Question

Numpy是否具有快速搜索2D数组中的元素并返回其索引的功能？意思是例如：

     var arrs = [];
    jQuery('.option').each(function(){
        arrs.push(jQuery(this).val());
    });
    var arrayLength = arrs.length;
    var retval = "";
    for (var i = 0; i < arrayLength; i++) {
        retval = arrs[i] + '|';
        //Do something
    }
    alert(retval);

所以同等价值将是a=54 array([[ 0, 1, 2, 3], [ 4, 5, 54, 7], [ 8, 9, 10, 11]])。当然，我可以使用简单的循环 - 但我想要类似的东西：

array[1][2]

Answer 1

In [4]: import numpy as np 

In [5]: my_array = np.array([[ 0,  1,  2,  3],
                             [ 4,  5,  54,  7],
                             [8, 54, 10, 54]])

In [6]: my_array
Out[6]: 
array([[ 0,  1,  2,  3],
       [ 4,  5, 54,  7],
       [ 8, 54, 10, 54]])

In [7]: np.where(my_array == 54) #indices of all elements equal to 54
Out[7]: (array([1, 2, 2]), array([2, 1, 3])) #(row_indices, col_indices)

In [10]: temp = np.where(my_array == 54)

In [11]: zip(temp[0], temp[1])   # maybe this format is what you want
Out[11]: [(1, 2), (2, 1), (2, 3)]

Python快速搜索numpy数组

1 个答案: