Average of all rows matching all unique rows

Question

I have a numpy array with two columns:

A = [[1,1,1,2,3,1,2,3],[0.1,0.2,0.2,0.1,0.3,0.2,0.2,0.1]]

for all uniques in the first column, I want the average to match it. for example

B = [[1,2,3], [0.175, 0.15, 0.2]]

Is there a pythonic way to do this?

Answer 1

I think the following standard numpy method for this kind of computation. A call np.uniquemay be missed if the records A[0]are small integers, but make the whole operation more reliable and independent of the actual data.

>>> A = [[1,1,1,2,3,1,2,3],[0.1,0.2,0.2,0.1,0.3,0.2,0.2,0.1]]
>>> unq, unq_idx = np.unique(A[0], return_inverse=True)
>>> unq_sum = np.bincount(unq_idx, weights=A[1])
>>> unq_counts = np.bincount(unq_idx)
>>> unq_avg = unq_sum / unq_counts
>>> unq
array([1, 2, 3])
>>> unq_avg
array([ 0.175,  0.15 ,  0.2  ])

Of course, you can compose both arrays, although this will convert unqto float dtype:

>>> np.vstack((unq, unq_avg))
array([[ 1.   ,  2.   ,  3.   ],
       [ 0.175,  0.15 ,  0.2  ]])

Answer 2

One possible solution:

In [37]: a=np.array([[1,1,1,2,3,1,2,3],[0.1,0.2,0.2,0.1,0.3,0.2,0.2,0.1]])
In [38]: np.array([list(set(a[0])), [np.average(np.compress(a[0]==i, a[1])) for i in set(a[0])]])
Out[38]:
array([[ 1.   ,  2.   ,  3.  ],
       [ 0.175,  0.15 ,  0.2 ]]) 

Answer 3

, , np.histogram, A[1], A[1], .

:

import numpy as np

A = np.array([[1,1,1,2,3,1,2,3],[0.1,0.2,0.2,0.1,0.3,0.2,0.2,0.1]])

# NB for n unique values in A[0] we want (n + 1) bin edges, such that
# A[0].max() < bin_edges[-1]
bin_edges = np.arange(A[0].min(), A[0].max()+2, dtype=np.int)

# the `weights` parameter means that the count for each bin is weighted
# by the corresponding value in A[1]
weighted_sums,_ = np.histogram(A[0], bins=bin_edges, weights=A[1])

# by calling `np.histogram` again without the `weights` parameter, we get
# the total number of occurrences of each unique index
index_counts,_ = np.histogram(A[0], bins=bin_edges)

# now just divide the weighted sums by the total occurrences
urow_avg = weighted_sums / index_counts

print urow_avg
# [ 0.175  0.15   0.2  ]

Answer 4

, reduceat:

A=np.array(zip(*[[1,1,1,2,3,1,2,3],[0.1,0.2,0.2,0.1,0.3,0.2,0.2,0.1]]),
        dtype=[('id','int64'),('value','float64')])
A.sort(order='id')
unique_ids,idxs = np.unique(A['id'],return_index=True)
avgs = np.add.reduceat(A['value'],idxs)
#divide by the number of samples to obtain the actual averages.
avgs[:-1]/=np.diff(idxs)
avgs[-1]/=A.size-idxs[-1]

Answer 5

:

values = {}

# get all values for each index
for index, value in zip(*A):
    if index not in values:
        values[index] = []
    values[index].append(value)

# create average for each index
for index in values:
    values[index] = sum(values[index]) / float(len(values[index]))

B = np.array(zip(*values.items()))

:

>>> B
array([[ 1.   ,  2.   ,  3.   ],
       [ 0.175,  0.15 ,  0.2  ]])

collections.defaultdict:

from collections import defaultdict

values = defaultdict(list)

for index, value in zip(*A):
    values[index].append(value)