get nth line of string in python

Question

How can you get the nth line of a string in Python 3? For example

getline(\"line1\\nline2\\nline3\",3)

Is there any way to do this

Answer 1

Since you brought up the point of memory efficiency, is this any better:

s = "line1\nline2\nline3"

# number of the line you want
line_number = 2

i = 0
line = ''
for c in s:
   if i > line_number:
     break
   else:
     if i == line_number-1 and c != '\n':
       line += c
     elif c == '\n':
       i += 1

Answer 2

`my_string.strip().split("\n")[-1]`

Answer 3

My solution (effecient and compact):

def getLine(data, line_no):
    index = -1
    for _ in range(line_no):index = data.index('\n',index+1)
    return data[index+1:data.index('\n',index+1)]

Answer 4

Wrote into two functions for readability

    string = "foo\nbar\nbaz\nfubar\nsnafu\n"

    def iterlines(string):
      word = ""
      for letter in string:
        if letter == '\n':
          yield word
          word = ""
          continue
        word += letter

    def getline(string, line_number):
      for index, word in enumerate(iterlines(string),1):
        if index == line_number:
          #print(word)
          return word

    print(getline(string, 4))

Answer 5

Use a string buffer:

import io    
def getLine(data, line_no):
    buffer = io.StringIO(data)
    for i in range(line_no - 1):
        try:
            next(buffer)
        except StopIteration:
            return '' #Reached EOF

    try:
        return next(buffer)
    except StopIteration:
        return '' #Reached EOF

Answer 6

Try the following:

s = "line1\nline2\nline3"
print s.splitlines()[2]