How do I get rid of "K" from a list of string which I want to be integer

Question

Not sure my title is clear enough. I have the following list :

[u'394', u'132.2k', u'6.8k', u'1', u'100', u'457', u'3.5k', u'3.5k', u'184', u'507', u'57']

I want to loop over this list, delete all the K, and when deleting them multiply the element by 1000. So I guess I need to convert each string as an interger before multiplying it.

I tried various recipe but I'm kind of lost. Here is what I've done until now :

  for item in stringlist:
    if 'k' in item:
      finallist.append(item)

however this doesn't work at all because it return me a list of k without any numbers. I also tried to use replace but couldn't make it work.

I know that's is really easy to convert strings into integers but that's not my main issue here...

First, write code that solves the problem for a single value in the list, then apply that code to each element. — Karl Knechtel, May 18 '23 at 23:27

asongtoruin · Accepted Answer · 2017-02-10T11:49:06.193

6

Try this:

start_list = [u'394', u'132.2k', u'6.8k', u'1', u'100', u'457', u'3.5k', u'3.5k', u'184', u'507', u'57']

end_list = [int(1000*float(x.replace('k', ''))) if 'k' in x else int(x) for x in start_list]

print(end_list)

prints:

[394, 132200, 6800, 1, 100, 457, 3500, 3500, 184, 507, 57]

EDIT: we can make this a little more robust, in case of empty elements, as so (spaced out for legibility):

start_list = [u'394', u'132.2k', u'6.8k', u'1', u'100', u'457', u'3.5k', u'3.5k', u'184', u'507', u'57', u'.', u'1.2']

end_list = [int(1000*float(x.replace('k', ''))) if 'k' in x
            else int(float('0' + x + '0')) if '.' in x
            else int(x)
            for x in start_list]

print(end_list)

by adding a '0' to either side of any element with a . (but no k) in it, we capture the empty case of u'.'. Note that this will round decimal values, with the output as follows:

[394, 132200, 6800, 1, 100, 457, 3500, 3500, 184, 507, 57, 0, 1]

edited Feb 10 '17 at 11:49

answered Feb 10 '17 at 11:07

asongtoruin

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I've tried on my real list of elements and I have this error, what does it means ? `invalid literal for int() with base 10: '.'` – Simon Breton Feb 10 '17 at 11:10
2

Does your list contain an element `u'.'`? – L3viathan Feb 10 '17 at 11:14
1

@SimonBreton Edited to try and catch possible issues. – asongtoruin Feb 10 '17 at 11:49
Thanks a lot. I'll keep working on this script anyway ! – Simon Breton Feb 10 '17 at 12:11

dnswlt · Answer 2 · 2017-02-10T11:39:34.697

You could use a regular expression to determine whether you have a regular number, or a "k" number (both "3k" and "3.0k" will yield 3000).

Update: as a bonus, this one also deals with "M" for Mega and "G" for Giga :-)

import re

ds = [u'394', u'132.2k', u'6.8k', u'7k', u'9.7G', u'18.1M']


factors = {'k': 1e3, 'M': 1e6, 'G': 1e9}
def f(s):
    mo = re.match(r'(\d+(?:\.\d+)?)([kMG])?', s)
    return int(float(mo.group(1)) * factors.get(mo.group(2), 1))

print([f(d) for d in ds])

How do I get rid of "K" from a list of string which I want to be integer

2 Answers2