Count the number of files in a folder that have certain strings - python-2.7

Question

I have a folder with 200 files. Each file has data like
VISITERM_90 VISITERM_0 VISITERM_34 ..... etc.
Each file does not have the same elements. So, I would like to count the number of files which have the elements from VISITERM_0 to VISITERM_99. That is I should get my output as:
VISITERM_0 200
VISTERM_1 140
VISITERM_2 150
and so on depending upon the numbers of files that has the specified elements. I want to run it in a loop from VISITERM_0 till VISITERM_99 and for each element I need to find the number of files.
My code is:
import os
vt = 'VISITERM_'
no = 0
while no < 10:
for doc in os.listdir('/home/krupa/Krupa/Mirellas_Image_Annotation_Data/Test/sample_codes/Files'):
doc2 = '/home/krupa/Krupa/Mirellas_Image_Annotation_Data/Test/sample_codes/Files/' + doc
c = vt + (repr(no))
with open (doc2, 'r') as inF:
for line in inF:
if c in line:
print c, doc2
else:
print "DOES NOT EXIST" , c, doc2
no = no + 1
This code is printing me each visiterm and each of the file that has it. I just want the VISITERMS_* and their corresponding number of files. Please help!

Answer 1

My python skills are a bit rusty, so bear with me. I think that you need a way to store the values while looping, I'll use a dictionary. This is not the complete solution, but it can help you figure out what you need to do:
dict={}
for doc in os.listdir('..'):
doc2 = '..'
with open (doc2, 'r') as inF:
for line in inF:
while no < 10:
c = vt + (repr(no))
if c in line:
numberOfElements = 0
if dict.has_key(c):
numberOfElements = dict[c]
numberOfElements += 1
else:
numberOfElements = 1
dict[c] = numberOfElements
no += 1
for key in dict.keys():
print key, dict[key]