搜索具有多个值的Numpy数组

问题描述：

我有具有重复值的numpy 2d数组.

I have numpy 2d array having duplicate values.

我正在搜索这样的数组.

I am searching the array like this.

In [104]: import numpy as np

In [105]: array = np.array

In [106]: a = array([[1, 2, 3],
     ...:            [1, 2, 3],
     ...:            [2, 5, 6],
     ...:            [3, 8, 9],
     ...:            [4, 8, 9],
     ...:            [4, 2, 3],
     ...:            [5, 2, 3])

In [107]: num_list = [1, 4, 5]

In [108]: for i in num_list :
     ...:     print(a[np.where(a[:,0] == num_list)])
     ...:
 [[1 2 3]
 [1 2 3]]
[[4 8 9]
 [4 2 3]]
[[5 2 3]]

输入是具有类似于列0值的数字的列表. 我想要的最终结果是以任何格式生成的行，例如数组，列表或元组

The input is list having number similar to column 0 values. The end result I want is the resulting rows in any format like array, list or tuple for example

array([[1, 2, 3],
       [1, 2, 3],
       [4, 8, 9],
       [4, 2, 3],
       [5, 2, 3]])

我的代码工作正常，但似乎不是pythonic.有没有更好的多值搜索策略?

My code works fine but doesn't seem pythonic. Is there any better searching strategy with multiple values?

类似于a[np.where(a[:,0] == l)]，其中仅执行一次查找即可获取所有值.

like a[np.where(a[:,0] == l)] where only one time lookup is done to get all the values.

我的真实数组很大

答

方法1:使用 方法2::使用 np.searchsorted -

num_arr = np.sort(num_list) # Sort num_list and get as array

# Get indices of occurrences of first column in num_list
idx = np.searchsorted(num_arr, a[:,0])

# Take care of out of bounds cases
idx[idx==len(num_arr)] = 0 

out = a[a[:,0] == num_arr[idx]]

搜索具有多个值的Numpy数组

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