如何计算字符串中的字符？ （蟒蛇）

Question

# -*- coding:UTF-8 -*-

str= "Green tree"
scr= "e"

cstr= len(str)
n=0
a=0

while n < cstr:
    if str[n] == scr:
        print(len(scr))
    n=n+1

I have to count "e" in -str- string, but when I run this script I get

1
1
1
1

instead of 4.

What's the problem?

Answer 1

Use the count method :

>>> st="Green tree"
>>> st.count('e')
4

If the count method is broken on your Python ;-), you can use a for loop:

st="Green tree"
tgt='e'
i=0
for c in st:
    if c==tgt: i+=1

print i 
# 4 

If you really want a while loop:

idx=0
i=0
while idx<len(st):
    if st[idx]==tgt: i+=1
    idx+=1

print i    

But, this being Python, a more 'Pythonic' approach if your count method broken is to use sum on a generator expression:

>>> sum(1 for c in st if c=='e')
4

Answer 2

First of all, don't use str as a variable name, it will mask the built-in name.

As for counting characters in a string, just use the str.count() method:

>>> s = "Green tree"
>>> s.count("e")
4

If you are just interested in understanding why your current code doesn't work, you are printing 1 four times because you will find four occurrences of 'e', and when an occurrence is found you are printing len(scr) which is always 1 .

Instead of printing len(scr) in your if block, you should be incrementing a counter that keeps track of the total number of occurrences found, it looks like you set up a variable a that you aren't using, so the smallest change to your code to get it to work would be the following (however as noted above, str.count() is a better approach):

str= "Green tree"
scr= "e"

cstr= len(str)
n=0
a=0

while n < cstr:
    if str[n] == scr:
        a+=1
    n=n+1
print(a)

Answer 3

scr= "e"
##
print(len(scr))

For why it's doing this, it's doing what you asked, and printing the length of the variable scr , which is always one.

You're best to use the str.count() method as others mentioned, or increment a counter yourself manually.