如何将具有键作为元组的python字典中的值相加

Question

I have the following dictionary with key as tuple

D_grpTagReport = { ('Tag1', '1'):345.56 ,
                   ('Tag1', '2'):45.6 ,
                   ('Tag2', '3'):3.1 ,
                   ('Tag3', '1'):21.56 ,
                   ('Tag2', '3'):1.56 ,
                 }

I would like to get the sum of values for each unique Tag. Is there any built in utility that can be used to process this dictionary.

Result Example:

Tag1 : 391.16     # (total of all Tag1)
Tag2 : 4.66       # (total of all Tag2) 
Tag3 : 21.56      # (total of all Tag3) 

Answer 1

You could use a defaultdict :

>>> from collections import defaultdict
>>> D_grpTagReport = { ('Tag1', '1'):345.56 ,
...                    ('Tag1', '2'):45.6 ,
...                    ('Tag2', '3'):3.1 ,
...                    ('Tag3', '1'):21.56 ,
...                    ('Tag2', '3'):1.56 ,
...                  }
>>> result = defaultdict(int)
>>> for t in D_grpTagReport:
...     result[t[0]] += D_grpTagReport[t]
...
>>> result
defaultdict(<class 'int'>, {'Tag3': 21.56, 'Tag1': 391.16, 'Tag2': 1.56})

Answer 2

Some good answers already, but here's a different (simpler IMO) approach.

d = {('A', '1'): 2, ('A', '2'): 5, ('B', '1'): 1}
keys={k[0]:0 for k in d.keys()}
for key in d:
    keys[key[0]] = keys[key[0]] + d[key]
print(keys)  # {'A': 7, 'B': 1}

Once more with the data from the OP:

d = { ('Tag1', '1'):345.56 ,
      ('Tag1', '2'):45.6 ,
      ('Tag2', '3'):3.1 ,
      ('Tag3', '1'):21.56 ,
      ('Tag2', '3'):1.56 ,
    }
keys={k[0]:0 for k in d.keys()}
for key in d:
    keys[key[0]] = keys[key[0]] + d[key]
print(keys)  # {'Tag1': 391.16, 'Tag2': 1.56, 'Tag3': 21.56}

Answer 3

You can use itertools.groupby for this , first take .items() for the dictionary and then sort this list based on the first element of the key and the use itertools.groupby on it with key as the first element of the key (first element of first element of the tuple of key,value ). Then for each group you can take sum() over the values to get what you want. Example -

from itertools import groupby
for key,group in groupby(sorted(D_grpTagReport.items()),key=lambda x:x[0][0]):
    total = sum(g[1] for g in group)
    print(key,':',total)

Please note you result is a bit wrong, and so is your dictionary, since dictionary cannot have same key with multiple values.

Demo -

>>> D_grpTagReport = { ('Tag1', '1'):345.56 ,
...                    ('Tag1', '2'):45.6 ,
...                    ('Tag2', '3'):3.1 ,
...                    ('Tag3', '1'):21.56 ,
...                    ('Tag2', '3'):1.56 ,
...                  }
>>>
>>>
>>> from itertools import groupby
>>> for key,group in groupby(sorted(D_grpTagReport.items()),key=lambda x:x[0][0]):
...     total = sum(g[1] for g in group)
...     print(key,':',total)
...
Tag1 : 391.16
Tag2 : 1.56
Tag3 : 21.56