Clash Royale CLAN TAG#URR8PPP

How to get file count as zero with recursive directory search?

I am trying to get a file count for a particular filename match and write to a dictionary. The below code works fine if all directories contain some files (match or no match). However, if there is an empty directory, it is not shown in the dictionary. Folder2 is empty and is not shown in the result.
I would also like to know if there is a way to print the result with one forward slash separator instead of combination of double back and forward slashes?
My code:

import os
import re
def file_count_search(root_dir,keyword):
dict=
for dirpath,dirnames,filenames in os.walk(root_dir,topdown=True):
matches = re.findall(keyword, str(filenames))
if keyword in matches:
dict[os.path.join(root_dir,dirpath)] = len(matches)
print dict
file_count_search("c://test","file")


My Result:


'c://test\folder3\subdir_folder3': 1,
'c://test': 1, 'c://test\folder1': 3,
'c://test\folder3': 1



Desired Result:


'c:/test/folder3/subdir_folder3': 1,
'c:/test': 1, 'c:/test/folder1': 3,
'c:/test/folder2': 0,
'c:/test/folder3': 1



dict

1 Answer
1

If there are no matching files in a given directory, matches will be an empty list, so keyword in matches will evaluate to False, and nothing will be added to dict.

matches

keyword in matches

False

dict

Try replacing this line:

if keyword in matches:
dict[os.path.join(root_dir,dirpath)] = len(matches)


with just this (also replacing dict with a non-reserved variable name per Dan Farrell's note):

dict

path_to_match_count[os.path.join(root_dir,dirpath)] = len(matches)


... for an update script like this:

import os
import re
def file_count_search(root_dir,keyword):
path_to_match_count=
for dirpath,dirnames,filenames in os.walk(root_dir,topdown=True):
matches = re.findall(keyword, str(filenames))
path_to_match_count[os.path.join(root_dir,dirpath)] = len(matches)
print path_to_match_count
file_count_search("c://test","file")


/

split

join

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