使用for循环汇总python中列表中的元素

Question

While working on a program that creates bars out of sets of numbers in a list, I found that adding up items in my list doesn't work. I thought the best way to do this is just to make a for loop.

Here's my list:

phonelist = [[12,14,16,17,18],[16,23,54,64,32]]

And then I try to add this up with a for loop

numphone = 0
for x in len(phonelist[0]):
    numphone = numphone + x

Yet I get this error:

TypeError: 'int' object is not iterable

What should I do?

Answer 1

>>> phonelist = [[12,14,16,17,18],[16,23,54,64,32]]
>>> [sum(li) for li in phonelist]
[77, 189]
>>> sum([sum(li) for li in phonelist])
266

or:

>>> sum(sum(li) for li in phonelist)    # generator expression...
266

If you are trying to create individual categories, you can use a dict:

data={'Bar {}'.format(i):sum(li) for i, li in enumerate(phonelist, 1)}
data['Total']=sum(data.values())

print data
{'Bar 2': 189, 'Bar 1': 77, 'Total': 266}

Then if you want to produce a simple bar graph:

for bar in sorted(data.keys()):
    print '{}: {}'.format(bar, int(round(25.0*data[bar]/data['Total']))*'*')

Prints:

Bar 1: *******
Bar 2: ******************
Total: *************************

Answer 2

len(phonelist[0]) is an int, so you can't loop over it. You can change that to

for x in phonelist[0]:

This way x will take on each value of phonelist[0] .

Answer 3

numphone = 0
for x in phonelist[0]:
    numphone = numphone + x

This should work. You iterate overt the list, not over the length of it, since the length is an integer and iterating over an integer doesn't make sense

Answer 4

The best solution to this is to use the sum() built-in function . One method, as given in other answers to this question, is to sum each sumlist, then sum those subtotals.

However, a better solution is to flatten the sublists - this is best achieved with itertools.chain.from_iterable() .

sum(itertools.chain.from_iterable(phonelist))

This is an optimal solution as it means that you do not have to perform a lot of separate sum operations, or create an intermediate list. It will also work lazily if your data is a lazy source, making your code more memory-efficient and spreading the processing load.