计算一个密钥和一个python（3.2）字典的多少值

Question

I am sure this is silly, but I simply cannot get around it. I have a dictionary, like this, with unequal number of values for each key:

'John greased ': ['axle', 'wheel', 'wheels', 'wheel', 'engine', ''], 
'Paul alleged ': ['truth', 'crime', 'facts', 'infidelity', 'incident', ''], 
'Tracy freed ': ['animals', 'fish', 'slaves', 'slaves', 'slaves', 'pizza'], 
'Lisa plowed ': ['field', 'field', '', '', '', ''],

I want to know how many values there are for each key, not each unique value but how many tokens there are per key, repeated or not. So I would have a result like:

John greased  5
Paul alleged  5
Tracy freed  6
Lisa plowed  2

I was trying to use this to work it out using the code bellow:

for key, value in sorted(result.items()):
         print(key, len(value)) 

But because of the missing values all the lengths turn out to be the same. Any ideas on how to solve this or where to find it out? Thanks a lot for any help.

Answer 1

One way to solve this, is by changing your last line:

print(key, len([item for item in value if item])) 

So your complete code:

ITEMS = {
    'John greased ': ['axle', 'wheel', 'wheels', 'wheel', 'engine', ''],
    'Paul alleged ': ['truth', 'crime', 'facts', 'infidelity', 'incident', ''],
    'Tracy freed ': ['animals', 'fish', 'slaves', 'slaves', 'slaves', 'pizza'],
    'Lisa plowed ': ['field', 'field', '', '', '', ''],
}

for key, value in ITEMS.items():
    #print value
    print(key, len([item for item in value if item]))

You can also use filter with bool :

print(key, len(filter(bool, value)))

So, the loop:

for key, value in ITEMS.items():
    #print value
    print(key, len(filter(bool, value)))

You need to apply list over filter like so print(key, len(list(filter(bool, value)))) in Python 3.

Answer 2

Use filter with None , it filters out all falsy values from the iterable passed to it.

In Python3 filter returns an iterator so you should call list() on it.:

>>> lis = ['field', 'field', '', '', '', '']
>>> list(filter(None, lis))
['field', 'field']
>>> len(list(filter(None, lis)))
2

Code:

>>> my_dict = {
    'John greased ': ['axle', 'wheel', 'wheels', 'wheel', 'engine', ''],
    'Paul alleged ': ['truth', 'crime', 'facts', 'infidelity', 'incident', ''],
    'Tracy freed ': ['animals', 'fish', 'slaves', 'slaves', 'slaves', 'pizza'],
    'Lisa plowed ': ['field', 'field', '', '', '', ''],
}
for k,v in my_dict.items():
    print (k, len(list(filter(None, v))))
...     
Paul alleged  5
Lisa plowed  2
John greased  5
Tracy freed  6

Timing comparision between filter(None,..) and list comprehension:

>>> lis = ['field', 'field', '', '', '', '']*100
>>> %timeit list(filter(None, lis))
10000 loops, best of 3: 22.2 us per loop
>>> %timeit [item for item in lis if item]
10000 loops, best of 3: 53.1 us per loop
>>> lis = ['field', 'field', '', '', '', '']*10000
>>> %timeit list(filter(None, lis))
100 loops, best of 3: 2.36 ms per loop
>>> %timeit [item for item in lis if item]
100 loops, best of 3: 5.22 ms per loop

Answer 3

Look at this:

>>> dct = {'John greased ': ['axle', 'wheel', 'wheels', 'wheel', 'engine', ''],
... 'Paul alleged ': ['truth', 'crime', 'facts', 'infidelity', 'incident', ''],
... 'Tracy freed ': ['animals', 'fish', 'slaves', 'slaves', 'slaves', 'pizza'],
... 'Lisa plowed ': ['field', 'field', '', '', '', '']}
>>>
>>> {k:sum(1 for x in v if x) for k,v in dct.items()}
{'Paul alleged ': 5, 'Lisa plowed ': 2, 'John greased ': 5, 'Tracy freed ': 6}
>>>
>>> for key,value in dct.items():
...     print(key, sum(1 for v in value if v))
...
Paul alleged  5
Lisa plowed  2
John greased  5
Tracy freed  6
>>>

Answer 4

data = {
    'John greased ': ['axle', 'wheel', 'wheels', 'wheel', 'engine', ''], 
    'Paul alleged ': ['truth', 'crime', 'facts', 'infidelity', 'incident', ''], 
    'Tracy freed ': ['animals', 'fish', 'slaves', 'slaves', 'slaves', 'pizza'], 
    'Lisa plowed ': ['field', 'field', '', '', '', '']
}

for each in data:
    i = 0
    print each
    for item in data[each]:
        if len(item) > 0:
            i =i +1
    print i