One line, probably pretty fast:
num_lines = sum[1 for line in open['myfile.txt']]
answered Jun 19, 2009 at 19:07
5
You can't get any better than that.
After all, any solution will have to read the entire file, figure out how many \n you have, and return that result.
Do you have a better way of doing that without reading the entire file? Not sure... The best solution will always be I/O-bound, best you can do is make sure you don't use unnecessary memory, but it looks like you have that covered.
answered May 10, 2009 at 10:37
Yuval AdamYuval Adam
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I believe that a memory mapped file will be the fastest solution. I tried four functions: the function posted by the OP [opcount]; a simple iteration over the lines in the file [simplecount]; readline with a memory-mapped filed [mmap] [mapcount]; and the buffer read solution offered by Mykola Kharechko [bufcount].
I ran each function five times, and calculated the average run-time for a 1.2 million-line text file.
Windows XP, Python 2.5, 2GB RAM, 2 GHz AMD processor
Here are my results:
mapcount : 0.465599966049
simplecount : 0.756399965286
bufcount : 0.546800041199
opcount : 0.718600034714
Edit: numbers for Python 2.6:
mapcount : 0.471799945831
simplecount : 0.634400033951
bufcount : 0.468800067902
opcount : 0.602999973297
So the buffer read strategy seems to be the fastest for Windows/Python 2.6
Here is the code:
from __future__ import with_statement
import time
import mmap
import random
from collections import defaultdict
def mapcount[filename]:
f = open[filename, "r+"]
buf = mmap.mmap[f.fileno[], 0]
lines = 0
readline = buf.readline
while readline[]:
lines += 1
return lines
def simplecount[filename]:
lines = 0
for line in open[filename]:
lines += 1
return lines
def bufcount[filename]:
f = open[filename]
lines = 0
buf_size = 1024 * 1024
read_f = f.read # loop optimization
buf = read_f[buf_size]
while buf:
lines += buf.count['\n']
buf = read_f[buf_size]
return lines
def opcount[fname]:
with open[fname] as f:
for i, l in enumerate[f]:
pass
return i + 1
counts = defaultdict[list]
for i in range[5]:
for func in [mapcount, simplecount, bufcount, opcount]:
start_time = time.time[]
assert func["big_file.txt"] == 1209138
counts[func].append[time.time[] - start_time]
for key, vals in counts.items[]:
print key.__name__, ":", sum[vals] / float[len[vals]]
answered May 12, 2009 at 2:49
Ryan GinstromRyan Ginstrom
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2
I had to post this on a similar question until my reputation score jumped a bit [thanks to whoever bumped me!].
All of these solutions ignore one way to make this run considerably faster, namely by using the unbuffered [raw] interface, using bytearrays, and doing your own buffering. [This only applies in Python 3. In Python 2, the raw interface may or may not be used by default, but in Python 3, you'll default into Unicode.]
Using a modified version of the timing tool, I believe the following code is faster [and marginally more pythonic] than any of the solutions offered:
def rawcount[filename]:
f = open[filename, 'rb']
lines = 0
buf_size = 1024 * 1024
read_f = f.raw.read
buf = read_f[buf_size]
while buf:
lines += buf.count[b'\n']
buf = read_f[buf_size]
return lines
Using a separate generator function, this runs a smidge faster:
def _make_gen[reader]:
b = reader[1024 * 1024]
while b:
yield b
b = reader[1024*1024]
def rawgencount[filename]:
f = open[filename, 'rb']
f_gen = _make_gen[f.raw.read]
return sum[ buf.count[b'\n'] for buf in f_gen ]
This can be done completely with generators expressions in-line using itertools, but it gets pretty weird looking:
from itertools import [takewhile,repeat]
def rawincount[filename]:
f = open[filename, 'rb']
bufgen = takewhile[lambda x: x, [f.raw.read[1024*1024] for _ in repeat[None]]]
return sum[ buf.count[b'\n'] for buf in bufgen ]
Here are my timings:
function average, s min, s ratio
rawincount 0.0043 0.0041 1.00
rawgencount 0.0044 0.0042 1.01
rawcount 0.0048 0.0045 1.09
bufcount 0.008 0.0068 1.64
wccount 0.01 0.0097 2.35
itercount 0.014 0.014 3.41
opcount 0.02 0.02 4.83
kylecount 0.021 0.021 5.05
simplecount 0.022 0.022 5.25
mapcount 0.037 0.031 7.46
answered Dec 17, 2014 at 4:32
Michael BaconMichael Bacon
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13
You could execute a subprocess and run wc -l filename
import subprocess
def file_len[fname]:
p = subprocess.Popen[['wc', '-l', fname], stdout=subprocess.PIPE,
stderr=subprocess.PIPE]
result, err = p.communicate[]
if p.returncode != 0:
raise IOError[err]
return int[result.strip[].split[][0]]
nosklo
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answered May 10, 2009 at 10:28
Ólafur WaageÓlafur Waage
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7
After a perfplot analysis, one has to recommend the buffered read solution
def buf_count_newlines_gen[fname]:
def _make_gen[reader]:
while True:
b = reader[2 ** 16]
if not b: break
yield b
with open[fname, "rb"] as f:
count = sum[buf.count[b"\n"] for buf in _make_gen[f.raw.read]]
return count
It's fast and memory-efficient. Most other solutions are about 20 times slower.
Code to reproduce the plot:
import mmap
import subprocess
from functools import partial
import perfplot
def setup[n]:
fname = "t.txt"
with open[fname, "w"] as f:
for i in range[n]:
f.write[str[i] + "\n"]
return fname
def for_enumerate[fname]:
i = 0
with open[fname] as f:
for i, _ in enumerate[f]:
pass
return i + 1
def sum1[fname]:
return sum[1 for _ in open[fname]]
def mmap_count[fname]:
with open[fname, "r+"] as f:
buf = mmap.mmap[f.fileno[], 0]
lines = 0
while buf.readline[]:
lines += 1
return lines
def for_open[fname]:
lines = 0
for _ in open[fname]:
lines += 1
return lines
def buf_count_newlines[fname]:
lines = 0
buf_size = 2 ** 16
with open[fname] as f:
buf = f.read[buf_size]
while buf:
lines += buf.count["\n"]
buf = f.read[buf_size]
return lines
def buf_count_newlines_gen[fname]:
def _make_gen[reader]:
b = reader[2 ** 16]
while b:
yield b
b = reader[2 ** 16]
with open[fname, "rb"] as f:
count = sum[buf.count[b"\n"] for buf in _make_gen[f.raw.read]]
return count
def wc_l[fname]:
return int[subprocess.check_output[["wc", "-l", fname]].split[][0]]
def sum_partial[fname]:
with open[fname] as f:
count = sum[x.count["\n"] for x in iter[partial[f.read, 2 ** 16], ""]]
return count
def read_count[fname]:
return open[fname].read[].count["\n"]
b = perfplot.bench[
setup=setup,
kernels=[
for_enumerate,
sum1,
mmap_count,
for_open,
wc_l,
buf_count_newlines,
buf_count_newlines_gen,
sum_partial,
read_count,
],
n_range=[2 ** k for k in range[27]],
xlabel="num lines",
]
b.save["out.png"]
b.show[]
answered Jul 14, 2021 at 22:19
Nico SchlömerNico Schlömer
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1
Here is a python program to use the multiprocessing library to distribute the line counting across machines/cores. My test improves counting a 20million line file from 26 seconds to 7 seconds using an 8 core windows 64 server. Note: not using memory mapping makes things much slower.
import multiprocessing, sys, time, os, mmap
import logging, logging.handlers
def init_logger[pid]:
console_format = 'P{0} %[levelname]s %[message]s'.format[pid]
logger = logging.getLogger[] # New logger at root level
logger.setLevel[ logging.INFO ]
logger.handlers.append[ logging.StreamHandler[] ]
logger.handlers[0].setFormatter[ logging.Formatter[ console_format, '%d/%m/%y %H:%M:%S' ] ]
def getFileLineCount[ queues, pid, processes, file1 ]:
init_logger[pid]
logging.info[ 'start' ]
physical_file = open[file1, "r"]
# mmap.mmap[fileno, length[, tagname[, access[, offset]]]
m1 = mmap.mmap[ physical_file.fileno[], 0, access=mmap.ACCESS_READ ]
#work out file size to divide up line counting
fSize = os.stat[file1].st_size
chunk = [fSize / processes] + 1
lines = 0
#get where I start and stop
_seedStart = chunk * [pid]
_seekEnd = chunk * [pid+1]
seekStart = int[_seedStart]
seekEnd = int[_seekEnd]
if seekEnd < int[_seekEnd + 1]:
seekEnd += 1
if _seedStart < int[seekStart + 1]:
seekStart += 1
if seekEnd > fSize:
seekEnd = fSize
#find where to start
if pid > 0:
m1.seek[ seekStart ]
#read next line
l1 = m1.readline[] # need to use readline with memory mapped files
seekStart = m1.tell[]
#tell previous rank my seek start to make their seek end
if pid > 0:
queues[pid-1].put[ seekStart ]
if pid < processes-1:
seekEnd = queues[pid].get[]
m1.seek[ seekStart ]
l1 = m1.readline[]
while len[l1] > 0:
lines += 1
l1 = m1.readline[]
if m1.tell[] > seekEnd or len[l1] == 0:
break
logging.info[ 'done' ]
# add up the results
if pid == 0:
for p in range[1,processes]:
lines += queues[0].get[]
queues[0].put[lines] # the total lines counted
else:
queues[0].put[lines]
m1.close[]
physical_file.close[]
if __name__ == '__main__':
init_logger[ 'main' ]
if len[sys.argv] > 1:
file_name = sys.argv[1]
else:
logging.fatal[ 'parameters required: file-name [processes]' ]
exit[]
t = time.time[]
processes = multiprocessing.cpu_count[]
if len[sys.argv] > 2:
processes = int[sys.argv[2]]
queues=[] # a queue for each process
for pid in range[processes]:
queues.append[ multiprocessing.Queue[] ]
jobs=[]
prev_pipe = 0
for pid in range[processes]:
p = multiprocessing.Process[ target = getFileLineCount, args=[queues, pid, processes, file_name,] ]
p.start[]
jobs.append[p]
jobs[0].join[] #wait for counting to finish
lines = queues[0].get[]
logging.info[ 'finished {} Lines:{}'.format[ time.time[] - t, lines ] ]
namit
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answered Jul 26, 2011 at 6:51
MartlarkMartlark
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7
A one-line bash solution similar to this answer, using the modern subprocess.check_output function:
def line_count[filename]:
return int[subprocess.check_output[['wc', '-l', filename]].split[][0]]
answered Apr 3, 2017 at 7:48
1''1''
25.7k32 gold badges136 silver badges197 bronze badges
2
I would use Python's file object method readlines, as follows:
with open[input_file] as foo:
lines = len[foo.readlines[]]
This opens the file, creates a list of lines in the file, counts the length of the list, saves that to a variable and closes the file again.
answered Oct 8, 2013 at 12:46
Daniel LeeDaniel Lee
1,9601 gold badge21 silver badges29 bronze badges
4
This is the fastest thing I have found using pure python. You can use whatever amount of memory you want by setting buffer, though 2**16 appears to be a sweet spot on my computer.
from functools import partial
buffer=2**16
with open[myfile] as f:
print sum[x.count['\n'] for x in iter[partial[f.read,buffer], '']]
I found the answer here Why is reading lines from stdin much slower in C++ than Python? and tweaked it just a tiny bit. Its a very good read to understand how to count lines quickly, though wc -l is still about 75% faster than anything else.
answered Dec 9, 2016 at 20:34
jeffpkampjeffpkamp
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def file_len[full_path]:
""" Count number of lines in a file."""
f = open[full_path]
nr_of_lines = sum[1 for line in f]
f.close[]
return nr_of_lines
answered May 10, 2009 at 10:33
pkitpkit
7,6936 gold badges34 silver badges36 bronze badges
2
Here is what I use, seems pretty clean:
import subprocess
def count_file_lines[file_path]:
"""
Counts the number of lines in a file using wc utility.
:param file_path: path to file
:return: int, no of lines
"""
num = subprocess.check_output[['wc', '-l', file_path]]
num = num.split[' ']
return int[num[0]]
UPDATE: This is marginally faster than using pure python but at the cost of memory usage. Subprocess will fork a new process with the same memory footprint as the parent process while it executes your command.
answered Jul 26, 2017 at 18:13
radtekradtek
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5
One line solution:
import os
os.system["wc -l filename"]
My snippet:
>>> os.system['wc -l *.txt']
0 bar.txt
1000 command.txt
3 test_file.txt
1003 total
kalehmann
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answered Jan 6, 2017 at 8:29
TheExorcistTheExorcist
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8
Kyle's answer
num_lines = sum[1 for line in open['my_file.txt']]
is probably best, an alternative for this is
num_lines = len[open['my_file.txt'].read[].splitlines[]]
Here is the comparision of performance of both
In [20]: timeit sum[1 for line in open['Charts.ipynb']]
100000 loops, best of 3: 9.79 µs per loop
In [21]: timeit len[open['Charts.ipynb'].read[].splitlines[]]
100000 loops, best of 3: 12 µs per loop
answered Oct 15, 2014 at 5:22
Chillar AnandChillar Anand
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I got a small [4-8%] improvement with this version which re-uses a constant buffer so it should avoid any memory or GC overhead:
lines = 0
buffer = bytearray[2048]
with open[filename] as f:
while f.readinto[buffer] > 0:
lines += buffer.count['\n']
You can play around with the buffer size and maybe see a little improvement.
answered Feb 25, 2013 at 19:31
Scott PersingerScott Persinger
3,5362 gold badges19 silver badges12 bronze badges
3
Just to complete the above methods I tried a variant with the fileinput module:
import fileinput as fi
def filecount[fname]:
for line in fi.input[fname]:
pass
return fi.lineno[]
And passed a 60mil lines file to all the above stated methods:
mapcount : 6.1331050396
simplecount : 4.588793993
opcount : 4.42918205261
filecount : 43.2780818939
bufcount : 0.170812129974
It's a little surprise to me that fileinput is that bad and scales far worse than all the other methods...
answered May 5, 2010 at 11:48
BandGapBandGap
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As for me this variant will be the fastest:
#!/usr/bin/env python
def main[]:
f = open['filename']
lines = 0
buf_size = 1024 * 1024
read_f = f.read # loop optimization
buf = read_f[buf_size]
while buf:
lines += buf.count['\n']
buf = read_f[buf_size]
print lines
if __name__ == '__main__':
main[]
reasons: buffering faster than reading line by line and string.count is also very fast
SilentGhost
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answered May 10, 2009 at 11:29
Mykola KharechkoMykola Kharechko
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6
This code is shorter and clearer. It's probably the best way:
num_lines = open['yourfile.ext'].read[].count['\n']
answered Feb 23, 2015 at 18:38
Texom512Texom512
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2
I have modified the buffer case like this:
def CountLines[filename]:
f = open[filename]
try:
lines = 1
buf_size = 1024 * 1024
read_f = f.read # loop optimization
buf = read_f[buf_size]
# Empty file
if not buf:
return 0
while buf:
lines += buf.count['\n']
buf = read_f[buf_size]
return lines
finally:
f.close[]
Now also empty files and the last line [without \n] are counted.
answered Nov 25, 2011 at 14:55
DummyDummy
411 bronze badge
2
A lot of answers already, but unfortunately most of them are just tiny economies on a barely optimizable problem...
I worked on several projects where line count was the core function of the software, and working as fast as possible with a huge number of files was of paramount importance.
The main bottleneck with line count is I/O access, as you need to read each line in order to detect the line return character, there is simply no way around. The second potential bottleneck is memory management: the more you load at once, the faster you can process, but this bottleneck is negligible compared to the first.
Hence, there are 3 major ways to reduce the processing time of a line count function, apart from tiny optimizations such as disabling gc collection and other micro-managing tricks:
Hardware solution: the major and most obvious way is non-programmatic: buy a very fast SSD/flash hard drive. By far, this is how you can get the biggest speed boosts.
Data preparation solution: if you generate or can modify how the files you process are generated, or if it's acceptable that you can pre-process them, first convert the line return to unix style [
\n] as this will save 1 character compared to Windows or MacOS styles [not a big save but it's an easy gain], and secondly and most importantly, you can potentially write lines of fixed length. If you need variable length, you can always pad smaller lines. This way, you can calculate instantly the number of lines from the total filesize, which is much faster to access. Often, the best solution to a problem is to pre-process it so that it better fits your end purpose.Parallelization + hardware solution: if you can buy multiple hard disks [and if possible SSD flash disks], then you can even go beyond the speed of one disk by leveraging parallelization, by storing your files in a balanced way [easiest is to balance by total size] among disks, and then read in parallel from all those disks. Then, you can expect to get a multiplier boost in proportion with the number of disks you have. If buying multiple disks is not an option for you, then parallelization likely won't help [except if your disk has multiple reading headers like some professional-grade disks, but even then the disk's internal cache memory and PCB circuitry will likely be a bottleneck and prevent you from fully using all heads in parallel, plus you have to devise a specific code for this hard drive you'll use because you need to know the exact cluster mapping so that you store your files on clusters under different heads, and so that you can read them with different heads after]. Indeed, it's commonly known that sequential reading is almost always faster than random reading, and parallelization on a single disk will have a performance more similar to random reading than sequential reading [you can test your hard drive speed in both aspects using CrystalDiskMark for example].
If none of those are an option, then you can only rely on micro-managing tricks to improve by a few percents the speed of your line counting function, but don't expect anything really significant. Rather, you can expect the time you'll spend tweaking will be disproportionated compared to the returns in speed improvement you'll see.
answered Nov 9, 2020 at 19:12
gaborousgaborous
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the result of opening a file is an iterator, which can be converted to a sequence, which has a length:
with open[filename] as f:
return len[list[f]]
this is more concise than your explicit loop, and avoids the enumerate.
answered May 10, 2009 at 11:35
Andrew JaffeAndrew Jaffe
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3
print open['file.txt', 'r'].read[].count["\n"] + 1
answered Mar 21, 2014 at 6:10
If one wants to get the line count cheaply in Python in Linux, I recommend this method:
import os
print os.popen["wc -l file_path"].readline[].split[][0]
file_path can be both abstract file path or relative path. Hope this may help.
answered Aug 28, 2014 at 9:09
Lerner ZhangLerner Zhang
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def count_text_file_lines[path]:
with open[path, 'rt'] as file:
line_count = sum[1 for _line in file]
return line_count
answered Dec 17, 2017 at 14:50
jciloajciloa
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2
Simple method:
1]
>>> f = len[open["myfile.txt"].readlines[]]
>>> f
430
>>> f = open["myfile.txt"].read[].count['\n']
>>> f
430
>>>
num_lines = len[list[open['myfile.txt']]]
M.Innat
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answered Sep 17, 2018 at 10:27
2
What about this
def file_len[fname]:
counts = itertools.count[]
with open[fname] as f:
for _ in f: counts.next[]
return counts.next[]
answered May 10, 2009 at 18:20
odwlodwl
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count = max[enumerate[open[filename]]][0]
answered Mar 11, 2011 at 21:09
pyanonpyanon
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2
How about this?
import fileinput
import sys
counter=0
for line in fileinput.input[[sys.argv[1]]]:
counter+=1
fileinput.close[]
print counter
answered Jul 19, 2011 at 15:55
leba-levleba-lev
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How about this one-liner:
file_length = len[open['myfile.txt','r'].read[].split['\n']]
Takes 0.003 sec using this method to time it on a 3900 line file
def c[]:
import time
s = time.time[]
file_length = len[open['myfile.txt','r'].read[].split['\n']]
print time.time[] - s
answered Oct 3, 2013 at 0:55
onetwopunchonetwopunch
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def line_count[path]:
count = 0
with open[path] as lines:
for count, l in enumerate[lines, start=1]:
pass
return count
answered Jun 2, 2014 at 21:45
mdwhatcottmdwhatcott
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