Create a List that contain each Line of a File

Question

I\'m trying to open a file and create a list with each line read from the file.

   i=0
   List=[\"\"]
   for Line in inFile:
      List[i]=Line.split(\",\")
         


        

Answer 1

A file is almost a list of lines. You can trivially use it in a for loop.

myFile= open( "SomeFile.txt", "r" )
for x in myFile:
    print x
myFile.close()

Or, if you want an actual list of lines, simply create a list from the file.

myFile= open( "SomeFile.txt", "r" )
myLines = list( myFile )
myFile.close()
print len(myLines), myLines

You can't do someList[i] to put a new item at the end of a list. You must do someList.append(i).

Also, never start a simple variable name with an uppercase letter. List confuses folks who know Python.

Also, never use a built-in name as a variable. list is an existing data type, and using it as a variable confuses folks who know Python.

Answer 2

f.readlines() returns a list that contains each line as an item in the list

if you want eachline to be split(",") you can use list comprehensions

[ list.split(",") for line in file ]

Answer 3

I am not sure about Python but most languages have push/append function for arrays.

Answer 4

I did it this way

lines_list = open('file.txt').read().splitlines()

Every line comes with its end of line characters (\n\r); this way the characters are removed.

Answer 5

Please read PEP8. You're swaying pretty far from python conventions.

If you want a list of lists of each line split by comma, I'd do this:

l = []
for line in in_file:
    l.append(line.split(','))

You'll get a newline on each record. If you don't want that:

l = []
for line in in_file:
    l.append(line.rstrip().split(','))

Answer 6

Assuming you also want to strip whitespace at beginning and end of each line, you can map the string strip function to the list returned by readlines:

map(str.strip, open('filename').readlines())