python reading files - python

I need to get a specific line number from a file that I am passing into a python program I wrote. I know that the line I want will be line 5, so is there a way I can just grab line 5, and not have to iterate through the file?

If you know how many bytes you have before the line you're interested in, you could seek to that point and read out a line. Otherwise, a "line" is not a first class construct (it's just a list of characters terminated by a character you're assigning a special meaning to - a newline). To find these newlines, you have to read the file in.
Practically speaking, you could use the readline method to read off 5 lines and then read your line.
Why are you trying to do this?

you can to use linecache
import linecache
get = linecache.getline
print(get(path_of_file, number_of_line))

I think following should do :
line_number=4
# Avoid reading the whole file
f = open('path/to/my/file','r')
count=1
for i in f.readline():
if count==line_number:
print i
break
count+=1
# By reading the whole file
f = open('path/to/my/file','r')
lines = f.read().splitlines()
print lines[line_number-1] # Index starts from 0
This should give you the 4th line in the file.

Related

How do i check for a keyword in a specific line of a text file? python [duplicate]

I want to go to line 34 in a .txt file and read it. How would you do that in Python?
Use Python Standard Library's linecache module:
line = linecache.getline(thefilename, 33)
should do exactly what you want. You don't even need to open the file -- linecache does it all for you!
This code will open the file, read the line and print it.
# Open and read file into buffer
f = open(file,"r")
lines = f.readlines()
# If we need to read line 33, and assign it to some variable
x = lines[33]
print(x)
A solution that will not read more of the file than necessary is
from itertools import islice
line_number = 34
with open(filename) as f:
# Adjust index since Python/islice indexes from 0 and the first
# line of a file is line 1
line = next(islice(f, line_number - 1, line_number))
A very straightforward solution is
line_number = 34
with open(filename) as f:
f.readlines()[line_number - 1]
There's two ways:
Read the file, line by line, stop when you've gotten to the line you want
Use f.readlines() which will read the entire file into memory, and return it as a list of lines, then extract the 34th item from that list.
Solution 1
Benefit: You only keep, in memory, the specific line you want.
code:
for i in xrange(34):
line = f.readline();
# when you get here, line will be the 34th line, or None, if there wasn't
# enough lines in the file
Solution 2
Benefit: Much less code
Downside: Reads the entire file into memory
Problem: Will crash if less than 34 elements are present in the list, needs error handling
line = f.readlines()[33]
You could just read all the lines and index the line your after.
line = open('filename').readlines()[33]
for linenum,line in enumerate(open("file")):
if linenum+1==34: print line.rstrip()
I made a thread about this and didn't receive help so I took matter into my own hands.
Not any complicated code here.
import linecache
#Simply just importing the linecache function to read our line of choosing
number = int(input("Enter a number from 1-10 for a random quote "))
#Asks the user for which number they would like to read(not necessary)
lines = linecache.getline("Quotes.txt", number)
#Create a new variable in order to grab the specific line, the variable
#integer can be replaced by any integer of your choosing.
print(lines)
#This will print the line of your choosing.
If you are completing this in python make sure you have both files (.py) and (.txt) in the same location otherwise python will not be able to retrieve this, unless you specify the file location. EG.
linecache.getline("C:/Directory/Folder/Quotes.txt
This is used when the file is in another folder than the .py file you are using.
Hope this helps!
Option that always closes the file and doesn't load the whole file into memory
with open('file.txt') as f:
for i, line in enumerate(f):
if i+1 == 34: break
print(line.rstrip())

editing a single .txt line in python 3.1

i have some data stored in a .txt file in this format:
----------|||||||||||||||||||||||||-----------|||||||||||
1029450386abcdefghijklmnopqrstuvwxy0293847719184756301943
1020414646canBeFollowedBySpaces 3292532113435532419963
don't ask...
i have many lines of this, and i need a way to add more digits to the end of a particular line.
i've written code to find the line i want, but im stumped as to how to add 11 characters to the end of it. i've looked around, this site has been helpful with some other issues i've run into, but i can't seem to find what i need for this.
it is important that the line retain its position in the file, and its contents in their current order.
using python3.1, how would you turn this:
1020414646canBeFollowedBySpaces 3292532113435532419963
into
1020414646canBeFollowedBySpaces 329253211343553241996301846372998
As a general principle, there's no shortcut to "inserting" new data in the middle of a text file. You will need to make a copy of the entire original file in a new file, modifying your desired line(s) of text on the way.
For example:
with open("input.txt") as infile:
with open("output.txt", "w") as outfile:
for s in infile:
s = s.rstrip() # remove trailing newline
if "target" in s:
s += "0123456789"
print(s, file=outfile)
os.rename("input.txt", "input.txt.original")
os.rename("output.txt", "input.txt")
Check out the fileinput module, it can do sort of "inplace" edits with files. though I believe temporary files are still involved in the internal process.
import fileinput
for line in fileinput.input('input.txt', inplace=1, backup='.orig'):
if line.startswith('1020414646canBeFollowedBySpaces'):
line = line.rstrip() + '01846372998' '\n'
print(line, end='')
The print now prints to the file instead of the console.
You might want to back up your original file before editing.
target_chain = '1020414646canBeFollowedBySpaces 3292532113435532419963'
to_add = '01846372998'
with open('zaza.txt','rb+') as f:
ch = f.read()
x = ch.find(target_chain)
f.seek(x + len(target_chain),0)
f.write(to_add)
f.write(ch[x + len(target_chain):])
In this method it's absolutely obligatory to open the file in binary mode 'b' for some reason linked to the treatment of the end of lines by Python (see Universal Newline, enabled by default)
The mode 'r+' is to allow the writing as well as the reading
In this method, what is before the target_chain in the file remains untouched. And what is after the target_chain is shifted ahead. As said by Greg Hewgill, there is no possibility to move apart bits on a hard drisk to insert new bits in the middle.
Evidently, if the file is very big, reading all of its content in ch could be too much memory consuming and the algorithm should then be changed: reading line after line until the line containing the target_chain, and then reading the next line before inserting, and then continuing to do "reading the next line - re-writing on the current line" until the end of the file in order to shift progressively the content from the line concerned with addition.
You see what I mean...
Copy the file, line by line, to another file. When you get to the line that needs extra chars then add them before writing.

How do you read a file into a list in Python? [duplicate]

This question already has answers here:
How to read a file line-by-line into a list?
(28 answers)
Closed 8 years ago.
I want to prompt a user for a number of random numbers to be generated and saved to a file. He gave us that part. The part we have to do is to open that file, convert the numbers into a list, then find the mean, standard deviation, etc. without using the easy built-in Python tools.
I've tried using open but it gives me invalid syntax (the file name I chose was "numbers" and it saved into "My Documents" automatically, so I tried open(numbers, 'r') and open(C:\name\MyDocuments\numbers, 'r') and neither one worked).
with open('C:/path/numbers.txt') as f:
lines = f.read().splitlines()
this will give you a list of values (strings) you had in your file, with newlines stripped.
also, watch your backslashes in windows path names, as those are also escape chars in strings. You can use forward slashes or double backslashes instead.
Two ways to read file into list in python (note these are not either or) -
use of with - supported from python 2.5 and above
use of list comprehensions
1. use of with
This is the pythonic way of opening and reading files.
#Sample 1 - elucidating each step but not memory efficient
lines = []
with open("C:\name\MyDocuments\numbers") as file:
for line in file:
line = line.strip() #or some other preprocessing
lines.append(line) #storing everything in memory!
#Sample 2 - a more pythonic and idiomatic way but still not memory efficient
with open("C:\name\MyDocuments\numbers") as file:
lines = [line.strip() for line in file]
#Sample 3 - a more pythonic way with efficient memory usage. Proper usage of with and file iterators.
with open("C:\name\MyDocuments\numbers") as file:
for line in file:
line = line.strip() #preprocess line
doSomethingWithThisLine(line) #take action on line instead of storing in a list. more memory efficient at the cost of execution speed.
the .strip() is used for each line of the file to remove \n newline character that each line might have. When the with ends, the file will be closed automatically for you. This is true even if an exception is raised inside of it.
2. use of list comprehension
This could be considered inefficient as the file descriptor might not be closed immediately. Could be a potential issue when this is called inside a function opening thousands of files.
data = [line.strip() for line in open("C:/name/MyDocuments/numbers", 'r')]
Note that file closing is implementation dependent. Normally unused variables are garbage collected by python interpreter. In cPython (the regular interpreter version from python.org), it will happen immediately, since its garbage collector works by reference counting. In another interpreter, like Jython or Iron Python, there may be a delay.
f = open("file.txt")
lines = f.readlines()
Look over here. readlines() returns a list containing one line per element. Note that these lines contain the \n (newline-character) at the end of the line. You can strip off this newline-character by using the strip()-method. I.e. call lines[index].strip() in order to get the string without the newline character.
As joaquin noted, do not forget to f.close() the file.
Converting strint to integers is easy: int("12").
The pythonic way to read a file and put every lines in a list:
from __future__ import with_statement #for python 2.5
with open('C:/path/numbers.txt', 'r') as f:
lines = f.readlines()
Then, assuming that each lines contains a number,
numbers =[int(e.strip()) for e in lines]
You need to pass a filename string to open. There's an extra complication when the string has \ in it, because that's a special string escape character to Python. You can fix this by doubling up each as \\ or by putting a r in front of the string as follows: r'C:\name\MyDocuments\numbers'.
Edit: The edits to the question make it completely different from the original, and since none of them was from the original poster I'm not sure they're warrented. However it does point out one obvious thing that might have been overlooked, and that's how to add "My Documents" to a filename.
In an English version of Windows XP, My Documents is actually C:\Documents and Settings\name\My Documents. This means the open call should look like:
open(r"C:\Documents and Settings\name\My Documents\numbers", 'r')
I presume you're using XP because you call it My Documents - it changed in Vista and Windows 7. I don't know if there's an easy way to look this up automatically in Python.
hdl = open("C:/name/MyDocuments/numbers", 'r')
milist = hdl.readlines()
hdl.close()
To summarize a bit from what people have been saying:
f=open('data.txt', 'w') # will make a new file or erase a file of that name if it is present
f=open('data.txt', 'r') # will open a file as read-only
f=open('data.txt', 'a') # will open a file for appending (appended data goes to the end of the file)
If you wish have something in place similar to a try/catch
with open('data.txt') as f:
for line in f:
print line
I think #movieyoda code is probably what you should use however
If you have multiple numbers per line and you have multiple lines, you can read them in like this:
#!/usr/bin/env python
from os.path import dirname
with open(dirname(__file__) + '/data/path/filename.txt') as input_data:
input_list= [map(int,num.split()) for num in input_data.readlines()]

taking a character input in python from a file?

in python , suppose i have file data.txt . which has 6 lines of data . I want to calculate the no of lines which i am planning to do by going through each character and finding out the number of '\n' in the file . How to take one character input from the file ? Readline takes the whole line .
I think the method you're looking for is readlines, as in
lines = open("inputfilex.txt", "r").readlines()
This will give you a list of each of the lines in the file. To find out how many lines, you can just do:
len(lines)
And then access it using indexes, like lines[3] or lines[-1] as you would any normal Python list.
You can use read(1) to read a single byte. help(file) says:
read(size) -> read at most size bytes, returned as a string.
If the size argument is negative or omitted, read until EOF is reached.
Notice that when in non-blocking mode, less data than what was requested
may be returned, even if no size parameter was given.
Note that reading a file a byte at a time is quite un-"Pythonic". This is par for the course in C, but Python can do a lot more work with far less code. For example, you can read the entire file into an array in one line of code:
lines = f.readlines()
You could then access by line number with a simple lines[lineNumber] lookup.
Or if you don't want to store the entire file in memory at once, you can iterate over it line-by-line:
for line in f:
# Do whatever you want.
That is much more readable and idiomatic.
It seems the simplest answer for you would be to do:
for line in file:
lines += 1
# do whatever else you need to do for each line
Or the equivalent construction explicitly using readline(). I'm not sure why you want to look at every character when you said above that readline() is correctly reading each line in its entirety.
To access a file based on its lines, make a list of its lines.
with open('myfile') as f:
lines = list(f)
then simply access lines[3] to get the fourth line and so forth. (Note that this will not strip the newline characters.)
The linecache module can also be useful for this.

Go to a specific line in Python?

I want to go to line 34 in a .txt file and read it. How would you do that in Python?
Use Python Standard Library's linecache module:
line = linecache.getline(thefilename, 33)
should do exactly what you want. You don't even need to open the file -- linecache does it all for you!
This code will open the file, read the line and print it.
# Open and read file into buffer
f = open(file,"r")
lines = f.readlines()
# If we need to read line 33, and assign it to some variable
x = lines[33]
print(x)
A solution that will not read more of the file than necessary is
from itertools import islice
line_number = 34
with open(filename) as f:
# Adjust index since Python/islice indexes from 0 and the first
# line of a file is line 1
line = next(islice(f, line_number - 1, line_number))
A very straightforward solution is
line_number = 34
with open(filename) as f:
f.readlines()[line_number - 1]
There's two ways:
Read the file, line by line, stop when you've gotten to the line you want
Use f.readlines() which will read the entire file into memory, and return it as a list of lines, then extract the 34th item from that list.
Solution 1
Benefit: You only keep, in memory, the specific line you want.
code:
for i in xrange(34):
line = f.readline();
# when you get here, line will be the 34th line, or None, if there wasn't
# enough lines in the file
Solution 2
Benefit: Much less code
Downside: Reads the entire file into memory
Problem: Will crash if less than 34 elements are present in the list, needs error handling
line = f.readlines()[33]
You could just read all the lines and index the line your after.
line = open('filename').readlines()[33]
for linenum,line in enumerate(open("file")):
if linenum+1==34: print line.rstrip()
I made a thread about this and didn't receive help so I took matter into my own hands.
Not any complicated code here.
import linecache
#Simply just importing the linecache function to read our line of choosing
number = int(input("Enter a number from 1-10 for a random quote "))
#Asks the user for which number they would like to read(not necessary)
lines = linecache.getline("Quotes.txt", number)
#Create a new variable in order to grab the specific line, the variable
#integer can be replaced by any integer of your choosing.
print(lines)
#This will print the line of your choosing.
If you are completing this in python make sure you have both files (.py) and (.txt) in the same location otherwise python will not be able to retrieve this, unless you specify the file location. EG.
linecache.getline("C:/Directory/Folder/Quotes.txt
This is used when the file is in another folder than the .py file you are using.
Hope this helps!
Option that always closes the file and doesn't load the whole file into memory
with open('file.txt') as f:
for i, line in enumerate(f):
if i+1 == 34: break
print(line.rstrip())

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