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)
I have a 1000 files in a folder named md_1.mdp, md_2.mdp, ..., md_1000.mdp and the 186th line of every file reads:
gen_seed = 35086
This value is different in every file and it is what I want to extract and print as the output.
I have written the following code but it is not displaying any output.
import numpy as np
idx = np.arange(1,1000)
for i in idx:
f = open('/home/abc/xyz/mdp_200/md_'+str(i)+'.mdp','r')
l = f.readlines()
l = l[185].split(" ")
flag = 0
for k in l:
if flag==1:
if k!='':
print(k)
flag=0
if k=="t=":
flag=1
f.close()
What should I add to this program so that it prints the required value for each file one by one in the order of md_1.mdp, md_2.mdp and so on?
you can use:
for i in range(1, 1001):
with open('/home/abc/xyz/mdp_200/md_'+ str(i)+ '.mdp') as fp:
l = fp.readlines()
print(l[185].split('=')[-1].strip())
or you can use linecache.getline:
import linecache
for i in range(1, 1001):
file = f'/home/abc/xyz/mdp_200/md_{i}.mdp'
line = linecache.getline(file, 185)
print(line.split('=')[-1].strip())
after you get your line the split is done by = character
I'm trying to read lines of a file into a list so every N lines will be in the same tuple. Assuming the file is valid so there are xN lines, how can I achive it?
The way I read the lines into the list:
def readFileIntoAList(file,N):
lines = list()
with open(file) as f:
lines = [line.rstrip('\n') for line in f]
return lines
What change I have to do with N so it will be a list of tuples so each tuple is of length N? For example I have the following file content:
ABC
abc xyz
123
XYZ
xyz abc
321
The output will be:
[("ABC","abc xyz","123"),("XYZ,"xyz abc",321")]
You could try using a chunking function:
def readFileIntoAList(file, n):
with open(file) as f:
lines = f.readlines()
return [lines[i:i + n] for i in range(0, len(lines), n)]
This will split the list of lines in the file into evenly sized chunks.
One way would be:
>>> data = []
>>> N = 3
>>> with open('/tmp/data') as f:
... while True:
... chunk = []
... for i in range(N):
... chunk.append(f.readline().strip('\n'))
... if any(True for c in chunk if not c):
... break
... data.append(tuple(chunk))
...
>>> print(data)
[('ABC', 'abc xyz', '123'), ('XYZ', 'xyz abc', '321')]
Note that this assumes the file has the right number of lines. Having the wrong number of lines in the above code can lead to infinite loop. A solution without that risk is:
data = []
N = 3
with open('/tmp/data') as f:
i = 0
chunk = []
for line in f:
chunk.append(line.strip('\n'))
i += 1
if i % N == 0 and i != 0:
data.append(tuple(chunk))
chunk = []
Both of these ways will not read the whole file in memory which should be more efficient when you process large datasets
You can use itertools.islice():
from itertools import islice
N = 3 # chunk size
with open("filename") as f:
lines = []
chunk = tuple(s.strip() for s in islice(f, N))
while chunk:
lines.append(chunk)
chunk = tuple(s.strip() for s in islice(f, N))
Also you can use map() if you prefer functional style:
chunk = tuple(map(str.strip, islice(f, N)))
import math
def readFileIntoAList(file,N):
lines= list()
lines1 = list()
with open(file) as f:
lines1 = [lineNew.rstrip("\n") for lineNew in f]
for a in range(math.ceil(len(lines1)/N)):
lines.append((*lines1[a*N:(a+1)*N],))
return lines
I used loop, I tried to make it easily.
I want to open a file, get numbers after the = sign, and put the result into a list. I did the first steps, but I'm stuck with assignment of the results into a list.
I tried to create a list and assign the result on it but when I print my list it shows me only the last results:
import cv2 as cv
import time
import numpy
from math import log
import csv
import re
statList = []
with open("C:\\ProgramData\\OutilTestObjets3D\\MaquetteCB-2019\\DataSet\\DEFAULT\\terrain\\3DObjects\\building\\house01.ive.stat.txt", 'r') as f:
#
statList = f.readlines()
statList = [x.strip() for x in statList]
for line in statList :
if (re.search("=" ,str(line))):
if (re.search('#IND',str(line))):
print("ERREUR")
else:
results = re.findall("=\s*?(\d+\.\d+|\d+)", str(line))
print ("result="+str(results))
statList.append(log(float(results[0])))
floatList = [str(results)]
print(floatList)
Its because you are overwriting results variable each time through your loop.
try
#
results = []
statList = f.readlines()
statList = [x.strip() for x in statList]
for line in statList :
if (re.search("=" ,str(line))):
if (re.search('#IND',str(line))):
print("ERREUR")
else:
results.extend(re.findall("=\s*?(\d+\.\d+|\d+)", str(line)))
print ("result="+str(results))
statList.append(log(float(results[0])))
floatList = [str(results)]
print(floatList)
The problem with your program is defining an empty list statList, then redefine it as statList = f.readlines() and append results to it. So, change the name of empty list, then you can use extend as long as results are list objects. And finally, use built-in map function to apply a function for every single item of your list:
from math import log
import re
final_result = []
with open("file.txt", 'r') as f:
#
statList = f.readlines()
statList = [x.strip() for x in statList]
for line in statList :
if (re.search("=" ,str(line))):
if (re.search('#IND',str(line))):
print("ERREUR")
else:
result = re.findall("=\s*?(\d+\.\d+|\d+)", str(line))
print("result=" + result[0])
final_result.extend(result)
# final_result.append(result[0])
floats_list = list(map(float, final_result))
logs_list = list(map(log, floats_list))
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)