find all index positions

[micklee74]

hi
say i have string like this
astring = 'abcd efgd 1234 fsdf gfds abcde 1234'
if i want to find which postion is 1234, how can i achieve this...? i
want to use index() but it only give me the first occurence. I want to
know the positions of both "1234"
thanks

[Tim Chase]

> say i have string like this

astring = 'abcd efgd 1234 fsdf gfds abcde 1234'
if i want to find which postion is 1234, how can i
achieve this...? i want to use index() but it only give
me the first occurence. I want to know the positions of
both "1234"

Well, I'm not sure how efficient it is, but the following
seemed to do it for me:

a = 'abcd efgd 1234 fsdf gfds abcde 1234'
thing = '1234'
offsets = [i for i in range(len(a)) if a.startswith(th ing, i)] print offsets

[10, 31]
HTH,

-tkc

[alisonken1]

mi*******@hotma il.com wrote:

hi
say i have string like this
astring = 'abcd efgd 1234 fsdf gfds abcde 1234'
if i want to find which postion is 1234, how can i achieve this...? i
want to use index() but it only give me the first occurence. I want to
know the positions of both "1234"
thanks

==========
def getAllIndex(aSt ring=None, aSub=None):
t=dict()
c=0
ndx=0
while True:
try:
ndx=aString.ind ex(aSub, ndx)
t[c]=ndx
ndx += 1
c += 1
except ValueError:
break
return t
===========

This will return a dictionary of what was found; i.e.,

getAllIndex('ab cd 1234 efgh 1234 ijkl', '1234')

{0: 5, 1: 15}

[Peter Otten]

alisonken1 wrote:

==========
def getAllIndex(aSt ring=None, aSub=None):
t=dict()
c=0
ndx=0
while True:
try:
ndx=aString.ind ex(aSub, ndx)
t[c]=ndx
ndx += 1
c += 1
except ValueError:
break
return t
===========

This will return a dictionary of what was found; i.e.,

getAllIndex('ab cd 1234 efgh 1234 ijkl', '1234')

{0: 5, 1: 15}

Consecutive integers starting at 0 as keys? Looks like you want a list
instead of a dict.

Peter

[Scott David Daniels]

mi*******@hotma il.com wrote:

astring = 'abcd efgd 1234 fsdf gfds abcde 1234'
<paraphrase> i want to find all positions of '1234' in astring.</paraphrase>

def positions(targe t, source):
'''Produce all positions of target in source'''
pos = -1
try:
while True:
pos = source.index(ta rget, pos + 1)
yield pos
except ValueError:
pass

print list(positions( '1234', 'abcd efgd 1234 fsdf gfds abcde 1234'))

prints:
[10, 31]

--Scott David Daniels
sc***********@a cm.org

[Paul Rubin]

mi*******@hotma il.com writes:

say i have string like this
astring = 'abcd efgd 1234 fsdf gfds abcde 1234'
if i want to find which postion is 1234, how can i achieve this...? i
want to use index() but it only give me the first occurence. I want to
know the positions of both "1234"

Most straightforward ly with re.findall -- see the docs.

[Gary Herron]

mi*******@hotma il.com wrote:

hi
say i have string like this
astring = 'abcd efgd 1234 fsdf gfds abcde 1234'
if i want to find which postion is 1234, how can i achieve this...? i
want to use index() but it only give me the first occurence. I want to
know the positions of both "1234"
thanks

The regular expression module (called re) has a function (named
finditer) that gives you what you want here.

The finditer function will find all matches of a pattern in a string and
return an iterator for them. You can loop through the iterator and do
what you want with each occurrence.

import re
astring = 'abcd efgd 1234 fsdf gfds abcde 1234'
pattern = '1234'
Perhaps just find the starting point of each match:
for match in re.finditer(pat tern,astring): .... print match.start()
....
10
31

Or the span of each match:
for match in re.finditer(pat tern,astring): .... print match.span()
....
(10, 14)
(31, 35)

Or use list comprehension to build a list of starting positions:
[match.start() for match in re.finditer(pat tern,astring)]

[10, 31]

And so on ....

Of course, if you wish, the re module can work with vastly more complex
patterns than just a constant string like your '1234'.

Gary Herron

[Gary Herron]

Paul Rubin wrote:

mi*******@hotm ail.com writes:

say i have string like this
astring = 'abcd efgd 1234 fsdf gfds abcde 1234'
if i want to find which postion is 1234, how can i achieve this...? i
want to use index() but it only give me the first occurence. I want to
know the positions of both "1234"

Most straightforward ly with re.findall -- see the docs.

Not quite. The findall will list the matching strings, not their
positions. -- He'll get ['1234','1234']. The finditer function will
work for his requirements. See my other post to this thread.

[vbgunz]

I thought this to be a great exercise so I went the extra length to
turn it into a function for my little but growing library. I hope you
enjoy :)
def indexer(string, target):
'''indexer(stri ng, target) ->[list of target indexes]

enter in a string and a target and indexer will either return a
list of all targeted indexes if at least one target is found or
indexer will return None if the target is not found in sequence.

indexer('a long long day is long', 'long') [2, 7, 19]
indexer('a long long day is long', 'day') [12]
indexer('a long long day is long', 'short')

None
'''

res = []

if string.count(ta rget) >= 1:
res.append(stri ng.find(target) )

if string.count(ta rget) >= 2:
for item in xrange(string.c ount(target) - 1):
res.append(stri ng.find(target, res[-1] + 1))

return res
if __name__ == '__main__':
print indexer('a long long day is long', 'long') # -> [2, 7, 19]
print indexer('a long long day is long', 'day') # -> [12]
print indexer('a long long day is long', 'short') # -> None

[alisonken1]

Scott David Daniels wrote:

mi*******@hotma il.com wrote: <SNIP> print list(positions( '1234', 'abcd efgd 1234 fsdf gfds abcde 1234'))

prints:
[10, 31]

Nicer than mine ;)
Shows I need to get a job where I use python more!