问题描述
numpy.take(array, indices)和numpy.choose(indices, array)似乎返回相同的东西:由indices索引的array的子集.
It seems that numpy.take(array, indices) and numpy.choose(indices, array) return the same thing: a subset of array indexed by indices.
两者之间只有细微的差别吗，还是我缺少更重要的东西?并且有理由偏爱另一个吗?
Are there only subtle differences between the two, or am I missing something more important? And is there a reason to prefer one over the other?
推荐答案
numpy.take(array, indices)和numpy.choose(indices, array)在一维数组上的行为类似，但这只是巧合.正如jonrsharpe指出的那样，它们在高维数组上的行为有所不同.
numpy.take(array, indices) and numpy.choose(indices, array) behave similarly on 1-D arrays, but this is just coincidence. As pointed out by jonrsharpe, they behave differently on higher-dimensional arrays.
numpy.take(array, indices)从平坦版本的array中挑选元素. (结果元素当然不一定来自同一行.)
numpy.take(array, indices) picks out elements from a flattened version of array. (The resulting elements are of course not necessarily from the same row.)
例如，
numpy.take([[1, 2], [3, 4]], [0, 3])
返回
array([1, 4])
numpy.choose
numpy.choose(indices, set_of_arrays)从数组indices[0]中抽出元素0，从数组indices[1]中抽出元素1，从数组indices[2]中抽出元素2，依此类推. (这里，array实际上是一组数组.)
numpy.choose
numpy.choose(indices, set_of_arrays) plucks out element 0 from array indices[0], element 1 from array indices[1], element 2 from array indices[2], and so on. (Here, array is actually a set of arrays.)
例如
numpy.choose([0, 1, 0, 0], [[1, 2, 3, 4], [4, 5, 6, 7]])
返回
array([1, 5, 3, 4])
因为元素0来自数组0，元素1来自数组1，元素2来自数组0，元素3来自数组0.
because element 0 comes from array 0, element 1 comes from array 1, element 2 comes from array 0, and element 3 comes from array 0.
这些描述得到了简化–完整的描述可以在这里找到: numpy.take ， numpy.choose .例如，当indices和array为一维时，numpy.take和numpy.choose的行为类似，因为numpy.choose首先广播array.
These descriptions are simplified – full descriptions can be found here: numpy.take, numpy.choose. For example, numpy.take and numpy.choose behave similarly when indices and array are 1-D because numpy.choose first broadcasts array.
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