如何在python上的列表中计算列表元素

Question

I sorted lists within list on python. but I need to count list elements too. following list:

fruit = [
    ['Apple', 'S+'], ['Apple', 'S+'], ['Apple', 'B+'],
    ['Grape', 'B+'], ['Grape', 'C+']
]

result:

{'Apple':{'total':3, 'S+':2, 'B+':1}, 'Grape':{'total':2, 'B+':1, 'C+':1}}

I got above result through several for and while. but I want simple way. Is there beautiful and simple way to get result above thing ?

Answer 1

itertools.groupby is fun.

>>> result = {}
>>> for k, v in groupby(fruit,lambda x:x[0]):
...     value = list(v)
...     result[k] = {'total':len(value)}
...     for i,j in groupby(value, lambda x:x[1]):
...         result[k].update({i:len(list(j))})

Output:

{'Grape': {'total': 2, 'C+': 1, 'B+': 1}, 'Apple': {'total': 3, 'S+': 2, 'B+': 1}}

NB

Though, not needed here, it is always wise to sort the collection before applying groupby. For this example:

fruit = sorted(fruit, key= lambda x:(x[0],x[1]))

Answer 2

Something approaching what you want, using collections.defaultdict and collections.Counter .

I tried to make it as pythonic as possible.

import collections

fruit = [
    ['Apple', 'S+'], ['Apple', 'S+'], ['Apple', 'B+'],
    ['Grape', 'B+'], ['Grape', 'C+']
]


d = collections.defaultdict(lambda : [collections.Counter(),0])

for k,v in fruit:
    d[k][0][v]+=1
    d[k][1]+=1

print(dict(d))  # convert to dict for readability when printing

result:

{'Grape': [Counter({'B+': 1, 'C+': 1}), 2], 'Apple': [Counter({'S+': 2, 'B+': 1}), 3]}

details:

• create a dictionary that defaults to creating a 2-element list when key doesn't exist. This element list is made of a collections.Counter object and an integer (for global count)
• loop on the "tuples", and count elements and total.

Answer 3

unique, counts = numpy.unique(fruits, return_counts=True)

在numpy 1.9.0 return_counts添加到唯一