Assigning value in a while comparison expression

Question

Hi I was wondering if it was possible to assign a value in the while comparison section of the code.

Here is an example of the code currently

startIndex = find(target, key, startIndex)
while( startIndex != -1):
    matchesFound += 1
    startIndex = find(target, key, startIndex + 1)
return matchesFound

What I want to do is move the startIndex = find(target, key, startIndex) into the while comparison expresion so it would look something like this

while( (startIndex = find(target, key, startIndex)) != -1):
    matchesFound += 1
    startIndex + 1
return matchesFound

if not, what would a better refactor be?

Thanks

edit I'm working through the MIT Open courseware 6.00 before I try out the famous 6.001 module

Answer 1

No, you can't do that in Python. I think the main reason Python disallows this, is to avoid the frequent errors resulting from confusing assignment and equality checking.

Python claims to have readable code as a main guideline, so I think your original code is fine. No need to refactor...

Answer 2

You're writing C in Python.

Try:

startIndex = -1
while True:
    startIndex = find(target, key, startIndex + 1)
    if startIndex < 0:
        break
    matchesFound += 1
return matchesFound

Or perhaps even:

return target.count(key)

Answer 3

i would do it like this

startIndex=0
while 1:
    startIndex = find(target, key, startIndex+1)
    if startIndex == -1: break
    matchesFound += 1

you can put in more conditions inside the while loop like that.

Edit: @OP, in future, to count match of string, just use count

>>> mystring = "abc defabc fgh ijkabc blah"
>>> mystring.count("abc")
3
>>>

Answer 4

Edit.

We refactor it like this.

matches = [ k for k in range(len(target)-len(key)) if target[k:].startswith(key) ]
matchesFound = len(matches)

We don't need C-style condition and assignment conflation.

Rarely do you simply want the count; the actual locations are available for free.

Answer 5

If for some reason you can't use a more appropriate method from whatever you're searching, this is equivalent to your working code from the question:

start = 0
count = 0
for match in iter(lambda: find(target, key, start), -1):
  count += 1
  start = match + 1
return count

However, you'll get the most mileage out of writing an iterator for finding these matches, "transforming" your current find function:

def findall(target, key, start=0):
  for match in iter(lambda: find(target, key, start), -1):
    yield match
    start = match + 1

Then count from that:

count = sum(1 for m in findall(target, key))