CODEWITHSUNDEEP
pythonlist
user1881957
user1881957

Reputation: 3388

How to sum the following list?

I have a list like this:

a = ['1 MB', '2 MB']

I want to sum all the elements in a.

I figured I first have to create integer list and use sum() function. How can I do that?

Upvotes: 0

Views: 107

Answers (4)

sotapme
sotapme

Reputation: 4903

This would work for differing multipliers: 

import re
pattern = "(\d+)\s*(\w+)?"

a = ['1 MB', '2 MB', '3 K', '250' ]

multiplier = { None: 1, 'MB' : 1000000, 'K' : 1000 }

def multi(m):
    if m:
        return int(m.group(1)) * multiplier[m.group(2)]
    else:
        return 0

r = map(multi, [ re.search(pattern, x) for x in a])
print r

print sum(r)

With the following output:

[1000000, 2000000, 3000, 250]
3003250

Upvotes: 0

StoryTeller - Unslander Monica
StoryTeller - Unslander Monica

Reputation: 170279

Use the built-in split to separate the number from the rest of the string 

>>> a = ['1 MB', '2 MB']
>>> sum([int(s.split(' ')[0]) for s in a])
3

s.split(' ') creates the list ['1', 'MB']

Or use a generator expression to avoid the unneeded list that the list comprehension creates:

>>> a = ['1 MB', '2 MB']
>>> sum(int(s.split(' ')[0]) for s in a)
3

Upvotes: 2

Ashwini Chaudhary
Ashwini Chaudhary

Reputation: 251186

you can use regex here:

In [19]: a = ['1 MB', '2 MB']

In [20]: sum(int(re.search(r'\d+',x).group()) for x in a)
Out[20]: 3

where re.search(r'\d+') returns something like:

In [23]: [re.search(r'\d+',x).group() for x in a]
Out[23]: ['1', '2']

Upvotes: 0

Volatility
Volatility

Reputation: 32308

Assuming all of the elements end in ' MB', you can do this:

sum(map(int, (x[:-3] for x in a)))

Breakdown:

  • (x[:-3] for x in a) takes all but the last three characters of the string.
  • map(int, iterable) 'maps' the int function to each element of the iterable.
  • sum(iterable) simply sums the elements of the iterable.

Upvotes: 1

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