I have a Numpy array with size [2000,6].now I have an array with size [1,6] and I want to know which row of the main Numpy array is the same as this [1,6] array. if it exists in the main array, return the index of the row. for example row 1. but I do not want to use for loops because it is really time-consuming. I do not know how I can find a vector in an array and extract its related index. please help me with this issue. Thanks

CodePudding user response：

Assuming you are using NumPy for this, your goal can be obtained by using the numpy.where method.

NumPy Solution

You can ignore this cell as I am just creating a simulated version of your problem:

# Simulate @david data
import numpy as np

target_row = np.array((7, 8, 9, 1, 2, 1))
simulated_array = np.zeros((2000, 6))

# I put the target row in index 3
simulated_array[3] = target_row

# Solution
import numpy as np

target_row = np.array((7, 8, 9, 1, 2, 1))
condition = simulated_array == target_row
where_result = np.where(condition)

# This is required since a row of indices is returned above
target_index = where_result[0]

Python List solution

# Simulate @david array

simulated_list = array.tolist()
target_list = [7, 8, 9, 1, 2, 1]

# Solution
target_list = [7, 8, 9, 1, 2, 1]
target_index = simulated_list.index(target_list)

Resources

numpy.where — https://numpy.org/doc/stable/reference/generated/numpy.where.html

CodePudding user response：

this is what I have, though I imagine theres some method in np.arrays that could do this for you

import numpy as np
arr = np.array([[1, 2 ,3, 4, 5, 6],[7, 8 ,9, 1, 2, 1],[ 1, 2, 3, 4, 5 ,6]])
    for index in arr:
        if index == [7, 8 ,9, 1, 2, 1]:
            print(index)