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Delete spaces in dictionary values python

I'm reading the data from an outsource. The data has "Name" and "Value with warnings" so I put those in a dictionary in a manner as
d[data[i:i+6]] = data[i+8:i+17], data[i+25:i+36]
Thus at the end I have my dict as;
{'GPT-P ': ('169 ', 'H '), 'GOT-P ': ('47 ', ' '), .....
As seen above both keys and values might have unnecessary spaces.
I was able to overcome spaces in keys with;
d = {x.replace(' ',''): v
for x, v in d.items()}
but can't seem to manage similar for values. I tried using d.values() but it trims the key name and also works only for 1 of the values.
Can you help me understand how I can remove space for several values (2 values in this particular case) and end up with something like;
{'GPT-P': ('169', 'H'), 'GOT-P ': ('47', ''), .....
Thanks. Stay safe and healthy

You will need to do the space replacement in your v values also but
it seems that in your case the values in your dictionary are tuples.
I guess you will want to remove spaces in all elements of each tuple so you will need a second iteration here. You can do something like this:
d = {'GPT-P ': ('169 ', 'H '), 'GOT-P ': ('47 ', ' ')}
{x.replace(' ', ''): tuple(w.replace(' ', '') for w in v) for x, v in d.items()}
Which returns:
{'GPT-P': ('169', 'H'), 'GOT-P': ('47', '')}
Notice that there is list (or tuple) comprehension tuple(w.replace(' ', '') for w in v) within the dictionary comprehension.

Given:
DoT={'GPT-P ': ('169 ', 'H '), 'GOT-P ': ('47 ', ' ')}
Since you have tuples of strings as your values, you need to apply .strip() to each string in the tuple:
>>> tuple(e.strip() for e in ('47 ', ' '))
('47', '')
Apply that to each key, value in a dict comprehension and there you are:
>>> {k.strip():tuple(e.strip() for e in t) for k,t in DoT.items()}
{'GPT-P': ('169', 'H'), 'GOT-P': ('47', '')}
You use .replace(' ','') in your attempt. That will replace ALL spaces:
>>> ' 1 2 3 '.replace(' ','')
'123'
It is more typical to use one of the .strips():
>>> ' 1 2 3 '.strip()
'1 2 3'
>>> ' 1 2 3 '.lstrip()
'1 2 3 '
>>> ' 1 2 3 '.rstrip()
' 1 2 3'
You can use .replace or any of the .strips() in the comprehensions that I used above.

How would I convert this string into a list where the spaces are also separate items?

code = ("1 2 3 4")
b = code.split()
print(b)
This code returns ['1','2','3','4'] when I want it to return ['1',' ','2', '3',' ','4']. How would I do that?

I suppose you want to split words, not characters (otherwise list(code) would decompose that for you).
use re.split preserving split char by wrapping the split regex between parentheses:
import re
code = "1 2 3 4"
print(re.split("( )",code))
result:
['1', ' ', '2', ' ', '3', ' ', '4']
make it re.split("(\s+)",code) if you want to match more than 1 space, tabs, etc...

>>> code.replace(' ', ', ,').split(',')
['1', ' ', '2', ' ', '3', ' ', '4']

Here's a readable and easy approach without importing any library
code = ("1 2 3 4")
b = []
for element in code.split():
b.append(element)
b.append(" ")
del b[-1]
print b

>>> [x for y in code.split() for x in (y, ' ')][:-1]
['1', ' ', '2', ' ', '3', ' ', '4']
Or
>>> from itertools import cycle,chain
>>> list(chain(*zip(code.split(), cycle(' '))))[:-1]
['1', ' ', '2', ' ', '3', ' ', '4']

Someone posted this and deleted it, but this works quite simply. list() will give you a list of an iterable.
code = ("1 2 3 4")
b = list(code)
print(b)
['1', ' ', '2', ' ', '3', ' ', '4']

Simply convert it to list
code = ("1 2 3 4")
b = list(code)
print(b)
Note:- if you have 2 digit no. It will also split those numbers

code = ("1 2 3 4")
list(code)

Align numbers in sublist

I have a set of numbers that I want to align considering the comma:
10 3
200 4000,222 3 1,5
200,21 0,3 2
30000 4,5 1
mylist = [['10', '3', '', ''],
['200', '4000,222', '3', '1,5'],
['200,21', '', '0,3', '2'],
['30000', '4,5', '1', '']]
What I want is to align this list considering the comma:
expected result:
mylist = [[' 10 ', ' 3 ', ' ', ' '],
[' 200 ', '4000,222', '3 ', '1,5'],
[' 200,21', ' ', '0,3', '2 '],
['30000 ', ' 4,5 ', '1 ', ' ']]
I tried to turn the list:
mynewlist = list(zip(*mylist))
and to find the longest part after the comma in every sublist:
for m in mynewlist:
max([x[::-1].find(',') for x in m]
and to use rjust and ljust but I don't know how to ljust after a comma and rjust before the comma, both in the same string.
How can I resolve this without using format()?
(I want to align with ljust and rjust)

Here's another approach that currently does the trick. Unfortunately, I can't see any simple way to make this work, maybe due to the time :-)
Either way, I'll explain it. r is the result list created before hand.
r = [[] for i in range(4)]
Then we loop through the values and also grab an index with enumerate:
for ind1, vals in enumerate(zip(*mylist)):
Inside the loop we grab the max length of the decimal digits present and the max length of the word (the word w/o the decimal digits):
l = max(len(v.partition(',')[2]) for v in vals) + 1
mw = max(len(v if ',' not in v else v.split(',')[0]) for v in vals)
Now we go through the values inside the tuple vals and build our results (yup, can't currently think of a way to avoid this nesting).
for ind2, v in enumerate(vals):
If it contains a comma, it should be formatted differently. Specifically, we rjust it based on the max length of a word mw and then add the decimal digits and any white-space needed:
if ',' in v:
n, d = v.split(',')
v = "".join((n.rjust(mw),',', d, " " * (l - 1 - len(d))))
In the opposite case, we simply .rjust and then add whitespace:
else:
v = "".join((v.rjust(mw) + " " * l))
finally, we append to r.
r[ind1].append(v)
All together:
r = [[] for i in range(4)]
for ind1, vals in enumerate(zip(*mylist)):
l = max(len(v.partition(',')[2]) for v in vals) + 1
mw = max(len(v if ',' not in v else v.split(',')[0]) for v in vals)
for ind2, v in enumerate(vals):
if ',' in v:
n, d = v.split(',')
v = "".join((n.rjust(mw),',', d, " " * (l - 1 - len(d))))
else:
v = "".join((v.rjust(mw) + " " * l))
r[ind1].append(v)
Now, we can print it out:
>>> print(*map(list,zip(*r)), sep='\n)
[' 10 ', ' 3 ', ' ', ' ']
[' 200 ', '4000,222', '3 ', '1,5']
[' 200,21', ' ', '0,3', '2 ']
['30000 ', ' 4,5 ', '1 ', ' ']

Here's a bit different solution that doesn't transpose my_list but instead iterates over it twice. On the first pass it generates a list of tuples, one for each column. Each tuple is a pair of numbers where first number is length before comma and second number is length of comma & everything following it. For example '4000,222' results to (4, 4). On the second pass it formats the data based on the formatting info generated on first pass.
from functools import reduce
mylist = [['10', '3', '', ''],
['200', '4000,222', '3', '1,5'],
['200,21', '', '0,3', '2'],
['30000', '4,5', '1', '']]
# Return tuple (left part length, right part length) for given string
def part_lengths(s):
left, sep, right = s.partition(',')
return len(left), len(sep) + len(right)
# Return string formatted based on part lengths
def format(s, first, second):
left, sep, right = s.partition(',')
return left.rjust(first) + sep + right.ljust(second - len(sep))
# Generator yielding part lengths row by row
parts = ((part_lengths(c) for c in r) for r in mylist)
# Combine part lengths to find maximum for each column
# For example data it looks like this: [[5, 3], [4, 4], [1, 2], [1, 2]]
sizes = reduce(lambda x, y: [[max(z) for z in zip(a, b)] for a, b in zip(x, y)], parts)
# Format result based on part lengths
res = [[format(c, *p) for c, p in zip(r, sizes)] for r in mylist]
print(*res, sep='\n')
Output:
[' 10 ', ' 3 ', ' ', ' ']
[' 200 ', '4000,222', '3 ', '1,5']
[' 200,21', ' ', '0,3', '2 ']
['30000 ', ' 4,5 ', '1 ', ' ']

This works for python 2 and 3. I didn't use ljust or rjust though, i just added as many spaces before and after the number as are missing to the maximum sized number in the column:
mylist = [['10', '3', '', ''],
['200', '4000,222', '3', '1,5'],
['200,21', '', '0,3', '2'],
['30000', '4,5', '1', '']]
transposed = list(zip(*mylist))
sizes = [[(x.index(",") if "," in x else len(x), len(x) - x.index(",") if "," in x else 0)
for x in l] for l in transposed]
maxima = [(max([x[0] for x in l]), max([x[1] for x in l])) for l in sizes]
withspaces = [
[' ' * (maxima[i][0] - sizes[i][j][0]) + number + ' ' * (maxima[i][1] - sizes[i][j][1])
for j, number in enumerate(l)] for i, l in enumerate(transposed)]
result = list(zip(*withspaces))
Printing the result in python3:
>>> print(*result, sep='\n')
(' 10 ', ' 3 ', ' ', ' ')
(' 200 ', '4000,222', '3 ', '1,5')
(' 200,21', ' ', '0,3', '2 ')
('30000 ', ' 4,5 ', '1 ', ' ')

Reversing list with strings in python

I have this list,
last_names = [
'Hag ', 'Hag ', 'Basmestad ', 'Grimlavaag ', 'Kleivesund ',
'Fintenes ', 'Svalesand ', 'Molteby ', 'Hegesen ']
and I want to print i reversed, so 'Hegesen' comes first, then ' Molteby' and at the end 'Hag'.
I have tried last_names.reverse(), but that returnes None..
Any help?

.reverse returns None because it reverses in-place:
>>> last_names = [
... 'Hag ', 'Hag ', 'Basmestad ', 'Grimlavaag ', 'Kleivesund ',
... 'Fintenes ', 'Svalesand ', 'Molteby ', 'Hegesen ']
>>> last_names.reverse()
>>> last_names
['Hegesen ', 'Molteby ', 'Svalesand ', 'Fintenes ', 'Kleivesund ', 'Grimlavaag ', 'Basmestad ', 'Hag ', 'Hag ']
To do this in an expression, do last_names[::-1].

As stated before, .reverse reverses the list in place, a more pythonic way to reverse a list and return it, is to use reversed:
>>> list(reversed([1,2,3]))
[3, 2, 1]

Finding the index and sum in a list of lists

Hello I have what I hope is an easy problem to solve. I am attempting to read a csv file and write a portion into a list. I need to determine the index and the value in each row and then summarize.
so the row will have 32 values...each value is a classification (class 0, class 1, etc.) with a number associated with it. I need a pythonic solution to make this work.
import os,sys,csv
csvfile=sys.argv[1]
f=open(csvfile,'rt')
reader=csv.reader(f)
classes=[]
for row in reader:
classes.append(row[60:92])
f.close()
classes = [' ', '1234', '645', '9897'], [' ', '76541', ' ', '8888']
how would i extract the index values from each list to get a sum for each?
for example: 0=(' ', ' ') 1=('1234', '76541') 2= ('645', ' ') 3= ('9897', '8888')
then find the sum of each
class 0 = 0
class 1 = 77775
class 2 = 645
class3 = 18785
Any assistance would be greatly appreciated

I find your use case a bit difficult to understand, but does this list comprehension give you some new ideas about how to solve your problem?
>>> classes = [' ', '1234', '645', '9897'], [' ', '76541', ' ', '8888']
>>> [sum(int(n) for n in x if n != ' ') for x in zip(*classes)]
[0, 77775, 645, 18785]

>>> classes = [[' ', '1234', '645', '9897'], [' ', '76541', ' ', '8888']]
>>> my_int = lambda s: int(s) if s.isdigit() else 0
>>> class_groups = dict(zip(range(32), zip(*classes)))
>>> class_groups[1]
('1234', '76541')
>>> class_sums = {}
>>> for class_ in class_groups:
... group_sum = sum(map(my_int, class_groups[class_]))
... class_sums[class_] = group_sum
...
>>> class_sums[1]
77775
>>> class_sums[3]
18785
>>>

You could sum as you go through the CSV file rows. (e.g. put the for class_ loop inside your rows loop) .. :
>>> classes
[[' ', '1234', '645', '9897'], [' ', '76541', ' ', '8888']]
>>> sums = {}
>>> for row in classes:
... for class_, num in enumerate(row):
... try:
... num = int(num)
... except ValueError:
... num = 0
... sums[class_] = sums.get(class_, 0) + num
...
>>> sums
{0: 0, 1: 77775, 2: 645, 3: 18785}