Python比较列表到Dict值

Question

Updated in an attempt to be more clear

I have three list of dictionaries that I want to merge into one based on a value.

The lists looks like this. They vary in how many dictionaries that they can have.

unplanned = [{'service__name': u'Email', 'service_sum': 4}, {'service__name': u'Peoplesoft', 'service_sum': 2}]
planned = [{'service__name': u'Email', 'service_sum': 2}, {'service__name': u'Gopher', 'service_sum': 2}, {'service__name': u'Peoplesoft', 'service_sum': 4}]
emerg = [{'service__name': u'Internet', 'service_sum': 1}]

I want to take the 3 lists and and create a new list that has the name's from all 3 lists and the values or 0 in a set order. So I am thinking something like this.

[(Email, (4, 2, 0)), (Peoplesoft, (2, 4, 0)), Gopher, (0, 2, 0)), Internet, (0, 0, 1))]

I thought I should create a list of the service__name's to compare against each list so I did that but I am not sure how to compare the 3 lists against this name list. I thought izip_longest would work but have no idea how to implement it. I am using 2.7.

Answer 1

Just use a dict, then convert it into a list afterwards:

some_list = [{'service__name': u'Email', 'service_sum': 4}, {'service__name': u'Email', 'service_sum': 1}, {'service__name': u'Network', 'service_sum': 0}]

def combine(list):
   combined = {}
   for item in list:
      if item['service__name'] not in combined:
         combined[item['service__name']] = []
      combined[item['service__name']].append(item['service_sum'])
   return combined.items()

combine(some_list)  # [(u'Email', [4, 1]), (u'Network', [0])]
combine(unplanned)
combine(emerg + planned)
.....

Here's the version of the function that uses defaultdict:

def combine(list):
   from collections import defaultdict
   combined = defaultdict(list)
   for item in list:
      combined[item['service__name']].append(item['service_sum'])
   return combined.items()

A little cleaner, but there's an unnecessary import, and a few other problems with it that may pop up in the future if the function definition is changed (see comments).

Answer 2

It seems like you could do something like:

output = []
for dicts in zip(unplanned,planned,emerg):
    output.append(('Email',tuple(d['service_sum'] if d['service__name'] == 'Email' else 0 for d in dicts)))

Answer 3

Try the following codes. You can give variables better name since you know better about the contexts.

def convert(unplanned, planned, emerg):
    chain = (unplanned, planned, emerg)
    names = map(lambda lst: [d['service__name'] for d in lst], chain)
    sums = map(lambda lst: [d['service_sum'] for d in lst], chain)
    ds = [dict(zip(n, s)) for n,s in zip(names, sums)]
    unique_names = set([])
    unique_names = reduce(unique_names.union,names)
    results = []
    for n in unique_names:
        s = []
        for i in range(3):
            s.append(ds[i].get(n,0))
        results.append((n, tuple(s)))

    return results

print convert(unplanned, planned, emerg)

The output at my machine is

[(u'Internet', (0, 0, 1)), (u'Peoplesoft', (2, 4, 0)), (u'Email', (4, 2, 0)), (u'Gopher', (0, 2, 0))]