find files that have specific word in a loop?

up vote
1
down vote

favorite

In that folder there are files called: a11.shp,a11.shx,u21.shp,u21.shx

import os 

words = ('a11','u21')

for root, dirs,files in os.walk(r'C:UsersuserDesktopfolder1'):

    for i in files: 

        if i in words:

            print(i)

TypeError: tuple indices must be integers or slices, not str

It requires some kind of an index for the tuple to run for all the items.

I want to use it as 'contains' that word and it doesn't work like contains the in in this case.

How can this be done when giving these exact words in the tuple. Without startswith.

edited Nov 22 at 14:44

asked Nov 22 at 14:35

user10680652

234

do you want to match the exact words, or just partial words? For instance can a11_xxxx.txt matches a11 ?
– Jean-François Fabre
Nov 22 at 14:50

add a comment |

up vote
1
down vote

favorite

In that folder there are files called: a11.shp,a11.shx,u21.shp,u21.shx

import os 

words = ('a11','u21')

for root, dirs,files in os.walk(r'C:UsersuserDesktopfolder1'):

    for i in files: 

        if i in words:

            print(i)

TypeError: tuple indices must be integers or slices, not str

It requires some kind of an index for the tuple to run for all the items.

I want to use it as 'contains' that word and it doesn't work like contains the in in this case.

How can this be done when giving these exact words in the tuple. Without startswith.

edited Nov 22 at 14:44

asked Nov 22 at 14:35

user10680652

234

do you want to match the exact words, or just partial words? For instance can a11_xxxx.txt matches a11 ?
– Jean-François Fabre
Nov 22 at 14:50

add a comment |

up vote
1
down vote

favorite

In that folder there are files called: a11.shp,a11.shx,u21.shp,u21.shx

import os 

words = ('a11','u21')

for root, dirs,files in os.walk(r'C:UsersuserDesktopfolder1'):

    for i in files: 

        if i in words:

            print(i)

TypeError: tuple indices must be integers or slices, not str

It requires some kind of an index for the tuple to run for all the items.

I want to use it as 'contains' that word and it doesn't work like contains the in in this case.

How can this be done when giving these exact words in the tuple. Without startswith.

edited Nov 22 at 14:44

asked Nov 22 at 14:35

user10680652

234

In that folder there are files called: a11.shp,a11.shx,u21.shp,u21.shx

import os 

words = ('a11','u21')

for root, dirs,files in os.walk(r'C:UsersuserDesktopfolder1'):

    for i in files: 

        if i in words:

            print(i)

TypeError: tuple indices must be integers or slices, not str

It requires some kind of an index for the tuple to run for all the items.

I want to use it as 'contains' that word and it doesn't work like contains the in in this case.

How can this be done when giving these exact words in the tuple. Without startswith.

python

edited Nov 22 at 14:44

asked Nov 22 at 14:35

user10680652

234

edited Nov 22 at 14:44

asked Nov 22 at 14:35

user10680652

234

edited Nov 22 at 14:44

asked Nov 22 at 14:35

user10680652

234

asked Nov 22 at 14:35

user10680652

234

asked Nov 22 at 14:35

user10680652

234

do you want to match the exact words, or just partial words? For instance can a11_xxxx.txt matches a11 ?
– Jean-François Fabre
Nov 22 at 14:50

add a comment |

do you want to match the exact words, or just partial words? For instance can a11_xxxx.txt matches a11 ?
– Jean-François Fabre
Nov 22 at 14:50

do you want to match the exact words, or just partial words? For instance can a11_xxxx.txt matches a11 ?
– Jean-François Fabre
Nov 22 at 14:50

add a comment |

2 Answers
2

active

oldest

votes

up vote
-1
down vote

accepted

I think the (working) idea of if words in i: would be if any(w in i for w in words), but that matches substrings (not exact strings) and is not very fast.

You have to do this the other way round. Test file_without_extension in words.

For instance like this:

for i in files: 

    # check if the filename (without extension) is in "words" iterable

    if os.path.splitext(i)[0] in words:

        print(i)

and if you have a lot of "words" make that a set instead of a tuple for faster lookup (words = {'a11','u21'}).

edited Nov 22 at 14:43

answered Nov 22 at 14:40

Jean-François Fabre

99.8k953109

Can you write an example of what you describe?
– user10680652
Nov 22 at 14:42

I did. Just added your code around now
– Jean-François Fabre
Nov 22 at 14:43

1

I'm more inclined to agree with if any(w in i for w in words) since filename in words would require a full match. It may be that OP is only looking for partial match (although the example shows a full one). Also might explain why they shy away from startswith.
– Idlehands
Nov 22 at 14:48

OP requests: "How can this be done when giving these exact words in the tuple. Without startswith" I may have understood it wrong.
– Jean-François Fabre
Nov 22 at 14:49

yes, that's what I thought using any. Nice, but I wrote it wrong. Or this os.path.basename(i).split('.')[0] in words.
– user10680652
Nov 22 at 14:53

|
show 1 more comment

up vote
0
down vote

I find that choosing descriptive names for variables makes the code easier to read. Examine my example below:

import os



words = ('a11','u21')



for root, dirs, files in os.walk(r'C:UsersuserDesktopfolder1'):

    for filename in files:

        for word in words:

            if word in filename:

                print('Match: "', word, '" is in "', file, '"', sep='')

This code produces the result I think you are aiming for.

answered Nov 22 at 14:54

figbeam

2,580137

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
-1
down vote

accepted

I think the (working) idea of if words in i: would be if any(w in i for w in words), but that matches substrings (not exact strings) and is not very fast.

You have to do this the other way round. Test file_without_extension in words.

For instance like this:

for i in files: 

    # check if the filename (without extension) is in "words" iterable

    if os.path.splitext(i)[0] in words:

        print(i)

and if you have a lot of "words" make that a set instead of a tuple for faster lookup (words = {'a11','u21'}).

edited Nov 22 at 14:43

answered Nov 22 at 14:40

Jean-François Fabre

99.8k953109

Can you write an example of what you describe?
– user10680652
Nov 22 at 14:42

I did. Just added your code around now
– Jean-François Fabre
Nov 22 at 14:43

1

I'm more inclined to agree with if any(w in i for w in words) since filename in words would require a full match. It may be that OP is only looking for partial match (although the example shows a full one). Also might explain why they shy away from startswith.
– Idlehands
Nov 22 at 14:48

OP requests: "How can this be done when giving these exact words in the tuple. Without startswith" I may have understood it wrong.
– Jean-François Fabre
Nov 22 at 14:49

yes, that's what I thought using any. Nice, but I wrote it wrong. Or this os.path.basename(i).split('.')[0] in words.
– user10680652
Nov 22 at 14:53

|
show 1 more comment

up vote
-1
down vote

accepted

I think the (working) idea of if words in i: would be if any(w in i for w in words), but that matches substrings (not exact strings) and is not very fast.

You have to do this the other way round. Test file_without_extension in words.

For instance like this:

for i in files: 

    # check if the filename (without extension) is in "words" iterable

    if os.path.splitext(i)[0] in words:

        print(i)

and if you have a lot of "words" make that a set instead of a tuple for faster lookup (words = {'a11','u21'}).

edited Nov 22 at 14:43

answered Nov 22 at 14:40

Jean-François Fabre

99.8k953109

Can you write an example of what you describe?
– user10680652
Nov 22 at 14:42

I did. Just added your code around now
– Jean-François Fabre
Nov 22 at 14:43

1

I'm more inclined to agree with if any(w in i for w in words) since filename in words would require a full match. It may be that OP is only looking for partial match (although the example shows a full one). Also might explain why they shy away from startswith.
– Idlehands
Nov 22 at 14:48

OP requests: "How can this be done when giving these exact words in the tuple. Without startswith" I may have understood it wrong.
– Jean-François Fabre
Nov 22 at 14:49

yes, that's what I thought using any. Nice, but I wrote it wrong. Or this os.path.basename(i).split('.')[0] in words.
– user10680652
Nov 22 at 14:53

|
show 1 more comment

up vote
-1
down vote

accepted

I think the (working) idea of if words in i: would be if any(w in i for w in words), but that matches substrings (not exact strings) and is not very fast.

You have to do this the other way round. Test file_without_extension in words.

For instance like this:

for i in files: 

    # check if the filename (without extension) is in "words" iterable

    if os.path.splitext(i)[0] in words:

        print(i)

and if you have a lot of "words" make that a set instead of a tuple for faster lookup (words = {'a11','u21'}).

edited Nov 22 at 14:43

answered Nov 22 at 14:40

Jean-François Fabre

99.8k953109

I think the (working) idea of if words in i: would be if any(w in i for w in words), but that matches substrings (not exact strings) and is not very fast.

You have to do this the other way round. Test file_without_extension in words.

For instance like this:

for i in files: 

    # check if the filename (without extension) is in "words" iterable

    if os.path.splitext(i)[0] in words:

        print(i)

and if you have a lot of "words" make that a set instead of a tuple for faster lookup (words = {'a11','u21'}).

edited Nov 22 at 14:43

answered Nov 22 at 14:40

Jean-François Fabre

99.8k953109

edited Nov 22 at 14:43

answered Nov 22 at 14:40

Jean-François Fabre

99.8k953109

answered Nov 22 at 14:40

Jean-François Fabre

99.8k953109

answered Nov 22 at 14:40

Jean-François Fabre

99.8k953109

Can you write an example of what you describe?
– user10680652
Nov 22 at 14:42

I did. Just added your code around now
– Jean-François Fabre
Nov 22 at 14:43

1

I'm more inclined to agree with if any(w in i for w in words) since filename in words would require a full match. It may be that OP is only looking for partial match (although the example shows a full one). Also might explain why they shy away from startswith.
– Idlehands
Nov 22 at 14:48

OP requests: "How can this be done when giving these exact words in the tuple. Without startswith" I may have understood it wrong.
– Jean-François Fabre
Nov 22 at 14:49

yes, that's what I thought using any. Nice, but I wrote it wrong. Or this os.path.basename(i).split('.')[0] in words.
– user10680652
Nov 22 at 14:53

|
show 1 more comment

Can you write an example of what you describe?
– user10680652
Nov 22 at 14:42

I did. Just added your code around now
– Jean-François Fabre
Nov 22 at 14:43

1

I'm more inclined to agree with if any(w in i for w in words) since filename in words would require a full match. It may be that OP is only looking for partial match (although the example shows a full one). Also might explain why they shy away from startswith.
– Idlehands
Nov 22 at 14:48

OP requests: "How can this be done when giving these exact words in the tuple. Without startswith" I may have understood it wrong.
– Jean-François Fabre
Nov 22 at 14:49

yes, that's what I thought using any. Nice, but I wrote it wrong. Or this os.path.basename(i).split('.')[0] in words.
– user10680652
Nov 22 at 14:53

Can you write an example of what you describe?
– user10680652
Nov 22 at 14:42

I did. Just added your code around now
– Jean-François Fabre
Nov 22 at 14:43

I'm more inclined to agree with if any(w in i for w in words) since filename in words would require a full match. It may be that OP is only looking for partial match (although the example shows a full one). Also might explain why they shy away from startswith.
– Idlehands
Nov 22 at 14:48

OP requests: "How can this be done when giving these exact words in the tuple. Without startswith" I may have understood it wrong.
– Jean-François Fabre
Nov 22 at 14:49

yes, that's what I thought using any. Nice, but I wrote it wrong. Or this os.path.basename(i).split('.')[0] in words.
– user10680652
Nov 22 at 14:53

|
show 1 more comment

up vote
0
down vote

I find that choosing descriptive names for variables makes the code easier to read. Examine my example below:

import os



words = ('a11','u21')



for root, dirs, files in os.walk(r'C:UsersuserDesktopfolder1'):

    for filename in files:

        for word in words:

            if word in filename:

                print('Match: "', word, '" is in "', file, '"', sep='')

This code produces the result I think you are aiming for.

answered Nov 22 at 14:54

figbeam

2,580137

add a comment |

up vote
0
down vote

I find that choosing descriptive names for variables makes the code easier to read. Examine my example below:

import os



words = ('a11','u21')



for root, dirs, files in os.walk(r'C:UsersuserDesktopfolder1'):

    for filename in files:

        for word in words:

            if word in filename:

                print('Match: "', word, '" is in "', file, '"', sep='')

This code produces the result I think you are aiming for.

answered Nov 22 at 14:54

figbeam

2,580137

add a comment |

up vote
0
down vote

I find that choosing descriptive names for variables makes the code easier to read. Examine my example below:

import os



words = ('a11','u21')



for root, dirs, files in os.walk(r'C:UsersuserDesktopfolder1'):

    for filename in files:

        for word in words:

            if word in filename:

                print('Match: "', word, '" is in "', file, '"', sep='')

This code produces the result I think you are aiming for.

answered Nov 22 at 14:54

figbeam

2,580137

I find that choosing descriptive names for variables makes the code easier to read. Examine my example below:

import os



words = ('a11','u21')



for root, dirs, files in os.walk(r'C:UsersuserDesktopfolder1'):

    for filename in files:

        for word in words:

            if word in filename:

                print('Match: "', word, '" is in "', file, '"', sep='')

This code produces the result I think you are aiming for.

answered Nov 22 at 14:54

figbeam

2,580137

answered Nov 22 at 14:54

figbeam

2,580137

answered Nov 22 at 14:54

figbeam

2,580137

answered Nov 22 at 14:54

figbeam

2,580137

add a comment |

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