Trying to count the number of times a letter appears using only a while loop

Question

I am trying to count how many o's there are in the raw input 'phrase' and it counts the first o and says it's done however there are many o's after that. How do I get it to count all of the o's with no for loop.

while loop_count<len(phrase):
    if "o" in phrase:
        loop_count += 1
        count2 += 1
      if loop_count>len(phrase):
          print loop_count
          break
      else:
          continue
  else:
    print loop_count
    continue

Answer 1

You can use sum with an iterator (in this case a generator expression ):

>>> sum(c=='o' for c in 'oompa loompa')
4

You can use a regex with len:

>>> re.findall('o', 'oompa loompa')
['o', 'o', 'o', 'o']
>>> len(re.findall('o', 'oompa loompa'))
4

You could use a Counter:

>>> from collections import Counter
>>> Counter('oompa loompa')['o']
4

Or just use the 'count' method of the string:

>>> 'oompa loompa'.count('o')
4

If you really want to use a while loop, use a list as a stack with the pop method:

s='oompa loompa'
tgt=list(s)
count=0
while tgt:
    if tgt.pop()=='o':
        count+=1

Or a 'for' loop -- more Pythonic:

count=0        
for c in s:
    if c=='o':
        count+=1 

Answer 2

you can use count function :

phrase.count('o')

but if you want to send a message after one match of 'o' for skip loop over the all of string just use 'in' look like this:

if 'o' in phrase :
 # show your message ex: print('this is false')

Answer 3

Lets try to dissect your code you understand what is going on. See my comments

while loop_count<len(phrase):
    if "o" in phrase: # See comment 1
        loop_count += 1 # Why are you incrementing both of these?
        count2 += 1
      if loop_count>len(phrase): # What is this statement for? It should never happen.
                ## A loop always exits once the condition at the while line is false
                ## It seems like you are manually trying to control your loop
                ##  Which you dont need to do
          print loop_count
          break # What does this accomplish?
      else:
          continue
  else:
    print loop_count
    continue # Pointless as we are already at end of loop. Consider removing

Comment 1: You are asking if an 'o' is anywhere in the phrase. You instead want to ask if the current letter is an o. Perhaps you ment to access a LETTER of the phrase with the index, such as if 'o' == phrase[loop_count] . If you do this you want to increment loop_count every time, but the o count only when the letter is an o.

You could re-write it like this:

loop_count, o_count = 0, 0
while loop_count<len(phrase):
    # loop_count represents how many times we have looped so far
    if "o" == phrase[loop_count].lower(): # We ask if the current letter is 'o'
        o_count += 1 # If it is an 'o', increment count
    loop_count += 1 # Increment this every time, it tracks our loop count

print o_count

Answer 4

使用理解

len([i for i in phrase if i=="o"])