如何获得字符串中重复出现的字符的位置？

Question

I have this string:

|C100|0|1|F120570|55|00|32|754929|35150456228356008116550320007549291371271334|17042015|20042015|2077,14|1|2062,48|0,00|3729,51|9|0,00|0,00|3,51|0,00|0,00|0,00|0,00|0,00|0,00|0,00|0,00|0,00|

And I need to find the index / position of the 23rd "|" using Python.

Since I have more than one "|", I don't know how to do that..

I've found this code to find the second occurrence

def get_second_index(input_string, sub_string):
    return input_string.index(sub_string, input_string.index(sub_string) + 1)

but this one isn't helping me with my problem.

I thought that maybe I can get that with a loop like while x < 23: but I can't figure it out exactly how...

Answer 1

The easiest I can think of is to loop through each character and keep the count of how many '|' you have found.

count = 0
for index, c in enumerate(string):
   if c == '|':
      count = count + 1
   if count == 23:
      print index
      break

Answer 2

You could use the following regular expression to match exactly 23 '|' characters:

text = "|C100|0|1|F120570|55|00|32|754929|35150456228356008116550320007549291371271334|17042015|20042015|2077,14|1|2062,48|0,00|3729,51|9|0,00|0,00|3,51|0,00|0,00|0,00|0,00|0,00|0,00|0,00|0,00|0,00|"

print re.match("(\|.*?){23}", text).end() - 1

This gives 154 for your example text.

Answer 3

x = 0
for index, char in enumerate(string):
    if char == "|" and x == 23:
        output_index = index
    elif char == "|":
        x += 1
print output_index

enumerate() returns the index and value of each element in a string, list, dict, etc.

Answer 4

You can use re.finditer to get a generator contain all groups contain | the use re.matchObject.start() method which return the start position of your match and get the 23th element with enumerate and a generator expression in next function :

>>> all_case=re.finditer(r'\|',s)
>>> next(j.start() for i,j in enumerate(all_case,1) if i==23)
154

Answer 5

You can use string.find() within a loop and adjust the start index as you go. Something like the following:

start = -1
end = len(yourstring)
x = 0
while x < 23:
  start = string.find(yourstring, "|", start + 1, end)\
  x +=1

Answer 6

>>> [i for i, j in enumerate(input_string) if j == '|'][22]
>>> 154

Generates a list containing the index of each occurrence of | and then returns the 23rd element (=22 because lists are zero-indexed)

Answer 7

Assuming s is input string and you need to find position of nth |:

def find_position(s, n, c):
  return len("".join(s.split(c, maxsplit = n)[:n])) + n - 1