
You can use Counter
from collections import Counter
def find_needle(n,h):
return Counter(h.split())[n]
i.e.:
n = "portugal"
h = 'lobito programmer from portugal hello fromportugal portugal'
print find_needle(n,h)
Output:
2
DEMO
Actually, when you say O(n) you are forgetting the fact that after matching the first letter, you have to match the remaining ones as well (match n from needle to sentence, then match e, then the next e...) You are essentially trying to replicate the functionality of grep, so you can look at the grep algorithm. You can do well by building a finite state machine. There are many links that can help you, for one you could start from How does grep run so fast?
This is still going to be O(n) but it uses the power of the re module and python's generator expressions.
import re
def find_needle(n,h):
g = re.finditer(r'\b%s\b'%n, h) # use regex word boundaries
return sum(1 for _ in g) # return the length of the iterator
Should use far less memory than .split for a relatively large 'haystack'.
Note that this is not exactly the same as the code in the OP because it will not only find 'needle' but also 'needle,' and 'needle.' It will not find 'needles' though.
If you are concerned with the time it takes (as distinct from time complexity) multiprocess it. Basically make n smaller. Here is an example to run it in 2 processes.
from multiprocessing import Process
def find(word, string):
return string.count(word)
def search_for_words(word, string):
full_length = len(string)
part1 = string[:full_length/2]
proc1 = Process(target=find, args=(word, part1,))
proc1.start()
part2 = string[full_lenght/2:]
proc2 = Process(target=find, args=(word, part2,))
proc2.start()
proc1.join()
proc2.join()
if its O(n) you are worried about - then, i'm not sure there is much you can do, unless it is possible to get the string in another data structure. like a set or something. (but putting it in that set is also O(n), you can save on time if you are already iterating over the string somewhere else, and then make this structure then. write once, read many.
In order to guarantee finding a needle in a haystack, you need to examine each piece of hay until you find the needle. This is O(n) no matter what, a tight lower bound.
def find_needle(haystack):
for item in haystack:
if item == 'needle':
haystack.append(item)
return 'found the needle at position ' + str(haystack.index(item))
here's my one.
def find_needle(haystack, needle):
return haystack.count(needele)
here, we simply use the built-in count method to count the number of needles in a haystack.
Related
I want to use a regex to find a substring, followed by a variable number of characters, followed by any of several substrings.
an re.findall of
"ATGTCAGGTAAGCTTAGGGCTTTAGGATT"
should give me:
['ATGTCAGGTAA', 'ATGTCAGGTAAGCTTAG', 'ATGTCAGGTAAGCTTAGGGCTTTAG']
I have tried all of the following without success:
import re
string2 = "ATGTCAGGTAAGCTTAGGGCTTTAGGATT"
re.findall('(ATG.*TAA)|(ATG.*TAG)', string2)
re.findall('ATG.*(TAA|TAG)', string2)
re.findall('ATG.*((TAA)|(TAG))', string2)
re.findall('ATG.*(TAA)|(TAG)', string2)
re.findall('ATG.*(TAA)|ATG.*(TAG)', string2)
re.findall('(ATG.*)(TAA)|(ATG.*)(TAG)', string2)
re.findall('(ATG.*)TAA|(ATG.*)TAG', string2)
What am I missing here?
This is not super-easy, because a) you want overlapping matches, and b) you want greedy and non-greedy and everything inbetween.
As long as the strings are fairly short, you can check every substring:
import re
s = "ATGTCAGGTAAGCTTAGGGCTTTAGGATT"
p = re.compile(r'ATG.*TA[GA]$')
for start in range(len(s)-6): # string is at least 6 letters long
for end in range(start+6, len(s)):
if p.match(s, pos=start, endpos=end):
print(s[start:end])
This prints:
ATGTCAGGTAA
ATGTCAGGTAAGCTTAG
ATGTCAGGTAAGCTTAGGGCTTTAG
Since you appear to work with DNA sequences or something like that, make sure to check out Biopython, too.
I like the accepted answer just fine :-) That is, I'm adding this for info, not looking for points.
If you have heavy need for this, trying a match on O(N^2) pairs of indices may soon become unbearably slow. One improvement is to use the .search() method to "leap" directly to the only starting indices that can possibly pay off. So the following does that.
It also uses the .fullmatch() method so that you don't have to artificially change the "natural" regexp (e.g., in your example, no need to add a trailing $ to the regexp - and, indeed, in the following code doing so would no longer work as intended). Note that .fullmatch() was added in Python 3.4, so this code also requires Python 3!
Finally, this intends to generalize the re module's finditer() function/method. While you don't need match objects (you just want strings), they're far more generally applicable, and returning a generator is often friendlier than returning a list too.
So, no, this doesn't do exactly what you want, but does things from which you can get what you want, in Python 3, faster:
def finditer_overlap(regexp, string):
start = 0
n = len(string)
while start <= n:
# don't know whether regexp will find shortest or
# longest match, but _will_ find leftmost match
m = regexp.search(string, start)
if m is None:
return
start = m.start()
for finish in range(start, n+1):
m = regexp.fullmatch(string, start, finish)
if m is not None:
yield m
start += 1
Then, e.g.,
import re
string2 = "ATGTCAGGTAAGCTTAGGGCTTTAGGATT"
pat = re.compile("ATG.*(TAA|TAG)")
for match in finditer_overlap(pat, string2):
print(match.group())
prints what you wanted in your example. The other ways you tried to write a regexp should also work. In this example it's faster because the second time around the outer loop start is 1, and regexp.search(string, 1) fails to find another match, so the generator exits at once (so skips checking O(N^2) other index pairs).
I found a nice question where one can search for multiple endings of a string using: endswith(tuple)
Check if string ends with one of the strings from a list
My question is, how can I return which value from the tuple is actually found to be the match? and what if I have multiple matches, how can I choose the best match?
for example:
str= "ERTYHGFYUUHGFREDFYAAAAAAAAAA"
endings = ('AAAAA', 'AAAAAA', 'AAAAAAA', 'AAAAAAAA', 'AAAAAAAAA')
str.endswith(endings) ## this will return true for all of values inside the tuple, but how can I get which one matches the best
In this case, multiple matches can be found from the tuple, how can I deal with this and return only the best (biggest) match, which in this case should be: 'AAAAAAAAA' which I want to remove at the end (which can be done with a regular expression or so).
I mean one could do this in a for loop, but maybe there is an easier pythonic way?
>>> s = "ERTYHGFYUUHGFREDFYAAAAAAAAAA"
>>> endings = ['AAAAA', 'AAAAAA', 'AAAAAAA', 'AAAAAAAA', 'AAAAAAAAA']
>>> max([i for i in endings if s.endswith(i)],key=len)
'AAAAAAAAA'
import re
str= "ERTYHGFYUUHGFREDFYAAAAAAAAAA"
endings = ['AAAAA', 'AAAAAA', 'AAAAAAA', 'AAAAAAAA', 'AAAAAAAAA']
print max([i for i in endings if re.findall(i+r"$",str)],key=len)
How about:
len(str) - len(str.rstrip('A'))
str.endswith(tuple) is (currently) implemented as a simple loop over tuple, repeatedly re- running the match, any similarities between the endings are not taken into account.
In the example case, a regular expression should compile into an automaton that essentially runs in linear time:
regexp = '(' + '|'.join(
re.escape(ending) for ending in sorted(endings, key=len, reverse=True
) + ')$'
Edit 1: As pointed out correctly by Martijn Pieters, Python's re does not return the longest overall match, but for alternates only matches the first matching subexpression:
https://docs.python.org/2/library/re.html#module-re:
When one pattern completely matches, that branch is accepted. This means that once A matches, B will not be tested further, even if it would produce a longer overall match.
(emphasis mine)
Hence, unfortunately the need for sorting by length.
Note that this makes Python's re different from POSIX regular expressions, which match the longest overall match.
I'm trying to write a function process(s,d) to replace abbreviations in a string with their full meaning by using a dictionary. where s is the string input and d is the dictionary. For example:
>>>d = {'ASAP':'as soon as possible'}
>>>s = "I will do this ASAP. Regards, X"
>>>process(s,d)
>>>"I will do this as soon as possible. Regards, X"
I have tried using the split function to separate the string and compare each part with the dictionary.
def process(s):
return ''.join(d[ch] if ch in d else ch for ch in s)
However, it returns me the same exact string. I have a suspicion that the code doesn't work because of the full stop behind ASAP in the original string. If so, how do I ignore the punctuation and get ASAP to be replaced?
Here is a way to do it with a single regex:
In [24]: d = {'ASAP':'as soon as possible', 'AFAIK': 'as far as I know'}
In [25]: s = 'I will do this ASAP, AFAIK. Regards, X'
In [26]: re.sub(r'\b' + '|'.join(d.keys()) + r'\b', lambda m: d[m.group(0)], s)
Out[26]: 'I will do this as soon as possible, as far as I know. Regards, X'
Unlike versions based on str.replace(), this observes word boundaries and therefore won't replace abbreviations that happen to appear in the middle of other words (e.g. "etc" in "fetch").
Also, unlike most (all?) other solutions presented thus far, it iterates over the input string just once, regardless of how many search terms there are in the dictionary.
You can do something like this:
def process(s,d):
for key in d:
s = s.replace(key,d[key])
return s
Here is a working solution: use re.split(), and split by word boundaries (preserving the interstitial characters):
''.join( d.get( word, word ) for word in re.split( '(\W+)', s ) )
One significant difference that this code has from Vaughn's or Sheena's answer is that this code takes advantage of the O(1) lookup time of the dictionary, while their solutions look at every key in the dictionary. This means that when s is short and d is very large, their code will take significantly longer to run. Furthermore, parts of words will still be replaced in their solutions: if d = { "lol": "laugh out loud" } and s="lollipop" their solutions will incorrectly produce "laugh out loudlipop".
use regular expressions:
re.sub(pattern,replacement,s)
In your application:
ret = s
for key in d:
ret = re.sub(r'\b'+key+r'\b',d[key],ret)
return ret
\b matches the beginning or end of a word. Thanks Paul for the comment
Instead of splitting by spaces, use:
split("\W")
It will split by anything that's not a character that would be part of a word.
python 3.2
[s.replace(i,v) for i,v in d.items()]
This is string replacement as well (+1 to #VaughnCato). This uses the reduce function to iterate through your dictionary, replacing any instances of the keys in the string with the values. s in this case is the accumulator, which is reduced (i.e. fed to the replace function) on every iteration, maintaining all past replacements (also, per #PaulMcGuire's point above, this replaces keys starting with the longest and ending with the shortest).
In [1]: d = {'ASAP':'as soon as possible', 'AFAIK': 'as far as I know'}
In [2]: s = 'I will do this ASAP, AFAIK. Regards, X'
In [3]: reduce(lambda x, y: x.replace(y, d[y]), sorted(d, key=lambda i: len(i), reverse=True), s)
Out[3]: 'I will do this as soon as possible, as far as I know. Regards, X'
As for why your function didn't return what you expected - when you iterate through s, you are actually iterating through the characters of the string - not the words. Your version could be tweaked by iterating over s.split() (which would be a list of the words), but you then run into an issue where the punctuation is causing words to not match your dictionary. You can get it to match by importing string and stripping out string.punctuation from each word, but that will remove the punctuation from the final string (so regex would be likely be the best option if replacement doesn't work).
Simple question to which I can't find any "nice" answer by myself:
Let's say I have the following condition:
if 'foo' in mystring or 'bar' in mystring or 'hello' in mystring:
# Do something
pass
Where the number of or statement can be quite longer depending on the situation.
Is there a "nicer" (more Pythonic) way of writing this, without sacrificing performance ?
If thought of using any() but it takes a list of boolean-like elements, so I would have to build that list first (giving-up short circuit evaluation in the process), so I guess it's less efficient.
Thank you very much.
A way could be
if any(s in mystring for s in ('foo', 'bar', 'hello')):
pass
The thing you iterate over is a tuple, which is built upon compilation of the function, so it shouldn't be inferior to your original version.
If you fear that the tuple will become too long, you could do
def mystringlist():
yield 'foo'
yield 'bar'
yield 'hello'
if any(s in mystring for s in mystringlist()):
pass
This sounds like a job for a regex.
import re
if re.search("(foo|bar|hello)", mystring):
# Do something
pass
It should be faster, too. Especially if you compile the regex ahead of time.
If you're generating the regular expression automatically, you could use re.escape() to make sure no special characters break your regex. For example, if words is a list of strings you wish to search for, you could generate your pattern like this:
pattern = "(%s)" % ("|".join(re.escape(word) for word in words), )
You should also note that if you have m words and your string has n characters, your original code has O(n*m) complexity, while the regular expression has O(n) complexity. Even though Python regexs are not really theoretical comp-sci regular expressions, and are not always O(n) complexity, in this simple case they are.
Since you are processing word-by-word against mystring, surely mystring can be used as a set. Then just take the intersection between the set containing the words in mystring and the target groups of words:
In [370]: mystring=set(['foobar','barfoo','foo'])
In [371]: mystring.intersection(set(['foo', 'bar', 'hello']))
Out[371]: set(['foo'])
Your logical 'or' is the members of the intersection of the two sets.
Using a set is also faster. Here are relative timing vs a generator and regular expression:
f1: generator to test against large string
f2: re to test against large string
f3: set intersection of two sets of words
rate/sec f2 f1 f3
f2 101,333 -- -95.0% -95.5%
f1 2,026,329 1899.7% -- -10.1%
f3 2,253,539 2123.9% 11.2% --
So a generator and the in operation is 19x faster than a regular expression and a set intersection is 21x faster than a regex and 11% faster than a generator.
Here is the code that generated the timing:
import re
with open('/usr/share/dict/words','r') as fin:
set_words={word.strip() for word in fin}
s_words=' '.join(set_words)
target=set(['bar','foo','hello'])
target_re = re.compile("(%s)" % ("|".join(re.escape(word) for word in target), ))
gen_target=(word for word in ('bar','foo','hello'))
def f1():
""" generator to test against large string """
if any(s in s_words for s in gen_target):
return True
def f2():
""" re to test against large string """
if re.search(target_re, s_words):
return True
def f3():
""" set intersection of two sets of words """
if target.intersection(set_words):
return True
funcs=[f1,f2,f3]
legend(funcs)
cmpthese(funcs)
If you have a known list of items to check against, you could also write it as
if mystring in ['foo', 'bar', 'hello']:
You may not get the benefits of ensuring the comparison order (I don't think Python is required to check the list elements left-to-right) but that's only a problem if you know that 'foo' is way more likely than 'bar'.
Hey there, I love regular expressions, but I'm just not good at them at all.
I have a list of some 400 shortened words such as lol, omg, lmao...etc. Whenever someone types one of these shortened words, it is replaced with its English counterpart ([laughter], or something to that effect). Anyway, people are annoying and type these short-hand words with the last letter(s) repeated x number of times.
examples:
omg -> omgggg, lol -> lollll, haha -> hahahaha, lol -> lololol
I was wondering if anyone could hand me the regex (in Python, preferably) to deal with this?
Thanks all.
(It's a Twitter-related project for topic identification if anyone's curious. If someone tweets "Let's go shoot some hoops", how do you know the tweet is about basketball, etc)
FIRST APPROACH -
Well, using regular expression(s) you could do like so -
import re
re.sub('g+', 'g', 'omgggg')
re.sub('l+', 'l', 'lollll')
etc.
Let me point out that using regular expressions is a very fragile & basic approach to dealing with this problem. You could so easily get strings from users which will break the above regular expressions. What I am trying to say is that this approach requires lot of maintenance in terms of observing the patterns of mistakes the users make & then creating case specific regular expressions for them.
SECOND APPROACH -
Instead have you considered using difflib module? It's a module with helpers for computing deltas between objects. Of particular importance here for you is SequenceMatcher. To paraphrase from official documentation-
SequenceMatcher is a flexible class
for comparing pairs of sequences of
any type, so long as the sequence
elements are hashable. SequenceMatcher
tries to compute a "human-friendly
diff" between two sequences. The
fundamental notion is the longest
contiguous & junk-free matching subsequence.
import difflib as dl
x = dl.SequenceMatcher(lambda x : x == ' ', "omg", "omgggg")
y = dl.SequenceMatcher(lambda x : x == ' ', "omgggg","omg")
avg = (x.ratio()+y.ratio())/2.0
if avg>= 0.6:
print 'Match!'
else:
print 'Sorry!'
According to documentation, any ratio() over 0.6 is a close match. You might need to explore tweak the ratio for your data needs. If you need more stricter matching I found any value over 0.8 serves well.
How about
\b(?=lol)\S*(\S+)(?<=\blol)\1*\b
(replace lol with omg, haha etc.)
This will match lol, lololol, lollll, lollollol etc. but fail lolo, lollllo, lolly and so on.
The rules:
Match the word lol completely.
Then allow any repetition of one or more characters at the end of the word (i. e. l, ol or lol)
So \b(?=zomg)\S*(\S+)(?<=\bzomg)\1*\b will match zomg, zomggg, zomgmgmg, zomgomgomg etc.
In Python, with comments:
result = re.sub(
r"""(?ix)\b # assert position at a word boundary
(?=lol) # assert that "lol" can be matched here
\S* # match any number of characters except whitespace
(\S+) # match at least one character (to be repeated later)
(?<=\blol) # until we have reached exactly the position after the 1st "lol"
\1* # then repeat the preceding character(s) any number of times
\b # and ensure that we end up at another word boundary""",
"lol", subject)
This will also match the "unadorned" version (i. e. lol without any repetition). If you don't want this, use \1+ instead of \1*.