Essentially I want to suck a line of text from a file, assign the characters to a list, and create a list of all the separate characters in a list -- a list of lists.

At the moment, I've tried this:

fO = open(filename, 'rU') fL = fO.readlines()

That's all I've got. I don't quite know how to extract the single characters and assign them to a new list.

The line I get from the file will be something like:

fL = 'FHFF HHXH XXXX HFHX'

I want to turn it into this list, with each single character on its own:

['F', 'H', 'F', 'F', 'H', ...]
0

10 Answers

You can do this using list:

new_list = list(fL)

Be aware that any spaces in the line will be included in this list, to the best of my knowledge.

2

I'm a bit late it seems to be, but...

a='hello' print list(a) # ['h','e','l','l', 'o']

Strings are iterable (just like a list).

I'm interpreting that you really want something like:

fd = open(filename,'rU') chars = [] for line in fd: for c in line: chars.append(c)

or

fd = open(filename, 'rU') chars = [] for line in fd: chars.extend(line)

or

chars = [] with open(filename, 'rU') as fd: map(chars.extend, fd)

chars would contain all of the characters in the file.

2

python >= 3.5

Version 3.5 onwards allows the use of PEP 448 - Extended Unpacking Generalizations:

>>> string = 'hello' >>> [*string] ['h', 'e', 'l', 'l', 'o']

This is a specification of the language syntax, so it is faster than calling list:

>>> from timeit import timeit >>> timeit("list('hello')") 0.3042821969866054 >>> timeit("[*'hello']") 0.1582647830073256

So to add the string hello to a list as individual characters, try this:

newlist = [] newlist[:0] = 'hello' print (newlist) ['h','e','l','l','o']

However, it is easier to do this:

splitlist = list(newlist) print (splitlist)
2
fO = open(filename, 'rU') lst = list(fO.read())

Or use a fancy list comprehension, which are supposed to be "computationally more efficient", when working with very very large files/lists

fd = open(filename,'r') chars = [c for line in fd for c in line if c is not " "] fd.close()

Btw: The answer that was accepted does not account for the whitespaces...

a='hello world' map(lambda x:x, a)

['h', 'e', 'l', 'l', 'o', ' ', 'w', 'o', 'r', 'l', 'd']

An easy way is using function “map()”.

In python many things are iterable including files and strings. Iterating over a filehandler gives you a list of all the lines in that file. Iterating over a string gives you a list of all the characters in that string.

charsFromFile = [] filePath = r'path\to\your\file.txt' #the r before the string lets us use backslashes for line in open(filePath): for char in line: charsFromFile.append(char) #apply code on each character here

or if you want a one liner

#the [0] at the end is the line you want to grab. #the [0] can be removed to grab all lines [list(a) for a in list(open('test.py'))][0]

.

.

Edit: as agf mentions you can use itertools.chain.from_iterable

His method is better, unless you want the ability to specify which lines to grab list(itertools.chain.from_iterable(open(filename, 'rU)))

This does however require one to be familiar with itertools, and as a result looses some readablity

If you only want to iterate over the chars, and don't care about storing a list, then I would use the nested for loops. This method is also the most readable.

Because strings are (immutable) sequences they can be unpacked similar to lists:

with open(filename, 'rU') as fd: multiLine = fd.read() *lst, = multiLine

When running map(lambda x: x, multiLine) this is clearly more efficient, but in fact it returns a map object instead of a list.

with open(filename, 'rU') as fd: multiLine = fd.read() list(map(lambda x: x, multiLine))

Turning the map object into a list will take longer than the unpacking method.