Question

我有一个有两列的numpy数组：

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]]

对于第一列中的所有唯一身份，我想要与其对应的值的平均值。例如

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

有没有pythonic的方法来做到这一点？

Answer 1

我认为以下是这种计算的标准numpy方法。如果np.unique的条目是小整数，则可以跳过对A[0]的调用，但它会使整个操作更加健壮并且与实际数据无关。

>>> 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  ])

当然你可以堆叠两个数组，虽然这会将unq转换为float dtype：

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

Answer 2

一种可能的解决方案是：

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

您可以使用np.histogram首先获取与A[1]中每个唯一索引对应的A[1]中的值的总和，然后使用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 ]的总出现次数来更有效地执行此操作每个独特的指数。

例如：

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Answer 4

另一个有效的numpy解决方案，使用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

您可以按如下方式处理：

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()))

对于你的例子，这给了我：

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

您可以使用collections.defaultdict轻微简化：

from collections import defaultdict

values = defaultdict(list)

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

与所有唯一行对应的所有行的平均值

5 个答案: