python - Pythonic way to calculate the mean and variance of values in Counters -

i'm wondering whether there pythonic way compute means , variances of counters?

for example, have 4 counters sharing same keys:

a = counter({1: 23, 2: 39, 3: 1}) b = counter({1: 28, 2: 39, 3: 1}) c = counter({1: 23, 2: 39, 3: 2}) d = counter({1: 23, 2: 22, 3: 1}) 

my way is:

each_key_val = {}  in a.keys():  # assumption here counters must share same keys     j in [a, b, c, d]:         try:             each_key_val[i].append(j[i])                except:             each_key_val[i] = [j[i]] 

i use following code find mean / variance each key:

 np.mean(each_key_val[i])  np.var(each_key_val[i]) 

is there easier way compute mean / variance each key compared way?

it's not think following more readable have, uses list comprehensions.

say have

cs = (a, b, c, d) 

then dictionary of mean can found with

m = {k: float(d) / len(cs) k, d in sum(cs).iteritems()} 

for variance, note that, definition of variance v[x] = e[x²] - (e[x])², so, if define:

p = sum([counter({k: ((float(d**2) / len(cs))) (k, d) in cn.iteritems()}) \      cn in cs]) 

then variance dictionary is

{k: p[k] - m[k]**2 k in m}