How to read first N lines of a file?

N = 10 with open("file.txt", "r") as file: for i in range(N): line = next(file).strip() print(line) 

If you want to read the first lines quickly and you don't care about performance you can use .readlines() which returns list object and then slice the list.

E.g. for the first 5 lines:

with open("pathofmyfileandfileandname") as myfile: firstNlines=myfile.readlines()[0:5] #put here the interval you want 

Note: the whole file is read so is not the best from the performance point of view but it is easy to use, fast to write and easy to remember so if you want just perform some one-time calculation is very convenient

print firstNlines 

One advantage compared to the other answers is the possibility to select easily the range of lines e.g. skipping the first 10 lines [10:30] or the lasts 10 [:-10] or taking only even lines [::2].

What I do is to call the N lines using pandas. I think the performance is not the best, but for example if N=1000:

import pandas as pd yourfile = pd.read_csv('path/to/your/file.csv',nrows=1000) 

There is no specific method to read number of lines exposed by file object.

I guess the easiest way would be following:

lines =[] with open(file_name) as f: lines.extend(f.readline() for i in xrange(N)) 

The two most intuitive ways of doing this would be:

1. Iterate on the file line-by-line, and break after N lines.

2. Iterate on the file line-by-line using the next() method N times. (This is essentially just a different syntax for what the top answer does.)

Here is the code:

# Method 1: with open("fileName", "r") as f: counter = 0 for line in f: print line counter += 1 if counter == N: break # Method 2: with open("fileName", "r") as f: for i in xrange(N): line = f.next() print line 

The bottom line is, as long as you don't use readlines() or enumerateing the whole file into memory, you have plenty of options.

Based on gnibbler top voted answer (Nov 20 '09 at 0:27): this class add head() and tail() method to file object.

class File(file): def head(self, lines_2find=1): self.seek(0) #Rewind file return [self.next() for x in xrange(lines_2find)] def tail(self, lines_2find=1): self.seek(0, 2) #go to end of file bytes_in_file = self.tell() lines_found, total_bytes_scanned = 0, 0 while (lines_2find+1 > lines_found and bytes_in_file > total_bytes_scanned): byte_block = min(1024, bytes_in_file-total_bytes_scanned) self.seek(-(byte_block+total_bytes_scanned), 2) total_bytes_scanned += byte_block lines_found += self.read(1024).count('\n') self.seek(-total_bytes_scanned, 2) line_list = list(self.readlines()) return line_list[-lines_2find:] 

Usage:

f = File('path/to/file', 'r') f.head(3) f.tail(3) 

most convinient way on my own:

LINE_COUNT = 3 print [s for (i, s) in enumerate(open('test.txt')) if i < LINE_COUNT] 

Solution based on List Comprehension The function open() supports an iteration interface. The enumerate() covers open() and return tuples (index, item), then we check that we're inside an accepted range (if i < LINE_COUNT) and then simply print the result.

Enjoy the Python. ;)

For first 5 lines, simply do:

N=5 with open("data_file", "r") as file: for i in range(N): print file.next() 

If you want something that obviously (without looking up esoteric stuff in manuals) works without imports and try/except and works on a fair range of Python 2.x versions (2.2 to 2.6):

def headn(file_name, n): """Like *x head -N command""" result = [] nlines = 0 assert n >= 1 for line in open(file_name): result.append(line) nlines += 1 if nlines >= n: break return result if __name__ == "__main__": import sys rval = headn(sys.argv[1], int(sys.argv[2])) print rval print len(rval) 

If you have a really big file, and assuming you want the output to be a numpy array, using np.genfromtxt will freeze your computer. This is so much better in my experience:

def load_big_file(fname,maxrows): '''only works for well-formed text file of space-separated doubles''' rows = [] # unknown number of lines, so use list with open(fname) as f: j=0 for line in f: if j==maxrows: break else: line = [float(s) for s in line.split()] rows.append(np.array(line, dtype = np.double)) j+=1 return np.vstack(rows) # convert list of vectors to array 

I would like to handle the file with less than n-lines by reading the whole file

def head(filename: str, n: int): try: with open(filename) as f: head_lines = [next(f).rstrip() for x in range(n)] except StopIteration: with open(filename) as f: head_lines = f.read().splitlines() return head_lines 

Credit go to John La Rooy and Ilian Iliev. Use the function for the best performance with exception handle

Revise 1: Thanks FrankM for the feedback, to handle file existence and read permission we can futher add

import errno import os def head(filename: str, n: int): if not os.path.isfile(filename): raise FileNotFoundError(errno.ENOENT, os.strerror(errno.ENOENT), filename) if not os.access(filename, os.R_OK): raise PermissionError(errno.EACCES, os.strerror(errno.EACCES), filename) try: with open(filename) as f: head_lines = [next(f).rstrip() for x in range(n)] except StopIteration: with open(filename) as f: head_lines = f.read().splitlines() return head_lines 

You can either go with second version or go with the first one and handle the file exception later. The check is quick and mostly free from performance standpoint

Elaborating on previous answer from G M:

If you want to read the first lines quickly and you care about performance you can use .readlines(n) which reads first n bytes and then slice the list [0:5],
the count of bytes to read is in the sizehint-argument (1024 in the example)

 with open("pathofmyfileandfileandname") as myfile: firstNlines=myfile.readlines(1024)[0:5] # (): max byte count to read from file # []: put here the interval you want 

Only some first part of the file (byte count rounded up to next buffer size) is read (which speeds up the process if you are skimming bigger data files).

Syntax Reference in Python File readlines() Method

Starting at Python 2.6, you can take advantage of more sophisticated functions in the IO base clase. So the top rated answer above can be rewritten as:

 with open("datafile") as myfile: head = myfile.readlines(N) print head 

(You don't have to worry about your file having less than N lines since no StopIteration exception is thrown.)

This worked for me

f = open("history_export.csv", "r") line= 5 for x in range(line): a = f.readline() print(a) 

This works for Python 2 & 3:

from itertools import islice with open('/tmp/filename.txt') as inf: for line in islice(inf, N, N+M): print(line) 

 fname = input("Enter file name: ") num_lines = 0 with open(fname, 'r') as f: #lines count for line in f: num_lines += 1 num_lines_input = int (input("Enter line numbers: ")) if num_lines_input <= num_lines: f = open(fname, "r") for x in range(num_lines_input): a = f.readline() print(a) else: f = open(fname, "r") for x in range(num_lines_input): a = f.readline() print(a) print("Don't have", num_lines_input, " lines print as much as you can") print("Total lines in the text",num_lines) 

Simply Convert your CSV file object to a list using list(file_data)

import csv; with open('your_csv_file.csv') as file_obj: file_data = csv.reader(file_obj); file_list = list(file_data) for row in file_list[:4]: print(row) 

Here's another decent solution with a list comprehension:

file = open('file.txt', 'r') lines = [next(file) for x in range(3)] # first 3 lines will be in this list file.close() 

An easy way to get first 10 lines:

with open('fileName.txt', mode = 'r') as file: list = [line.rstrip('\n') for line in file][:10] print(list) 

#!/usr/bin/python import subprocess p = subprocess.Popen(["tail", "-n 3", "passlist"], stdout=subprocess.PIPE) output, err = p.communicate() print output 

This Method Worked for me