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问题:
Assume that I have a such set of pair datas where index 0 is the value and the index 1 is the type:
input = [ ('11013331', 'KAT'), ('9085267', 'NOT'), ('5238761', 'ETH'), ('5349618', 'ETH'), ('11788544', 'NOT'), ('962142', 'ETH'), ('7795297', 'ETH'), ('7341464', 'ETH'), ('9843236', 'KAT'), ('5594916', 'ETH'), ('1550003', 'ETH') ]
I want to group them by their type(by the 1st indexed string) as such:
result = [ { type:'KAT', items: ['11013331', '9843236'] }, { type:'NOT', items: ['9085267', '11788544'] }, { type:'ETH', items: ['5238761', '962142', '7795297', '7341464', '5594916', '1550003'] } ]
How can I achieve this in an efficient way?
Thanks
回答1:
Do it in 2 steps. First, create a dictionary.
>>> 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) ...
Then, convert that dictionary into the expected format.
>>> [{'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'}]
It is also possible with itertools.groupby but it requires the input to be sorted first.
>>> 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'}]
Note both of these do not respect the original order of the keys. You need an OrderedDict if you need to keep the order.
>>> 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'}]
回答2:
Python's built-in itertools module actually has a groupby function that you could use, but the elements to be grouped must first be sorted such that the elements to be grouped are contiguous in the list:
sortkeyfn = key=lambda s:s[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)
Now input looks like:
[('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)}
回答3:
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.
回答4:
I also liked pandas simple grouping. it's powerful, simple and most adequate for large data set
result = pandas.DataFrame(input).groupby(1).groups
