how to make a variable change from the text "1m" into "1000000" in python

Question

I have variables with values like 1.7m 1.8k and 1.2b how can I convert them to a real number value for example

1.7m = 1700000
1.8k = 1800
1.2b = 1200000000

-1: No Sample Code for stuff you've already tried. – S.Lott Mar 15 '10 at 19:36 — S.Lott, Mar 15 '10 at 19:36

score 11 · Answer 1 · edited Mar 15 '10 at 19:43

11

I would define a dictionary:

tens = dict(k=10e3, m=10e6, b=10e9)

then

x='1.7m'
factor, exp = x[0:-1], x[-1].lower()
ans = int(float(factor) * tens[exp])

edited Mar 15 '10 at 19:43

S.Lott

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answered Mar 15 '10 at 19:33

Vicki Laidler

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+1: Allows someone to split the hair of `m` meaning 1e6 or 1024*1024. – S.Lott Mar 15 '10 at 19:37
The dictionary approach is problematic due to where I intend to implement this. but thanks. I'll update what I went with in a little bit – Pevo Mar 15 '10 at 19:56
2

"problematic"? That's meaningless. Could you explain why a simple solution won't work? – S.Lott Mar 15 '10 at 21:00
@Pevo, The usage of a dictionary here is exactly the same as the usage in the accepted answer. – Mike Graham Mar 16 '10 at 02:43
1

Dude isn't the python 10e3 == 10000?? Shouldn't it be 1e3, 1e6 .. ?? – Mazeryt Mar 25 '20 at 20:49

score 1 · Answer 2 · answered Mar 15 '10 at 22:58

1

You might be interested in a units library like quantities or unum.

answered Mar 15 '10 at 22:58

Mike Graham

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score 1 · Answer 3 · answered Mar 16 '10 at 14:38

1

Using lambda:

>>> tens = {'k': 10e3, 'm': 10e6, 'b': 10e9}
>>> f = lambda x: int(float(x[:-1])*tens[x[-1]])
>>> f('1.7m')
17000000
>>> f('1.8k')
18000
>>> f('1.2b')
12000000000

answered Mar 16 '10 at 14:38

riza

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score 0 · Accepted Answer · answered Mar 15 '10 at 19:46

0

Here is an example using re:

input = '17k, 14.05m, 1.235b'

multipliers = { 'k': 1e3,
                'm': 1e6,
                'b': 1e9,
              }

pattern = r'([0-9.]+)([bkm])'

for number, suffix in re.findall(pattern, input):
    number = float(number)
    print number * multipliers[suffix]

answered Mar 15 '10 at 19:46

Joe Koberg

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Or even `pattern = r'([0-9.]+)(%s)' % '|'.join(multipliers)` to avoid the minor repetition of units in there. – Will McCutchen Mar 15 '10 at 20:05
@Will, being that `[bkm]` will only match one character, I don't see how there could be a "minor repetition of units in there". I do like your suggestion of doing a join on multipliers, but not for the reason you stated. – tgray Mar 15 '10 at 21:25
1

He is saying you can use the already-specified dictionary keys to fill out the RE rather than retyping 'bkm'. – Joe Koberg Mar 15 '10 at 21:29
But i considered that less easy to understand in this simple case. If the dict had all the SI prefixes in it or somesuch, I would definitely have done that. – Joe Koberg Mar 15 '10 at 21:30
Using regex here is quite unnecessary. – Mike Graham Mar 15 '10 at 23:08
Mike: Please feel free to provide a simpler or more effective answer. – Joe Koberg Mar 15 '10 at 23:39
@Joe, Vicki Laidler's answer uses the same technique as yours except that it grabs the last character directly rather than based on matching a regular expression. I suspect most people would find that solution more readable; I know I did. – Mike Graham Mar 16 '10 at 02:42
@Mike: There are more than one way to solve a problem, and I've learned that it's good to adopt the unix philosophy of being forgiving in your input and precise in output.... This example shows that the regex can pull out the formatted numbers even with noise characters such as ' ' and ',', and even extracts many such numbers from the input with very little additional work. The other solution solves the problem but is more "brittle". – Joe Koberg Mar 16 '10 at 15:34

how to make a variable change from the text "1m" into "1000000" in python

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