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How to put a group of integers in a row in a text file into a list?

I have a text file composed mostly of numbers something like this:
3 011236547892X
9 02321489764 Q
4 031246547873B
I would like to extract each of the following (spaces 5 to 14 (counting from zero)) into a list:
1236547892
321489764
1246547873
(Please note: each "number" is 10 "characters" long - the second row has a space at the end.)
and then perform analysis on the contents of each list.
I have umpteen versions, however I think I am closest with:
with open('k_d_m.txt') as f:
for line in f:
range = line.split()
num_lst = [x for x in range(3,10)]
print(num_lst)
However I have: TypeError: 'list' object is not callable
What is the best way forward?
What I want to do with num_lst is, amongst other things, as follows:
num_lst = list(map(int, str(num)))
print(num_lst)
nth = 2
odd_total = sum(num_lst[0::nth])
even_total = sum(num_lst[1::nth])
print(odd_total)
print(even_total)
if odd_total - even_total == 0 or odd_total - even_total == 11:
print("The number is ok")
else:
print("The number is not ok")

Use a simple slice:
with open('k_d_m.txt') as f:
num_lst = [x[5:15] for x in f]
Response to comment:
with open('k_d_m.txt') as f:
for line in f:
num_lst = list(line[5:15])
print(num_lst)

First of all, you shouldn't name your variable range, because that is already taken for the range() function. You can easily get the 5 to 14th chars of a string using string[5:15]. Try this:
num_lst = []
with open('k_d_m.txt') as f:
for line in f:
num_lst.append(line[5:15])
print(num_lst)

Extract text in string until certain new line ("\n") [duplicate]

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 do you sort by ascending order from a txt file in python?

I am new to python and have a hopefully simple question. I have a txt file with road names and lengths. i.e. Main street 56 miles. I am stumped on how to call on that file in Python and sort all the road lengths in ascending order. Thanks for your help.

Let's assume that there is a "miles" after every number. (This is an untested code so you can get it edited but I think the idea is right).
EDIT: THIS IS TESTED
import collections
originaldict = {}
newdict = collections.OrderedDict()
def isnum(string):
try:
if string is ".":
return True
float(string)
return True
except Exception:
return False
for line in open(input_file, "r"):
string = line[:line.find("miles") - 1]
print string
curnum = ""
for c in reversed(string):
if not isnum(c):
break
curnum = c + curnum
originaldict[float(curnum)] = []
originaldict[float(curnum)].append(line)
for num in sorted(originaldict.iterkeys()):
newdict[num] = []
newdict[num].append(originaldict[num][0])
del originaldict[num][0]
with open(output_file, "a") as o:
for value in newdict.values():
for line in value:
o.write(line + "\n")

Reason for two similar codes giving different result and different approaches to this task

The question is
def sum_numbers_in_file(filename):
"""
Return the sum of the numbers in the given file (which only contains
integers separated by whitespace).
>>> sum_numbers_in_file("numbers.txt")
19138
"""
this is my first code:
rtotal = 0
myfile = open(filename,"r")
num = myfile.readline()
num_list = []
while num:
number_line = ""
number_line += (num[:-1])
num_list.append(number_line.split(" "))
num = myfile.readline()
for item in num_list:
for item2 in item:
if item2!='':
rtotal+= int(item2)
return rtotal
this is my second code:
f = open(filename)
m = f.readline()
n = sum([sum([int(x) for x in line.split()]) for line in f])
f.close()
return n
however the first one returns 19138 and the second one 18138
numbers.txt contains the following:
1000
15000
2000
1138

Because m = f.readLine() already reads 1 line from f and then you do the operation with the rest of the lines. If you delete that statement the 2 outputs will be the same. (I think :))

I'd say that m = f.readline() in the second snippet skips the first line (which contains 1000), that's why you get a wrong result.

As requested.. another approach to the question:
import re
def sum(filename):
return sum(int(x.group()) for x in re.finditer(r'\d+',open(filename).read()))

As said by answers, you are skipping first line because f.readline(). But a shorter approach would be:
n=sum((int(line[:-1]) for line in open("numbers.txt") if line[0].isnumeric()))

How to read first N lines of a file?

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)