在Python的标准max函数中,我可以传入key参数:

s = numpy.array(['one','two','three'])
max(s) # 'two' (lexicographically last)
max(s, key=len) # 'three' (longest string)


对于较大的(多维)数组,我们不能再使用max,但是可以使用numpy.amax ...,不幸的是,它没有提供key参数。

t = numpy.array([['one','two','three'],
                 ['four','five','six']],
                dtype='object')
numpy.amax(t) # 'two` (max of the flat array)
numpy.amax(t, axis=1) # array([two, six], dtype=object) (max of first row, followed by max of second row)


我想要做的是:

amax2(t, key=len) # 'three'
amax2(t, key=len, axis=1) # array([three, four], dtype=object)


有内置的方法可以做到这一点吗?

注意:第一次尝试写此问题时,我找不到amax working in this toy example

最佳答案

这是一种非内置方式(使用out时缺少features of keepdimamaxkey参数),这似乎很长:

def amax2(x, *args, **kwargs):
    if 'key' not in kwargs:
        return numpy.amax(x,*args,**kwargs)
    else:
        key = kwargs.pop('key') # e.g. len, pop so no TypeError: unexpected keyword
        x_key = numpy.vectorize(key)(x) # apply key to x element-wise
        axis = kwargs.get('axis') # either None or axis is set in kwargs
        if len(args)>=2: # axis is set in args
            axis = args[1]

        # The following is kept verbose, but could be made more efficient/shorter
        if axis is None: # max of flattened
            max_flat_index = numpy.argmax(x_key, axis=axis)
            max_tuple_index = numpy.unravel_index(max_flat_index, x.shape)
            return x[max_tuple_index]
        elif axis == 0: # max in each column
            max_indices = numpy.argmax(x_key, axis=axis)
            return numpy.array(
                 [ x[max_i, i] # reorder for col
                     for i, max_i in enumerate(max_indices) ],
                 dtype=x.dtype)
        elif axis == 1: # max in each row
            max_indices = numpy.argmax(x_key, axis=axis)
            return numpy.array(
                 [ x[i, max_i]
                     for i, max_i in enumerate(max_indices) ],
                 dtype=x.dtype)


此功能的思想从@PeterSobot's answer的第二部分扩展到我之前的问题。

关于python - numpy.amax中的键,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/12655525/

10-13 00:47