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Iterate over the lines of a string

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How do you iterate through text lines in Python?

To iterate through lines in a file using Python, you can loop over each line in a file with a simple for loop. When reading files, the ability to read files sequentially line by line can be very useful. Reading text from a file is easy with the Python open() function.

Can you iterate over a string Python?

You can traverse a string as a substring by using the Python slice operator ([]). It cuts off a substring from the original string and thus allows to iterate over it partially. To use this method, provide the starting and ending indices along with a step value and then traverse the string.

Can we iterate a string?

Given string str of length N, the task is to traverse the string and print all the characters of the given string. Methods: Using Naive Approach.


Here are three possibilities:

foo = """
this is 
a multi-line string.
"""

def f1(foo=foo): return iter(foo.splitlines())

def f2(foo=foo):
    retval = ''
    for char in foo:
        retval += char if not char == '\n' else ''
        if char == '\n':
            yield retval
            retval = ''
    if retval:
        yield retval

def f3(foo=foo):
    prevnl = -1
    while True:
      nextnl = foo.find('\n', prevnl + 1)
      if nextnl < 0: break
      yield foo[prevnl + 1:nextnl]
      prevnl = nextnl

if __name__ == '__main__':
  for f in f1, f2, f3:
    print list(f())

Running this as the main script confirms the three functions are equivalent. With timeit (and a * 100 for foo to get substantial strings for more precise measurement):

$ python -mtimeit -s'import asp' 'list(asp.f3())'
1000 loops, best of 3: 370 usec per loop
$ python -mtimeit -s'import asp' 'list(asp.f2())'
1000 loops, best of 3: 1.36 msec per loop
$ python -mtimeit -s'import asp' 'list(asp.f1())'
10000 loops, best of 3: 61.5 usec per loop

Note we need the list() call to ensure the iterators are traversed, not just built.

IOW, the naive implementation is so much faster it isn't even funny: 6 times faster than my attempt with find calls, which in turn is 4 times faster than a lower-level approach.

Lessons to retain: measurement is always a good thing (but must be accurate); string methods like splitlines are implemented in very fast ways; putting strings together by programming at a very low level (esp. by loops of += of very small pieces) can be quite slow.

Edit: added @Jacob's proposal, slightly modified to give the same results as the others (trailing blanks on a line are kept), i.e.:

from cStringIO import StringIO

def f4(foo=foo):
    stri = StringIO(foo)
    while True:
        nl = stri.readline()
        if nl != '':
            yield nl.strip('\n')
        else:
            raise StopIteration

Measuring gives:

$ python -mtimeit -s'import asp' 'list(asp.f4())'
1000 loops, best of 3: 406 usec per loop

not quite as good as the .find based approach -- still, worth keeping in mind because it might be less prone to small off-by-one bugs (any loop where you see occurrences of +1 and -1, like my f3 above, should automatically trigger off-by-one suspicions -- and so should many loops which lack such tweaks and should have them -- though I believe my code is also right since I was able to check its output with other functions').

But the split-based approach still rules.

An aside: possibly better style for f4 would be:

from cStringIO import StringIO

def f4(foo=foo):
    stri = StringIO(foo)
    while True:
        nl = stri.readline()
        if nl == '': break
        yield nl.strip('\n')

at least, it's a bit less verbose. The need to strip trailing \ns unfortunately prohibits the clearer and faster replacement of the while loop with return iter(stri) (the iter part whereof is redundant in modern versions of Python, I believe since 2.3 or 2.4, but it's also innocuous). Maybe worth trying, also:

    return itertools.imap(lambda s: s.strip('\n'), stri)

or variations thereof -- but I'm stopping here since it's pretty much a theoretical exercise wrt the strip based, simplest and fastest, one.


I'm not sure what you mean by "then again by the parser". After the splitting has been done, there's no further traversal of the string, only a traversal of the list of split strings. This will probably actually be the fastest way to accomplish this, so long as the size of your string isn't absolutely huge. The fact that python uses immutable strings means that you must always create a new string, so this has to be done at some point anyway.

If your string is very large, the disadvantage is in memory usage: you'll have the original string and a list of split strings in memory at the same time, doubling the memory required. An iterator approach can save you this, building a string as needed, though it still pays the "splitting" penalty. However, if your string is that large, you generally want to avoid even the unsplit string being in memory. It would be better just to read the string from a file, which already allows you to iterate through it as lines.

However if you do have a huge string in memory already, one approach would be to use StringIO, which presents a file-like interface to a string, including allowing iterating by line (internally using .find to find the next newline). You then get:

import StringIO
s = StringIO.StringIO(myString)
for line in s:
    do_something_with(line)

You can iterate over "a file", which produces lines, including the trailing newline character. To make a "virtual file" out of a string, you can use StringIO:

import io  # for Py2.7 that would be import cStringIO as io

for line in io.StringIO(foo):
    print(repr(line))

If I read Modules/cStringIO.c correctly, this should be quite efficient (although somewhat verbose):

from cStringIO import StringIO

def iterbuf(buf):
    stri = StringIO(buf)
    while True:
        nl = stri.readline()
        if nl != '':
            yield nl.strip()
        else:
            raise StopIteration

Regex-based searching is sometimes faster than generator approach:

RRR = re.compile(r'(.*)\n')
def f4(arg):
    return (i.group(1) for i in RRR.finditer(arg))