I'm trying to enumerate the time the program finds a specific string in a longer text but I'm stuck, the program shows the length of the response string instead of the times ultrias is present in response. Here's the code:
ultrias = "am 17"
response = "Hi i am 17, did you know I am 17"
num = 0
for ultrias in response:
num += 1
print (num)
Others have answered with a few different ways to achieve your requirement, however, none explained why the output of your code is the length of the input string.
for loops have a loop variable that is assigned values from the object that is being iterated over. In the case of the object being a string, the for loop will iterate in order over each character in the string, assigning the character to the loop variable e.g.
>>> for i in "Hi there":
... print i
...
H
i
t
h
e
r
e
So you can see that i is assigned in turn a character from the string "Hi there". The same thing is happening in your code: ultrias is the loop variable and is being assigned consecutive characters from the string "Hi i am 17, did you know I am 17".
ultrias = "am 17"
response = "Hi i am 17, did you know I am 17"
num = 0
for ultrias in response:
print(ultrias)
num += 1
print (num)
The output will be:
H
i
i
a
m
1
7
.
.
.
3
Also note that the value of ultrias after the loop has terminated is the last value assigned it by the loop - not "am 17" but "7".
Python strings have a built-in str.count() method that can do that:
ultrias = "am 17"
response = "Hi i am 17, did you know I am 17"
print(response.count(ultrias))
output
2
If you aren't permitted to use the str.count() method, you could do something like this:
ultrias = "am 17"
response = "Hi i am 17, did you know I am 17"
count = 0
data = response[:]
while True:
start = data.find(ultrias)
if start < 0:
break
data = data[start + len(ultrias):]
count += 1
print(count)
But I wouldn't recommend it.
Neither of these algorithms handle overlapping matches, but the second example can be easily modified to do that:
count = 0
data = response[:]
while True:
start = data.find(ultrias)
if start < 0:
break
data = data[start + 1:]
count += 1
print(count)
Use string.count(string) to get what you need.
>>> find = "am 17"
>>> string = "Hi I am 17, did you know I am 17"
>>> string.count(find)
2
Through re module.
>>> ultrias = "am 17"
>>> response = "Hi i am 17, did you know I am 17"
>>> print(len(re.findall(ultrias, response)))
2
171in17171? One or two ?