Related
We have a large raw data file that we would like to trim to a specified size.
How would I go about getting the first N lines of a text file in python? Will the OS being used have any effect on the implementation?
Python 3:
with open(path_to_file) as input_file:
head = [next(input_file) for _ in range(lines_number)]
print(head)
Python 2:
with open(path_to_file) as input_file:
head = [next(input_file) for _ in xrange(lines_number)]
print head
Here's another way (both Python 2 & 3):
from itertools import islice
with open(path_to_file) as input_file:
head = list(islice(path_to_file, lines_number))
print(head)
N = 10
with open("file.txt", "a") as file: # the a opens it in append mode
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:
Iterate on the file line-by-line, and break after N lines.
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
This worked for me
f = open("history_export.csv", "r")
line= 5
for x in range(line):
a = f.readline()
print(a)
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
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 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)
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
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)
How can I do something like this?
with open(r'C:\some_list.txt') as f:
list = f.readlines()
for line in list:
if line: #has more than x characters
delete line
If the file is reasonably small, the easiest way is to read it all in, filter, then write it all out.
with open(r'C:\some_list.txt') as f:
lines = f.readlines()
# Keep lines <= 10 chars long with a list comprehension
filtered_lines = [line for line in lines if len(line) > 10]
# Do what you like with the lines, e.g. write them out into another file:
with open(r'C:\filtered_list.txt', 'w') as f:
for line in filtered_lines:
f.write(line)
If you want to stream the matching lines into another file, that's even easier:
with open(r'C:\some_list.txt') as in_file, open(r'C:\filtered_list.txt', 'w') as out_file:
for line in in_file:
if len(line) <= 10:
out_file.write(line)
You can read the file line by line, write the line in a new file if it pass the constrain (abandon other lines). For large files, its so efficient in terms of memory usage:
with open('file_r.txt', 'r') as file_r, open('file_w.txt', 'w') as file_w:
thresh = 3
for line in file_r:
if len(line) < thresh:
file_w.write(line)
Try (I do 3 as an example):
with open(r'C:\some_list.txt') as f:
l = [i for i in f if len(i) > 3]
I renamed list to l since list is a builtin.
Conversely, it could be done like this:
# fname : file name
# x : number of characters or length
def delete_lines(fname = 'test.txt', x = 8):
with open(fname, "r") as f:
lines = f.readlines()
with open(fname, "w") as f:
for line in lines:
if len(line) <= x:
f.write(line)
delete_lines()
Certainly, there are better ways of doing this.
first save the lines in a list which will not be deleted by reading one by one:
the_list = []
with open(r'C:\some_list.txt', "r") as f:
for line in f:
#print(len(line))
if (len(line)) < 50:#here I used 50 charecters
the_list.append(line)
then write the list into your file:
with open(r'C:\some_list.txt', 'w') as f:
for line in the_list:
f.write(line)
if you don't want to use a list or the file is too big then try:
with open(r'C:\some_list.txt', "r") as f, open('new.txt', 'a') as fw:
for line in f:
if (len(line)) < 50:
fw.write(line)
replace output.txt according to your need. Above code will read line by line from some_list.txt and then write in 'output.txt' if the line has less than 50 characters
I know it's not completely finished, but I'm very confused as how to format the save inventory function so it prints like the original text file. During the add_item function, it shows that the item has been added to the lists. But when going to write nothing is there or updated.
Example of how the text file needs to look
def save_inventory(inventoryFile, descriptionArray, quantityArray, priceArray, intrecords):
outfile = open(inventoryFile, "w")
with open('inventory1.txt', 'r') as f:
count = -1
for line in f:
count+=1
if count % 3 == 0: #this is the remainder operator
outfile.write(descriptionArray)
print(descriptionArray)
with open('inventory1.txt', 'r') as f:
count = -2
for line in f:
count+=1
if count % 3 == 0: #this is the remainder operator
outfile.write(str(quantityArray))
print(quantityArray)
with open('inventory1.txt', 'r') as f:
count = -3
for line in f:
count+=1
if count % 3 == 0: #this is the remainder operator
outfile.write(str(priceArray))
print(priceArray)
outfile.close()
You are only writing to the file when you have read a line. If your text file is empty you will never write to the file.
What I would do is zip the lists together and loop through them. Then write three lines to the file for each pass through the loop. You can print a carriage return with '\n'
with open(inventoryFile, 'w') as f:
for d, q, p in zip(descriptionArray, quantityArray, priceArray):
f.write('%s\n%s\n%s\n' % (d, q, p))
I was trying to extract even lines from a text file and output to a new file. But with my codes python warns me "list index out of range". Anyone can help me? THANKS~
Code:
f = open('input.txt', 'r')
i = 0
j = 0
num_lines = sum(1 for line in f)
newline = [0] * num_lines
print (num_lines)
for i in range(1, num_lines):
if i % 2 == 0:
newline[i] = f.readlines()[i]
print i, newline[i]
i = i + 1
f.close()
f = open('output.txt', 'w')
for j in range(0,num_lines):
if j % 2 == 0:
f.write(newline[j] + '\n')
j = j + 1
f.close()
Output:
17
Traceback (most recent call last):
File "./5", line 10, in <module>
a = f.readlines()[1]
IndexError: list index out of range
After
num_lines = sum(1 for line in f)
The file pointer in f is at the end of the file. Therefore any subsequent call of f.readlines() gives an empty list. The minimal fix is to use f.seek(0) to return to the start of the file.
However, a better solution would be to read through the file only once, e.g. using enumerate to get the line and its index i:
newline = []
for i, line in enumerate(f):
if i % 2 == 0:
newline.append(line)
In your original script you read the file once to scan the number of lines, then you (try to) read the lines in memory, you needlessly create a list for the full size instead of just extending it with list.append, you initialize the list with zeroes which does not make sense for a list containing strings, etc.
Thus, this script does what your original idea was, but better and simpler and faster:
with open('input.txt', 'r') as inf, open('output.txt', 'w') as outf:
for lineno, line in enumerate(inf, 1):
if lineno % 2 == 0:
outf.write(line)
Specifically
open the files with with statement so that they are automatically closed when
the block is exited.
write as they are read
as lines are numbered 1-based, use the enumerate with the start value 1 so that you truly get the even numbered lines.
You've also got the itertools.islice approach available:
from itertools import islice
with open('input') as fin, open('output', 'w') as fout:
fout.writelines(islice(fin, None, None, 2))
This saves the modulus operation and puts the line writing to system level.
We have a large raw data file that we would like to trim to a specified size.
How would I go about getting the first N lines of a text file in python? Will the OS being used have any effect on the implementation?
Python 3:
with open(path_to_file) as input_file:
head = [next(input_file) for _ in range(lines_number)]
print(head)
Python 2:
with open(path_to_file) as input_file:
head = [next(input_file) for _ in xrange(lines_number)]
print head
Here's another way (both Python 2 & 3):
from itertools import islice
with open(path_to_file) as input_file:
head = list(islice(path_to_file, lines_number))
print(head)
N = 10
with open("file.txt", "a") as file: # the a opens it in append mode
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:
Iterate on the file line-by-line, and break after N lines.
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
This worked for me
f = open("history_export.csv", "r")
line= 5
for x in range(line):
a = f.readline()
print(a)
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
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 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)
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
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)