Count multiple files in a directory with the same name - python-3.x

Question

I'm relatively new to Python and was working on a project where the user can navigate to a folder, after which the program does a count of all the files in that folder with a specific name.
The problem is that I have a folder with over 5000 files many of them sharing the same name but different extensions. I wrote code that somewhat does what I want the final version to do but its VERY redundant and I can't see myself doing this for over 600 file names.
Wanted to ask if it is possible to make this program "automated" or less redundant where I don't have to manually type out the names of 600 files to return data for.
Sample code I currently have:
import os, sys
print(sys.version)
file_counting1 = 0
file_counting2 = 0
filepath = input("Enter file path here: ")
if os.path.exists(filepath):
for file in os.listdir(filepath):
if file.startswith('expressmail'):
file_counting1 += 1
print('expressmail')
print('Total files found:', file_counting1)
for file in os.listdir(filepath):
if file.startswith('prioritymail'):
file_counting2 += 1
print('prioritymail')
print('Total files found:', file_counting2)
Sample Output:
expressmail
Total files found: 3
prioritymail
Total files found: 1

Answer 1

The following script will count occurrences of files with the same name. If the file does not have an extension, the whole filename is treated as the name. It also does not traverse subdirectories, since the original question just asks about files in the given folder.
import os
dir_name = "."
files = next(os.walk(dir_name))[2] # get all the files directly in the directory
names = [f[:f.rindex(".")] for f in files if "." in f] # drop the extensions
names += [f for f in files if "." not in f] # add those without extensions
for name in set(names): # for each unique name-
print("{}\nTotal files found: {}".format(name, names.count(name)))
If you want to support files in subdirectories, you could use something like
files = [os.path.join(r,file) for r,d,f in os.walk(dir_name) for file in f]
If you don't want to consider files without extensions, just remove the line:
names += [f for f in files if "." not in f]

Answer 2

There are a number of ways you can do what you're trying to do. Partly it depends on whether or not you need to recover the list of extension for a given duplicated file.
Counter, from the collections module - use this for a simple count of file. Ignore the extensions when building the count.
Use the filename without extension as a dictionary key, add a list of items as the key-value, where the list of items is each occurrence of the file.
Here's an example using the Counter class:
import os, sys, collections
c = collections.Counter()
for root, dirs,files in os.walk('/home/myname/hg/2018/'):
# discard any path data and just use filename
for names in files:
name, ext = os.path.splitext(names)
# discard any extension
c[name] += 1
# Counter.most_common() gives the values in the form of (entry, count)
# Counter.most_common(x) - pass a value to display only the top x counts
# e.g. Counter.most_common(2) = top 2
for x in c.most_common():
print(x[0] + ': ' + str(x[1]))

Answer 3

you can use regular expressions:
import os, sys, re
print(sys.version)
filepath = input("Enter file path here: ")
if os.path.exists(filepath):
allfiles = "\n".join(os.listdir(filepath))
file_counting1 = len(re.findall("^expressmail",allfiles,re.M))
print('expressmail')
print('Total files found:', file_counting1)
file_counting2 = len(re.findall("^prioritymail",allfiles,re.M))
print('prioritymail')
print('Total files found:', file_counting2)