파이썬 그룹

파이썬 그룹

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인덱스 0 이 값이고 인덱스 1 이 유형 인 데이터 쌍 세트가 있다고 가정하십시오 .

input = [
          ('11013331', 'KAT'), 
          ('9085267',  'NOT'), 
          ('5238761',  'ETH'), 
          ('5349618',  'ETH'), 
          ('11788544', 'NOT'), 
          ('962142',   'ETH'), 
          ('7795297',  'ETH'), 
          ('7341464',  'ETH'), 
          ('9843236',  'KAT'), 
          ('5594916',  'ETH'), 
          ('1550003',  'ETH')
        ]

다음과 같이 유형별로 (첫 번째 색인 문자열로) 그룹화하고 싶습니다.

result = [ 
           { 
             type:'KAT', 
             items: ['11013331', '9843236'] 
           },
           {
             type:'NOT', 
             items: ['9085267', '11788544'] 
           },
           {
             type:'ETH', 
             items: ['5238761', '962142', '7795297', '7341464', '5594916', '1550003'] 
           }
         ] 

효율적인 방법으로이를 달성하려면 어떻게해야합니까?

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2 단계로 수행하십시오. 먼저 사전을 만듭니다.

>>> input = [('11013331', 'KAT'), ('9085267', 'NOT'), ('5238761', 'ETH'), ('5349618', 'ETH'), ('11788544', 'NOT'), ('962142', 'ETH'), ('7795297', 'ETH'), ('7341464', 'ETH'), ('9843236', 'KAT'), ('5594916', 'ETH'), ('1550003', 'ETH')]
>>> from collections import defaultdict
>>> res = defaultdict(list)
>>> for v, k in input: res[k].append(v)
...

그런 다음 해당 사전을 예상 형식으로 변환하십시오.

>>> [{'type':k, 'items':v} for k,v in res.items()]
[{'items': ['9085267', '11788544'], 'type': 'NOT'}, {'items': ['5238761', '5349618', '962142', '7795297', '7341464', '5594916', '1550003'], 'type': 'ETH'}, {'items': ['11013331', '9843236'], 'type': 'KAT'}]

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itertools.groupby에서도 가능하지만 입력을 먼저 정렬해야합니다.

>>> sorted_input = sorted(input, key=itemgetter(1))
>>> groups = groupby(sorted_input, key=itemgetter(1))
>>> [{'type':k, 'items':[x[0] for x in v]} for k, v in groups]
[{'items': ['5238761', '5349618', '962142', '7795297', '7341464', '5594916', '1550003'], 'type': 'ETH'}, {'items': ['11013331', '9843236'], 'type': 'KAT'}, {'items': ['9085267', '11788544'], 'type': 'NOT'}]

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이 두 가지 모두 원래 키 순서를 따르지 않습니다. 주문을 유지하려면 OrderedDict가 필요합니다.

>>> from collections import OrderedDict
>>> res = OrderedDict()
>>> for v, k in input:
...   if k in res: res[k].append(v)
...   else: res[k] = [v]
... 
>>> [{'type':k, 'items':v} for k,v in res.items()]
[{'items': ['11013331', '9843236'], 'type': 'KAT'}, {'items': ['9085267', '11788544'], 'type': 'NOT'}, {'items': ['5238761', '5349618', '962142', '7795297', '7341464', '5594916', '1550003'], 'type': 'ETH'}]

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파이썬의 내장 itertools모듈에는 실제로 groupby함수가 있지만 그룹화 할 요소는 목록에서 연속되도록 그룹화 할 요소를 먼저 정렬해야합니다.

from operator import itemgetter
sortkeyfn = itemgetter(1)
input = [('11013331', 'KAT'), ('9085267', 'NOT'), ('5238761', 'ETH'), 
 ('5349618', 'ETH'), ('11788544', 'NOT'), ('962142', 'ETH'), ('7795297', 'ETH'), 
 ('7341464', 'ETH'), ('9843236', 'KAT'), ('5594916', 'ETH'), ('1550003', 'ETH')] 
input.sort(key=sortkeyfn)

이제 입력은 다음과 같습니다.

[('5238761', 'ETH'), ('5349618', 'ETH'), ('962142', 'ETH'), ('7795297', 'ETH'),
 ('7341464', 'ETH'), ('5594916', 'ETH'), ('1550003', 'ETH'), ('11013331', 'KAT'),
 ('9843236', 'KAT'), ('9085267', 'NOT'), ('11788544', 'NOT')]

groupby returns a sequence of 2-tuples, of the form (key, values_iterator). What we want is to turn this into a list of dicts where the 'type' is the key, and 'items' is a list of the 0'th elements of the tuples returned by the values_iterator. Like this:

from itertools import groupby
result = []
for key,valuesiter in groupby(input, key=sortkeyfn):
    result.append(dict(type=key, items=list(v[0] for v in valuesiter)))

Now result contains your desired dict, as stated in your question.

You might consider, though, just making a single dict out of this, keyed by type, and each value containing the list of values. In your current form, to find the values for a particular type, you'll have to iterate over the list to find the dict containing the matching 'type' key, and then get the 'items' element from it. If you use a single dict instead of a list of 1-item dicts, you can find the items for a particular type with a single keyed lookup into the master dict. Using groupby, this would look like:

result = {}
for key,valuesiter in groupby(input, key=sortkeyfn):
    result[key] = list(v[0] for v in valuesiter)

result now contains this dict (this is similar to the intermediate res defaultdict in @KennyTM's answer):

{'NOT': ['9085267', '11788544'], 
 'ETH': ['5238761', '5349618', '962142', '7795297', '7341464', '5594916', '1550003'], 
 'KAT': ['11013331', '9843236']}

(If you want to reduce this to a one-liner, you can:

result = dict((key,list(v[0] for v in valuesiter)
              for key,valuesiter in groupby(input, key=sortkeyfn))

or using the newfangled dict-comprehension form:

result = {key:list(v[0] for v in valuesiter)
              for key,valuesiter in groupby(input, key=sortkeyfn)}

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The following function will quickly (no sorting required) group tuples of any length by a key having any index:

# given a sequence of tuples like [(3,'c',6),(7,'a',2),(88,'c',4),(45,'a',0)],
# returns a dict grouping tuples by idx-th element - with idx=1 we have:
# if merge is True {'c':(3,6,88,4),     'a':(7,2,45,0)}
# if merge is False {'c':((3,6),(88,4)), 'a':((7,2),(45,0))}
def group_by(seqs,idx=0,merge=True):
    d = dict()
    for seq in seqs:
        k = seq[idx]
        v = d.get(k,tuple()) + (seq[:idx]+seq[idx+1:] if merge else (seq[:idx]+seq[idx+1:],))
        d.update({k:v})
    return d

In the case of your question, the index of key you want to group by is 1, therefore:

group_by(input,1)

gives

{'ETH': ('5238761','5349618','962142','7795297','7341464','5594916','1550003'),
 'KAT': ('11013331', '9843236'),
 'NOT': ('9085267', '11788544')}

which is not exactly the output you asked for, but might as well suit your needs.

---

I also liked pandas simple grouping. it's powerful, simple and most adequate for large data set

result = pandas.DataFrame(input).groupby(1).groups

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result = []
# Make a set of your "types":
input_set = set([tpl[1] for tpl in input])
>>> set(['ETH', 'KAT', 'NOT'])
# Iterate over the input_set
for type_ in input_set:
    # a dict to gather things:
    D = {}
    # filter all tuples from your input with the same type as type_
    tuples = filter(lambda tpl: tpl[1] == type_, input)
    # write them in the D:
    D["type"] = type_
    D["itmes"] = [tpl[0] for tpl in tuples]
    # append D to results:
    result.append(D)

result
>>> [{'itmes': ['9085267', '11788544'], 'type': 'NOT'}, {'itmes': ['5238761', '5349618', '962142', '7795297', '7341464', '5594916', '1550003'], 'type': 'ETH'}, {'itmes': ['11013331', '9843236'], 'type': 'KAT'}]

참고URL : https://stackoverflow.com/questions/3749512/python-group-by