How to count word “test” in file on Python?

Question

I have a file consists of many strings. Looks like

sdfsdf sdfsdfsdf sdfsdfsdf test gggg uff test test fffffffff sdgsdgsdgsdg sdgsdgsdgsdg uuuttt 5

Answer 1

If you want to find all matches:

with open("file") as f:
    numtest = f.read().count("test")

If you want to find only word matches:

with open("file") as f:
    numtest = f.read().split().count("test")

Answer 2

This should work.

   from collections import Counter
   with open('myfile.txt', 'r') as f:
       words = f.read().split()
       counts = Counter(words)

   print counts["test"] #counts just of exact string "test"
   #find all strings containing test (e.g 'atest', 'mytest')
   print sum([val for key,val in counts.iteritems() if "test" in key])

Answer 3

You can use regular expressions:

import re

with open('myfile.txt', 'r') as f:
    txt = f.read()

cnt = len(re.findall(r'\btest\b', txt))

If you don't care about case sensitivity (also match Test or TEST)

cnt = len(re.findall(r'\btest\b', txt, flags=re.I))

Answer 4

One-liner:

s.split().count('test')

Answer 5

It will count number of tests in the whole file:

f = open('my_file.txt', 'r')
num_tests = len([word for word in f.read().split() if word == 'test'])
f.close()

Note that it will NOT match words like tester, tested, testing, etc.... If you want to also match them, use instead:

f = open('my_file.txt', 'r')
num_tests = len([word for word in f.read().split() if 'test' in word])
f.close()