9

在 numpy 中是否有另一种方法来实现 scipy.stats.mode 函数以沿轴获取 ndarrays 中最频繁的值？（不导入其他模块）即

import numpy as np
from scipy.stats import mode

a = np.array([[[ 0,  1,  2,  3,  4],
                  [ 5,  6,  7,  8,  9],
                  [10, 11, 12, 13, 14],
                  [15, 16, 17, 18, 19]],

                 [[ 0,  1,  2,  3,  4],
                  [ 5,  6,  7,  8,  9],
                  [10, 11, 12, 13, 14],
                  [15, 16, 17, 18, 19]],

                 [[40, 40, 42, 43, 44],
                  [45, 46, 47, 48, 49],
                  [50, 51, 52, 53, 54],
                  [55, 56, 57, 58, 59]]])

mode= mode(data, axis=0)
mode = mode[0]
print mode
>>>[ 0,  1,  2,  3,  4],
   [ 5,  6,  7,  8,  9],
   [10, 11, 12, 13, 14],
   [15, 16, 17, 18, 19]
4

2 回答 2

14

该scipy.stats.mode函数是用这段代码定义的，它只依赖于numpy：

def mode(a, axis=0):
    scores = np.unique(np.ravel(a))       # get ALL unique values
    testshape = list(a.shape)
    testshape[axis] = 1
    oldmostfreq = np.zeros(testshape)
    oldcounts = np.zeros(testshape)

    for score in scores:
        template = (a == score)
        counts = np.expand_dims(np.sum(template, axis),axis)
        mostfrequent = np.where(counts > oldcounts, score, oldmostfreq)
        oldcounts = np.maximum(counts, oldcounts)
        oldmostfreq = mostfrequent

    return mostfrequent, oldcounts

来源：https ://github.com/scipy/scipy/blob/master/scipy/stats/stats.py#L609

于 2012-09-13T03:48:44.537 回答
1

如果您知道没有很多不同的值（相对于输入“itemArray”的大小），这样的事情可能会很有效：

uniqueValues = np.unique(itemArray).tolist()
uniqueCounts = [len(np.nonzero(itemArray == uv)[0])
                for uv in uniqueValues]

modeIdx = uniqueCounts.index(max(uniqueCounts))
mode = itemArray[modeIdx]

# All counts as a map
valueToCountMap = dict(zip(uniqueValues, uniqueCounts))
于 2012-12-24T17:51:33.080 回答