Python 2.7：在多值字典的列表之间的大小差异中添加零并将其求和

Question

I have a multi-valued dictionary and I want to sum whatever the values are there which can be done with

for key, value in nai_dictionary.items():
    nai_sum_rate.setdefault('%s' %(key), []).append(np.sum(rate_nai[key], axis=0))

for any nai_sum_rate dictionary, as suggested in this webpage . However, often times the length of the value lists is not the same.

This is the test data on which I'm testing right now

time = np.array([1,2,3,4,5,6])
test_dict['a'] = np.array([1,2,3,4,5,6]), np.array([1,2,3,4]), np.array([1,2,3,4,5,6])

What I want to do now, is to put a 0 in place of missing values for (in this case) the second array where there is no value against the time variable values and add them together.

In my previous testings, before the nai_sum_rate was a list instead of a dictionary I used the answer provided in this link to sort out the problem with the reference variable time . I've been trying with the dictionary but to no avail.

The expected sum is

3, 6, 9, 12, 10, 12

Answer 1

I solved the problem myself, here is the code that I came up with

for key, value in nai_dictionary.items():
    i=0
    while i < len(rate_nai[key]):
        for key, value in nai_dictionary.items():
            for index, values in enumerate(rate_nai[key]):
                if len(time) < len(values):
                    diff = len(values) - len(time)
                    rate_nai[values] = values[0:np.array(values).size-diff]
                elif len(time) > len(values):
                    diff = len(time) - len(values)
                    rate_nai['%s' %(key)].pop(index)
                    rate_nai.setdefault(
                        '%s' %(key)).append(np.append(np.array(values), np.zeros(diff)))
        i+=1

It can tackle the problem with any number of values that have a lesser length than the reference variable time .