Break string into list of characters in Python
Question:
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', ...]
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.
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.
fO = open(filename, 'rU')
lst = list(fO.read())
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'pathtoyourfile.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.
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…
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)
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()”.
I’m a bit late it seems to be, but…
a='hello'
print list(a)
# ['h','e','l','l', 'o']
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.
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
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', ...]
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.
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.
fO = open(filename, 'rU')
lst = list(fO.read())
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'pathtoyourfile.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.
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…
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)
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()”.
I’m a bit late it seems to be, but…
a='hello'
print list(a)
# ['h','e','l','l', 'o']
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.
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