I'm wondering whether there is a Pythonic way to compute the means and variances of Counters?
For example, I have four Counters sharing the 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 to do that is:
each_key_val = {}
for i in a.keys(): # The assumption here is that all Counters must share the same keys
for j in [a, b, c, d]:
try:
each_key_val[i].append(j[i])
except:
each_key_val[i] = [j[i]]
I could use the following code to find the mean / variance for each key:
np.mean(each_key_val[i])
np.var(each_key_val[i])
Is there an easier way to compute the mean / variance for each key compared to my way?
It's not that I think the following is more readable than what you have, but it only uses list comprehensions.
Say you have
cs = (a, b, c, d)
Then a dictionary of the mean can be found with
m = {k: float(d) / len(cs) for k, d in sum(cs).iteritems()}
For the variance, note that, by the definition of variance V[X] = E[x2] - (E[X])2, so, if you define:
p = sum([Counter({k: ((float(d**2) / len(cs))) for (k, d) in cn.iteritems()}) \
for cn in cs])
then the variance dictionary is
{k: p[k] - m[k]**2 for k in m}
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