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Python has string.find() and string.rfind() to get the index of a substring in a string.
I'm wondering whether there is something like string.find_all() which can return all found indexes (not only the first from the beginning or the first from the end).
For example:
string = "test test test test"
print string.find('test') # 0
print string.rfind('test') # 15
#this is the goal
print string.find_all('test') # [0,5,10,15]
For counting the occurrences, see Count number of occurrences of a substring in a string.
There is no simple built-in string function that does what you're looking for, but you could use the more powerful regular expressions:
import re
[m.start() for m in re.finditer('test', 'test test test test')]
#[0, 5, 10, 15]
If you want to find overlapping matches, lookahead will do that:
[m.start() for m in re.finditer('(?=tt)', 'ttt')]
#[0, 1]
If you want a reverse find-all without overlaps, you can combine positive and negative lookahead into an expression like this:
search = 'tt'
[m.start() for m in re.finditer('(?=%s)(?!.{1,%d}%s)' % (search, len(search)-1, search), 'ttt')]
#[1]
re.finditer returns a generator, so you could change the [] in the above to () to get a generator instead of a list which will be more efficient if you're only iterating through the results once.
>>> help(str.find)
Help on method_descriptor:
find(...)
S.find(sub [,start [,end]]) -> int
Thus, we can build it ourselves:
def find_all(a_str, sub):
start = 0
while True:
start = a_str.find(sub, start)
if start == -1: return
yield start
start += len(sub) # use start += 1 to find overlapping matches
list(find_all('spam spam spam spam', 'spam')) # [0, 5, 10, 15]
No temporary strings or regexes required.
Here's a (very inefficient) way to get all (i.e. even overlapping) matches:
>>> string = "test test test test"
>>> [i for i in range(len(string)) if string.startswith('test', i)]
[0, 5, 10, 15]
Use re.finditer:
import re
sentence = input("Give me a sentence ")
word = input("What word would you like to find ")
for match in re.finditer(word, sentence):
print (match.start(), match.end())
For word = "this" and sentence = "this is a sentence this this" this will yield the output:
(0, 4)
(19, 23)
(24, 28)
Again, old thread, but here's my solution using a generator and plain str.find.
def findall(p, s):
'''Yields all the positions of
the pattern p in the string s.'''
i = s.find(p)
while i != -1:
yield i
i = s.find(p, i+1)
Example
x = 'banananassantana'
[(i, x[i:i+2]) for i in findall('na', x)]
returns
[(2, 'na'), (4, 'na'), (6, 'na'), (14, 'na')]
You can use re.finditer() for non-overlapping matches.
>>> import re
>>> aString = 'this is a string where the substring "is" is repeated several times'
>>> print [(a.start(), a.end()) for a in list(re.finditer('is', aString))]
[(2, 4), (5, 7), (38, 40), (42, 44)]
but won't work for:
In [1]: aString="ababa"
In [2]: print [(a.start(), a.end()) for a in list(re.finditer('aba', aString))]
Output: [(0, 3)]
Come, let us recurse together.
def locations_of_substring(string, substring):
"""Return a list of locations of a substring."""
substring_length = len(substring)
def recurse(locations_found, start):
location = string.find(substring, start)
if location != -1:
return recurse(locations_found + [location], location+substring_length)
else:
return locations_found
return recurse([], 0)
print(locations_of_substring('this is a test for finding this and this', 'this'))
# prints [0, 27, 36]
No need for regular expressions this way.
If you're just looking for a single character, this would work:
string = "dooobiedoobiedoobie"
match = 'o'
reduce(lambda count, char: count + 1 if char == match else count, string, 0)
# produces 7
Also,
string = "test test test test"
match = "test"
len(string.split(match)) - 1
# produces 4
My hunch is that neither of these (especially #2) is terribly performant.
this is an old thread but i got interested and wanted to share my solution.
def find_all(a_string, sub):
result = []
k = 0
while k < len(a_string):
k = a_string.find(sub, k)
if k == -1:
return result
else:
result.append(k)
k += 1 #change to k += len(sub) to not search overlapping results
return result
It should return a list of positions where the substring was found.
Please comment if you see an error or room for improvment.
This does the trick for me using re.finditer
import re
text = 'This is sample text to test if this pythonic '\
'program can serve as an indexing platform for '\
'finding words in a paragraph. It can give '\
'values as to where the word is located with the '\
'different examples as stated'
# find all occurances of the word 'as' in the above text
find_the_word = re.finditer('as', text)
for match in find_the_word:
print('start {}, end {}, search string \'{}\''.
format(match.start(), match.end(), match.group()))
This thread is a little old but this worked for me:
numberString = "onetwothreefourfivesixseveneightninefiveten"
testString = "five"
marker = 0
while marker < len(numberString):
try:
print(numberString.index("five",marker))
marker = numberString.index("five", marker) + 1
except ValueError:
print("String not found")
marker = len(numberString)
You can try :
>>> string = "test test test test"
>>> for index,value in enumerate(string):
if string[index:index+(len("test"))] == "test":
print index
0
5
10
15
You can try :
import re
str1 = "This dress looks good; you have good taste in clothes."
substr = "good"
result = [_.start() for _ in re.finditer(substr, str1)]
# result = [17, 32]
When looking for a large amount of key words in a document, use flashtext
from flashtext import KeywordProcessor
words = ['test', 'exam', 'quiz']
txt = 'this is a test'
kwp = KeywordProcessor()
kwp.add_keywords_from_list(words)
result = kwp.extract_keywords(txt, span_info=True)
Flashtext runs faster than regex on large list of search words.
This function does not look at all positions inside the string, it does not waste compute resources. My try:
def findAll(string,word):
all_positions=[]
next_pos=-1
while True:
next_pos=string.find(word,next_pos+1)
if(next_pos<0):
break
all_positions.append(next_pos)
return all_positions
to use it call it like this:
result=findAll('this word is a big word man how many words are there?','word')
src = input() # we will find substring in this string
sub = input() # substring
res = []
pos = src.find(sub)
while pos != -1:
res.append(pos)
pos = src.find(sub, pos + 1)
Whatever the solutions provided by others are completely based on the available method find() or any available methods.
What is the core basic algorithm to find all the occurrences of a
substring in a string?
def find_all(string,substring):
"""
Function: Returning all the index of substring in a string
Arguments: String and the search string
Return:Returning a list
"""
length = len(substring)
c=0
indexes = []
while c < len(string):
if string[c:c+length] == substring:
indexes.append(c)
c=c+1
return indexes
You can also inherit str class to new class and can use this function
below.
class newstr(str):
def find_all(string,substring):
"""
Function: Returning all the index of substring in a string
Arguments: String and the search string
Return:Returning a list
"""
length = len(substring)
c=0
indexes = []
while c < len(string):
if string[c:c+length] == substring:
indexes.append(c)
c=c+1
return indexes
Calling the method
newstr.find_all('Do you find this answer helpful? then upvote
this!','this')
This is solution of a similar question from hackerrank. I hope this could help you.
import re
a = input()
b = input()
if b not in a:
print((-1,-1))
else:
#create two list as
start_indc = [m.start() for m in re.finditer('(?=' + b + ')', a)]
for i in range(len(start_indc)):
print((start_indc[i], start_indc[i]+len(b)-1))
Output:
aaadaa
aa
(0, 1)
(1, 2)
(4, 5)
Here's a solution that I came up with, using assignment expression (new feature since Python 3.8):
string = "test test test test"
phrase = "test"
start = -1
result = [(start := string.find(phrase, start + 1)) for _ in range(string.count(phrase))]
Output:
[0, 5, 10, 15]
I think the most clean way of solution is without libraries and yields:
def find_all_occurrences(string, sub):
index_of_occurrences = []
current_index = 0
while True:
current_index = string.find(sub, current_index)
if current_index == -1:
return index_of_occurrences
else:
index_of_occurrences.append(current_index)
current_index += len(sub)
find_all_occurrences(string, substr)
Note: find() method returns -1 when it can't find anything
The pythonic way would be:
mystring = 'Hello World, this should work!'
find_all = lambda c,s: [x for x in range(c.find(s), len(c)) if c[x] == s]
# s represents the search string
# c represents the character string
find_all(mystring,'o') # will return all positions of 'o'
[4, 7, 20, 26]
>>>
if you only want to use numpy here is a solution
import numpy as np
S= "test test test test"
S2 = 'test'
inds = np.cumsum([len(k)+len(S2) for k in S.split(S2)[:-1]])- len(S2)
print(inds)
if you want to use without re(regex) then:
find_all = lambda _str,_w : [ i for i in range(len(_str)) if _str.startswith(_w,i) ]
string = "test test test test"
print( find_all(string, 'test') ) # >>> [0, 5, 10, 15]
please look at below code
#!/usr/bin/env python
# coding:utf-8
'''黄哥Python'''
def get_substring_indices(text, s):
result = [i for i in range(len(text)) if text.startswith(s, i)]
return result
if __name__ == '__main__':
text = "How much wood would a wood chuck chuck if a wood chuck could chuck wood?"
s = 'wood'
print get_substring_indices(text, s)
def find_index(string, let):
enumerated = [place for place, letter in enumerate(string) if letter == let]
return enumerated
for example :
find_index("hey doode find d", "d")
returns:
[4, 7, 13, 15]
Not exactly what OP asked but you could also use the split function to get a list of where all the substrings don't occur. OP didn't specify the end goal of the code but if your goal is to remove the substrings anyways then this could be a simple one-liner. There are probably more efficient ways to do this with larger strings; regular expressions would be preferable in that case
# Extract all non-substrings
s = "an-example-string"
s_no_dash = s.split('-')
# >>> s_no_dash
# ['an', 'example', 'string']
# Or extract and join them into a sentence
s_no_dash2 = ' '.join(s.split('-'))
# >>> s_no_dash2
# 'an example string'
Did a brief skim of other answers so apologies if this is already up there.
def count_substring(string, sub_string):
c=0
for i in range(0,len(string)-2):
if string[i:i+len(sub_string)] == sub_string:
c+=1
return c
if __name__ == '__main__':
string = input().strip()
sub_string = input().strip()
count = count_substring(string, sub_string)
print(count)
I runned in the same problem and did this:
hw = 'Hello oh World!'
list_hw = list(hw)
o_in_hw = []
while True:
o = hw.find('o')
if o != -1:
o_in_hw.append(o)
list_hw[o] = ' '
hw = ''.join(list_hw)
else:
print(o_in_hw)
break
Im pretty new at coding so you can probably simplify it (and if planned to used continuously of course make it a function).
All and all it works as intended for what i was doing.
Edit: Please consider this is for single characters only, and it will change your variable, so you have to create a copy of the string in a new variable to save it, i didnt put it in the code cause its easy and its only to show how i made it work.
By slicing we find all the combinations possible and append them in a list and find the number of times it occurs using count function
s=input()
n=len(s)
l=[]
f=input()
print(s[0])
for i in range(0,n):
for j in range(1,n+1):
l.append(s[i:j])
if f in l:
print(l.count(f))
To find all the occurence of a character in a give string and return as a dictionary
eg: hello
result :
{'h':1, 'e':1, 'l':2, 'o':1}
def count(string):
result = {}
if(string):
for i in string:
result[i] = string.count(i)
return result
return {}
or else you do like this
from collections import Counter
def count(string):
return Counter(string)
I need to insert a string (character by character) into another string at every 3rd position
For example:- string_1:-wwwaabkccgkll
String_2:- toadhp
Now I need to insert string2 char by char into string1 at every third position
So the output must be wwtaaobkaccdgkhllp
Need in Python.. even Java is ok
So i tried this
Test_str="hiimdumbiknow"
challenge="toadh"
new_st=challenge [k]
Last=list(test_str)
K=0
For i in range(Len(test_str)):
if(i%3==0):
last.insert(i,new_st)
K+=1
and the output i get
thitimtdutmbtiknow
You can split test_str into sub-strings to length 2, and then iterate merging them with challenge:
def concat3(test_str, challenge):
chunks = [test_str[i:i+2] for i in range(0,len(test_str),2)]
result = []
i = j = 0
while i<len(chunks) or j<len(challenge):
if i<len(chunks):
result.append(chunks[i])
i += 1
if j<len(challenge):
result.append(challenge[j])
j += 1
return ''.join(result)
test_str = "hiimdumbiknow"
challenge = "toadh"
print(concat3(test_str, challenge))
# hitimoduambdikhnow
This method works even if the lengths of test_str and challenge are mismatching. (The remaining characters in the longest string will be appended at the end.)
You can split Test_str in to groups of two letters and then re-join with each letter from challenge in between as follows;
import itertools
print(''.join(f'{two}{letter}' for two, letter in itertools.zip_longest([Test_str[i:i+2] for i in range(0,len(Test_str),2)], challenge, fillvalue='')))
Output:
hitimoduambdikhnow
*edited to split in to groups of two rather than three as originally posted
you can try this, make an iter above the second string and iterate over the first one and select which character should be part of the final string according the position
def add3(s1, s2):
def n():
try:
k = iter(s2)
for i,j in enumerate(s1):
yield (j if (i==0 or (i+1)%3) else next(k))
except:
try:
yield s1[i+1:]
except:
pass
return ''.join(n())
def insertstring(test_str,challenge):
result = ''
x = [x for x in test_str]
y = [y for y in challenge]
j = 0
for i in range(len(x)):
if i % 2 != 0 or i == 0:
result += x[i]
else:
if j < 5:
result += y[j]
result += x[i]
j += 1
get_last_element = x[-1]
return result + get_last_element
print(insertstring(test_str,challenge))
#output: hitimoduambdikhnow
This question already has answers here:
Split string every nth character?
(19 answers)
Splitting a string into 2-letter segments [duplicate]
(6 answers)
Closed 2 years ago.
I want to divide text into pairs.
Input: text = "abcde"
Goal Output: result = ["ab", "cd", "e_"]
Current Output: result = ['ab', 'abcd']
My current code looks like this. But I do not know how I do that now. Anyone has a tip for me?
def split_pairs(text):
result = []
if text is None or not text:
return []
pair = ""
for i in range(len(text)):
if i % 2 == 0:
pair += text[i]
pair += text[i+1]
else:
result.append(pair)
return result
You could use a list comprehension to zip together the even values with the corresponding odd values. And using itertools.zip_longest you can use the fillvalue argument to provide a "fill in" if there is a length mismatch.
>>> from itertools import zip_longest
>>> s = 'abcde'
>>> pairs = [i+j for i,j in zip_longest(s[::2], s[1::2], fillvalue='_')]
>>> pairs
['ab', 'cd', 'e_']
You should reset your "pair" variable once appended to "result"
def split_pairs(text):
result = []
if text is None or not text:
return []
pair = ""
for i in range(len(text)):
if i % 2 == 0:
pair += text[i]
pair += text[i+1]
else:
result.append(pair)
pair = ""
return result
You could also use a list comprehension over a range with 3rd step parameter and add ljust to add _. This will also work nicely for more than just pairs:
>>> s = "abcde"
>>> k = 2
>>> [s[i:i+k].ljust(k, "_") for i in range(0, len(s), k)]
['ab', 'cd', 'e_']
I am not if your code needed to be in the format you originally wrote it in, but I wrote the below code that gets the job done.
def split_pairs(text):
if len(text) % 2 == 0:
result = [text[i:i+2] for i in range(0, len(text), 2)]
else:
result = [text[i:i+2] for i in range(0, len(text), 2)]
result[-1]+="_"
return result
The issue here is that the "pair" variable is never reinitialized to "".
Make sure you make it an empty string in your else block.
def split_pairs(text):
result = []
if text is None or not text:
return []
pair = ""
for i in range(len(text)):
if i % 2 == 0:
pair += text[i]
pair += text[i+1]
else:
result.append(pair)
pair = "" # Make sure you reset it
return result
If you want to have a "_" at the end (in case of an odd number of character), you could do like the following:
def split_pairs(text):
result = []
if text is None or not text:
return []
pair = "__" # Setting pair to "__" by default
for i in range(len(text)):
if i % 2 == 0:
pair[0] = text[i]
if i < len(text): # Avoiding overflow
pair[1] = text[i+1]
else:
result.append(pair)
pair = "__" # Make sure you reset it
if pair != "__": # Last bit
result.append(pair)
return result
Python has string.find() and string.rfind() to get the index of a substring in a string.
I'm wondering whether there is something like string.find_all() which can return all found indexes (not only the first from the beginning or the first from the end).
For example:
string = "test test test test"
print string.find('test') # 0
print string.rfind('test') # 15
#this is the goal
print string.find_all('test') # [0,5,10,15]
For counting the occurrences, see Count number of occurrences of a substring in a string.
There is no simple built-in string function that does what you're looking for, but you could use the more powerful regular expressions:
import re
[m.start() for m in re.finditer('test', 'test test test test')]
#[0, 5, 10, 15]
If you want to find overlapping matches, lookahead will do that:
[m.start() for m in re.finditer('(?=tt)', 'ttt')]
#[0, 1]
If you want a reverse find-all without overlaps, you can combine positive and negative lookahead into an expression like this:
search = 'tt'
[m.start() for m in re.finditer('(?=%s)(?!.{1,%d}%s)' % (search, len(search)-1, search), 'ttt')]
#[1]
re.finditer returns a generator, so you could change the [] in the above to () to get a generator instead of a list which will be more efficient if you're only iterating through the results once.
>>> help(str.find)
Help on method_descriptor:
find(...)
S.find(sub [,start [,end]]) -> int
Thus, we can build it ourselves:
def find_all(a_str, sub):
start = 0
while True:
start = a_str.find(sub, start)
if start == -1: return
yield start
start += len(sub) # use start += 1 to find overlapping matches
list(find_all('spam spam spam spam', 'spam')) # [0, 5, 10, 15]
No temporary strings or regexes required.
Here's a (very inefficient) way to get all (i.e. even overlapping) matches:
>>> string = "test test test test"
>>> [i for i in range(len(string)) if string.startswith('test', i)]
[0, 5, 10, 15]
Use re.finditer:
import re
sentence = input("Give me a sentence ")
word = input("What word would you like to find ")
for match in re.finditer(word, sentence):
print (match.start(), match.end())
For word = "this" and sentence = "this is a sentence this this" this will yield the output:
(0, 4)
(19, 23)
(24, 28)
Again, old thread, but here's my solution using a generator and plain str.find.
def findall(p, s):
'''Yields all the positions of
the pattern p in the string s.'''
i = s.find(p)
while i != -1:
yield i
i = s.find(p, i+1)
Example
x = 'banananassantana'
[(i, x[i:i+2]) for i in findall('na', x)]
returns
[(2, 'na'), (4, 'na'), (6, 'na'), (14, 'na')]
You can use re.finditer() for non-overlapping matches.
>>> import re
>>> aString = 'this is a string where the substring "is" is repeated several times'
>>> print [(a.start(), a.end()) for a in list(re.finditer('is', aString))]
[(2, 4), (5, 7), (38, 40), (42, 44)]
but won't work for:
In [1]: aString="ababa"
In [2]: print [(a.start(), a.end()) for a in list(re.finditer('aba', aString))]
Output: [(0, 3)]
Come, let us recurse together.
def locations_of_substring(string, substring):
"""Return a list of locations of a substring."""
substring_length = len(substring)
def recurse(locations_found, start):
location = string.find(substring, start)
if location != -1:
return recurse(locations_found + [location], location+substring_length)
else:
return locations_found
return recurse([], 0)
print(locations_of_substring('this is a test for finding this and this', 'this'))
# prints [0, 27, 36]
No need for regular expressions this way.
If you're just looking for a single character, this would work:
string = "dooobiedoobiedoobie"
match = 'o'
reduce(lambda count, char: count + 1 if char == match else count, string, 0)
# produces 7
Also,
string = "test test test test"
match = "test"
len(string.split(match)) - 1
# produces 4
My hunch is that neither of these (especially #2) is terribly performant.
this is an old thread but i got interested and wanted to share my solution.
def find_all(a_string, sub):
result = []
k = 0
while k < len(a_string):
k = a_string.find(sub, k)
if k == -1:
return result
else:
result.append(k)
k += 1 #change to k += len(sub) to not search overlapping results
return result
It should return a list of positions where the substring was found.
Please comment if you see an error or room for improvment.
This does the trick for me using re.finditer
import re
text = 'This is sample text to test if this pythonic '\
'program can serve as an indexing platform for '\
'finding words in a paragraph. It can give '\
'values as to where the word is located with the '\
'different examples as stated'
# find all occurances of the word 'as' in the above text
find_the_word = re.finditer('as', text)
for match in find_the_word:
print('start {}, end {}, search string \'{}\''.
format(match.start(), match.end(), match.group()))
This thread is a little old but this worked for me:
numberString = "onetwothreefourfivesixseveneightninefiveten"
testString = "five"
marker = 0
while marker < len(numberString):
try:
print(numberString.index("five",marker))
marker = numberString.index("five", marker) + 1
except ValueError:
print("String not found")
marker = len(numberString)
You can try :
>>> string = "test test test test"
>>> for index,value in enumerate(string):
if string[index:index+(len("test"))] == "test":
print index
0
5
10
15
You can try :
import re
str1 = "This dress looks good; you have good taste in clothes."
substr = "good"
result = [_.start() for _ in re.finditer(substr, str1)]
# result = [17, 32]
When looking for a large amount of key words in a document, use flashtext
from flashtext import KeywordProcessor
words = ['test', 'exam', 'quiz']
txt = 'this is a test'
kwp = KeywordProcessor()
kwp.add_keywords_from_list(words)
result = kwp.extract_keywords(txt, span_info=True)
Flashtext runs faster than regex on large list of search words.
This function does not look at all positions inside the string, it does not waste compute resources. My try:
def findAll(string,word):
all_positions=[]
next_pos=-1
while True:
next_pos=string.find(word,next_pos+1)
if(next_pos<0):
break
all_positions.append(next_pos)
return all_positions
to use it call it like this:
result=findAll('this word is a big word man how many words are there?','word')
src = input() # we will find substring in this string
sub = input() # substring
res = []
pos = src.find(sub)
while pos != -1:
res.append(pos)
pos = src.find(sub, pos + 1)
Whatever the solutions provided by others are completely based on the available method find() or any available methods.
What is the core basic algorithm to find all the occurrences of a
substring in a string?
def find_all(string,substring):
"""
Function: Returning all the index of substring in a string
Arguments: String and the search string
Return:Returning a list
"""
length = len(substring)
c=0
indexes = []
while c < len(string):
if string[c:c+length] == substring:
indexes.append(c)
c=c+1
return indexes
You can also inherit str class to new class and can use this function
below.
class newstr(str):
def find_all(string,substring):
"""
Function: Returning all the index of substring in a string
Arguments: String and the search string
Return:Returning a list
"""
length = len(substring)
c=0
indexes = []
while c < len(string):
if string[c:c+length] == substring:
indexes.append(c)
c=c+1
return indexes
Calling the method
newstr.find_all('Do you find this answer helpful? then upvote
this!','this')
This is solution of a similar question from hackerrank. I hope this could help you.
import re
a = input()
b = input()
if b not in a:
print((-1,-1))
else:
#create two list as
start_indc = [m.start() for m in re.finditer('(?=' + b + ')', a)]
for i in range(len(start_indc)):
print((start_indc[i], start_indc[i]+len(b)-1))
Output:
aaadaa
aa
(0, 1)
(1, 2)
(4, 5)
Here's a solution that I came up with, using assignment expression (new feature since Python 3.8):
string = "test test test test"
phrase = "test"
start = -1
result = [(start := string.find(phrase, start + 1)) for _ in range(string.count(phrase))]
Output:
[0, 5, 10, 15]
I think the most clean way of solution is without libraries and yields:
def find_all_occurrences(string, sub):
index_of_occurrences = []
current_index = 0
while True:
current_index = string.find(sub, current_index)
if current_index == -1:
return index_of_occurrences
else:
index_of_occurrences.append(current_index)
current_index += len(sub)
find_all_occurrences(string, substr)
Note: find() method returns -1 when it can't find anything
The pythonic way would be:
mystring = 'Hello World, this should work!'
find_all = lambda c,s: [x for x in range(c.find(s), len(c)) if c[x] == s]
# s represents the search string
# c represents the character string
find_all(mystring,'o') # will return all positions of 'o'
[4, 7, 20, 26]
>>>
if you only want to use numpy here is a solution
import numpy as np
S= "test test test test"
S2 = 'test'
inds = np.cumsum([len(k)+len(S2) for k in S.split(S2)[:-1]])- len(S2)
print(inds)
if you want to use without re(regex) then:
find_all = lambda _str,_w : [ i for i in range(len(_str)) if _str.startswith(_w,i) ]
string = "test test test test"
print( find_all(string, 'test') ) # >>> [0, 5, 10, 15]
please look at below code
#!/usr/bin/env python
# coding:utf-8
'''黄哥Python'''
def get_substring_indices(text, s):
result = [i for i in range(len(text)) if text.startswith(s, i)]
return result
if __name__ == '__main__':
text = "How much wood would a wood chuck chuck if a wood chuck could chuck wood?"
s = 'wood'
print get_substring_indices(text, s)
def find_index(string, let):
enumerated = [place for place, letter in enumerate(string) if letter == let]
return enumerated
for example :
find_index("hey doode find d", "d")
returns:
[4, 7, 13, 15]
Not exactly what OP asked but you could also use the split function to get a list of where all the substrings don't occur. OP didn't specify the end goal of the code but if your goal is to remove the substrings anyways then this could be a simple one-liner. There are probably more efficient ways to do this with larger strings; regular expressions would be preferable in that case
# Extract all non-substrings
s = "an-example-string"
s_no_dash = s.split('-')
# >>> s_no_dash
# ['an', 'example', 'string']
# Or extract and join them into a sentence
s_no_dash2 = ' '.join(s.split('-'))
# >>> s_no_dash2
# 'an example string'
Did a brief skim of other answers so apologies if this is already up there.
def count_substring(string, sub_string):
c=0
for i in range(0,len(string)-2):
if string[i:i+len(sub_string)] == sub_string:
c+=1
return c
if __name__ == '__main__':
string = input().strip()
sub_string = input().strip()
count = count_substring(string, sub_string)
print(count)
I runned in the same problem and did this:
hw = 'Hello oh World!'
list_hw = list(hw)
o_in_hw = []
while True:
o = hw.find('o')
if o != -1:
o_in_hw.append(o)
list_hw[o] = ' '
hw = ''.join(list_hw)
else:
print(o_in_hw)
break
Im pretty new at coding so you can probably simplify it (and if planned to used continuously of course make it a function).
All and all it works as intended for what i was doing.
Edit: Please consider this is for single characters only, and it will change your variable, so you have to create a copy of the string in a new variable to save it, i didnt put it in the code cause its easy and its only to show how i made it work.
By slicing we find all the combinations possible and append them in a list and find the number of times it occurs using count function
s=input()
n=len(s)
l=[]
f=input()
print(s[0])
for i in range(0,n):
for j in range(1,n+1):
l.append(s[i:j])
if f in l:
print(l.count(f))
To find all the occurence of a character in a give string and return as a dictionary
eg: hello
result :
{'h':1, 'e':1, 'l':2, 'o':1}
def count(string):
result = {}
if(string):
for i in string:
result[i] = string.count(i)
return result
return {}
or else you do like this
from collections import Counter
def count(string):
return Counter(string)
This question already has answers here:
How to find all occurrences of a substring?
(32 answers)
Closed 12 months ago.
How do I find multiple occurrences of a string within a string in Python? Consider this:
>>> text = "Allowed Hello Hollow"
>>> text.find("ll")
1
>>>
So the first occurrence of ll is at 1 as expected. How do I find the next occurrence of it?
Same question is valid for a list. Consider:
>>> x = ['ll', 'ok', 'll']
How do I find all the ll with their indexes?
Using regular expressions, you can use re.finditer to find all (non-overlapping) occurences:
>>> import re
>>> text = 'Allowed Hello Hollow'
>>> for m in re.finditer('ll', text):
print('ll found', m.start(), m.end())
ll found 1 3
ll found 10 12
ll found 16 18
Alternatively, if you don't want the overhead of regular expressions, you can also repeatedly use str.find to get the next index:
>>> text = 'Allowed Hello Hollow'
>>> index = 0
>>> while index < len(text):
index = text.find('ll', index)
if index == -1:
break
print('ll found at', index)
index += 2 # +2 because len('ll') == 2
ll found at 1
ll found at 10
ll found at 16
This also works for lists and other sequences.
I think what you are looking for is string.count
"Allowed Hello Hollow".count('ll')
>>> 3
Hope this helps
NOTE: this only captures non-overlapping occurences
For the list example, use a comprehension:
>>> l = ['ll', 'xx', 'll']
>>> print [n for (n, e) in enumerate(l) if e == 'll']
[0, 2]
Similarly for strings:
>>> text = "Allowed Hello Hollow"
>>> print [n for n in xrange(len(text)) if text.find('ll', n) == n]
[1, 10, 16]
this will list adjacent runs of "ll', which may or may not be what you want:
>>> text = 'Alllowed Hello Holllow'
>>> print [n for n in xrange(len(text)) if text.find('ll', n) == n]
[1, 2, 11, 17, 18]
FWIW, here are a couple of non-RE alternatives that I think are neater than poke's solution.
The first uses str.index and checks for ValueError:
def findall(sub, string):
"""
>>> text = "Allowed Hello Hollow"
>>> tuple(findall('ll', text))
(1, 10, 16)
"""
index = 0 - len(sub)
try:
while True:
index = string.index(sub, index + len(sub))
yield index
except ValueError:
pass
The second tests uses str.find and checks for the sentinel of -1 by using iter:
def findall_iter(sub, string):
"""
>>> text = "Allowed Hello Hollow"
>>> tuple(findall_iter('ll', text))
(1, 10, 16)
"""
def next_index(length):
index = 0 - length
while True:
index = string.find(sub, index + length)
yield index
return iter(next_index(len(sub)).next, -1)
To apply any of these functions to a list, tuple or other iterable of strings, you can use a higher-level function —one that takes a function as one of its arguments— like this one:
def findall_each(findall, sub, strings):
"""
>>> texts = ("fail", "dolly the llama", "Hello", "Hollow", "not ok")
>>> list(findall_each(findall, 'll', texts))
[(), (2, 10), (2,), (2,), ()]
>>> texts = ("parallellized", "illegally", "dillydallying", "hillbillies")
>>> list(findall_each(findall_iter, 'll', texts))
[(4, 7), (1, 6), (2, 7), (2, 6)]
"""
return (tuple(findall(sub, string)) for string in strings)
For your list example:
In [1]: x = ['ll','ok','ll']
In [2]: for idx, value in enumerate(x):
...: if value == 'll':
...: print idx, value
0 ll
2 ll
If you wanted all the items in a list that contained 'll', you could also do that.
In [3]: x = ['Allowed','Hello','World','Hollow']
In [4]: for idx, value in enumerate(x):
...: if 'll' in value:
...: print idx, value
...:
...:
0 Allowed
1 Hello
3 Hollow
This code might not be the shortest/most efficient but it is simple and understandable
def findall(f, s):
l = []
i = -1
while True:
i = s.find(f, i+1)
if i == -1:
return l
l.append(s.find(f, i))
findall('test', 'test test test test')
# [0, 5, 10, 15]
For the first version, checking a string:
def findall(text, sub):
"""Return all indices at which substring occurs in text"""
return [
index
for index in range(len(text) - len(sub) + 1)
if text[index:].startswith(sub)
]
print(findall('Allowed Hello Hollow', 'll'))
# [1, 10, 16]
No need to import re. This should run in linear time, as it only loops through the string once (and stops before the end, once there aren't enough characters left to fit the substring). I also find it quite readable, personally.
Note that this will find overlapping occurrences:
print(findall('aaa', 'aa'))
# [0, 1]
>>> for n,c in enumerate(text):
... try:
... if c+text[n+1] == "ll": print n
... except: pass
...
1
10
16
This version should be linear in length of the string, and should be fine as long as the sequences aren't too repetitive (in which case you can replace the recursion with a while loop).
def find_all(st, substr, start_pos=0, accum=[]):
ix = st.find(substr, start_pos)
if ix == -1:
return accum
return find_all(st, substr, start_pos=ix + 1, accum=accum + [ix])
bstpierre's list comprehension is a good solution for short sequences, but looks to have quadratic complexity and never finished on a long text I was using.
findall_lc = lambda txt, substr: [n for n in xrange(len(txt))
if txt.find(substr, n) == n]
For a random string of non-trivial length, the two functions give the same result:
import random, string; random.seed(0)
s = ''.join([random.choice(string.ascii_lowercase) for _ in range(100000)])
>>> find_all(s, 'th') == findall_lc(s, 'th')
True
>>> findall_lc(s, 'th')[:4]
[564, 818, 1872, 2470]
But the quadratic version is about 300 times slower
%timeit find_all(s, 'th')
1000 loops, best of 3: 282 µs per loop
%timeit findall_lc(s, 'th')
10 loops, best of 3: 92.3 ms per loop
Brand new to programming in general and working through an online tutorial. I was asked to do this as well, but only using the methods I had learned so far (basically strings and loops). Not sure if this adds any value here, and I know this isn't how you would do it, but I got it to work with this:
needle = input()
haystack = input()
counter = 0
n=-1
for i in range (n+1,len(haystack)+1):
for j in range(n+1,len(haystack)+1):
n=-1
if needle != haystack[i:j]:
n = n+1
continue
if needle == haystack[i:j]:
counter = counter + 1
print (counter)
The following function finds all the occurrences of a string inside another while informing the position where each occurrence is found.
You can call the function using the test cases in the table below. You can try with words, spaces and numbers all mixed up.
The function works well with overlapping characters.
theString
aString
"661444444423666455678966"
"55"
"661444444423666455678966"
"44"
"6123666455678966"
"666"
"66123666455678966"
"66"
Calling examples:
1. print("Number of occurrences: ", find_all("123666455556785555966", "5555"))
output:
Found in position: 7
Found in position: 14
Number of occurrences: 2
2. print("Number of occurrences: ", find_all("Allowed Hello Hollow", "ll "))
output:
Found in position: 1
Found in position: 10
Found in position: 16
Number of occurrences: 3
3. print("Number of occurrences: ", find_all("Aaa bbbcd$###abWebbrbbbbrr 123", "bbb"))
output:
Found in position: 4
Found in position: 21
Number of occurrences: 2
def find_all(theString, aString):
count = 0
i = len(aString)
x = 0
while x < len(theString) - (i-1):
if theString[x:x+i] == aString:
print("Found in position: ", x)
x=x+i
count=count+1
else:
x=x+1
return count
#!/usr/local/bin python3
#-*- coding: utf-8 -*-
main_string = input()
sub_string = input()
count = counter = 0
for i in range(len(main_string)):
if main_string[i] == sub_string[0]:
k = i + 1
for j in range(1, len(sub_string)):
if k != len(main_string) and main_string[k] == sub_string[j]:
count += 1
k += 1
if count == (len(sub_string) - 1):
counter += 1
count = 0
print(counter)
This program counts the number of all substrings even if they are overlapped without the use of regex. But this is a naive implementation and for better results in worst case it is advised to go through either Suffix Tree, KMP and other string matching data structures and algorithms.
Here is my function for finding multiple occurrences. Unlike the other solutions here, it supports the optional start and end parameters for slicing, just like str.index:
def all_substring_indexes(string, substring, start=0, end=None):
result = []
new_start = start
while True:
try:
index = string.index(substring, new_start, end)
except ValueError:
return result
else:
result.append(index)
new_start = index + len(substring)
A simple iterative code which returns a list of indices where the substring occurs.
def allindices(string, sub):
l=[]
i = string.find(sub)
while i >= 0:
l.append(i)
i = string.find(sub, i + 1)
return l
You can split to get relative positions then sum consecutive numbers in a list and add (string length * occurence order) at the same time to get the wanted string indexes.
>>> key = 'll'
>>> text = "Allowed Hello Hollow"
>>> x = [len(i) for i in text.split(key)[:-1]]
>>> [sum(x[:i+1]) + i*len(key) for i in range(len(x))]
[1, 10, 16]
>>>
Maybe not so Pythonic, but somewhat more self-explanatory. It returns the position of the word looked in the original string.
def retrieve_occurences(sequence, word, result, base_counter):
indx = sequence.find(word)
if indx == -1:
return result
result.append(indx + base_counter)
base_counter += indx + len(word)
return retrieve_occurences(sequence[indx + len(word):], word, result, base_counter)
I think there's no need to test for length of text; just keep finding until there's nothing left to find. Like this:
>>> text = 'Allowed Hello Hollow'
>>> place = 0
>>> while text.find('ll', place) != -1:
print('ll found at', text.find('ll', place))
place = text.find('ll', place) + 2
ll found at 1
ll found at 10
ll found at 16
You can also do it with conditional list comprehension like this:
string1= "Allowed Hello Hollow"
string2= "ll"
print [num for num in xrange(len(string1)-len(string2)+1) if string1[num:num+len(string2)]==string2]
# [1, 10, 16]
I had randomly gotten this idea just a while ago. Using a While loop with string splicing and string search can work, even for overlapping strings.
findin = "algorithm alma mater alison alternation alpines"
search = "al"
inx = 0
num_str = 0
while True:
inx = findin.find(search)
if inx == -1: #breaks before adding 1 to number of string
break
inx = inx + 1
findin = findin[inx:] #to splice the 'unsearched' part of the string
num_str = num_str + 1 #counts no. of string
if num_str != 0:
print("There are ",num_str," ",search," in your string.")
else:
print("There are no ",search," in your string.")
I'm an amateur in Python Programming (Programming of any language, actually), and am not sure what other issues it could have, but I guess it's working fine?
I guess lower() could be used somewhere in it too if needed.