How to find the row and column of element in 2d array in python?

Question

I have this 2d array :

   import numpy as np

R = int(input("Enter the number of rows:")) //4
C = int(input("Enter the number of columns:")) //5

randnums= np.random.randint(1,100, size=(R,C))

print(randnums)
    [[98 25 33  9 41]
     [67 32 67 27 85]
     [38 79 52 40 58]
     [84 76 44  9  2]]

Now, i want to happen is that i will search an element and the output will be its column and rows example.

enter number to search : 40

Number 40 found in row 3 column 4

enter number : 100

number not found

something like this ? thanks in advance

Use nested loops over the row and column indexes. When you find the matching element, return the current indexes. — Barmar
– Barmar, Commented Jun 12, 2020 at 6:28
Does this answer your question? Finding indices of values in 2D numpy array — T Tse
– T Tse, Commented Jun 12, 2020 at 6:35
I understand, and thankyou for that @Barmar, i will update the question and next time wont be again like this. appologies im new here for asking — Manjay
– Manjay, Commented Jun 12, 2020 at 6:46
and also, this is my very first time trying to learn python. as in first time, im confuse about syntaxes thats why i asked. but then, yeah i will update. thanks again Mr @Barmar — Manjay
– Manjay, Commented Jun 12, 2020 at 6:59

Kuldeep Singh Sidhu · Accepted Answer · 2020-06-12 06:31:04Z

1

l = [[98, 25, 33,  9, 41],
[67, 32, 67, 27, 85],
[38, 79, 52, 40, 58],
[84, 76, 44,  9,  2]]

def fnd(l,value):
    for i,v in enumerate(l):
        if value in v:
            return {'row':i+1,'col':v.index(value)+1}
    return {'row':-1,'col':-1}
print(fnd(l,40))

{'row': 3, 'col': 4}

answered Jun 12, 2020 at 6:31

Kuldeep Singh Sidhu

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Comments

Abhishek J · Accepted Answer · 2020-06-12 06:37:26Z

1

If the number of columns will be constant as shown in the example, you can search using below code.

a = [[98, 25, 33, 9, 41],
     [67, 32, 67, 27, 85],
     [38, 79, 52, 40, 58],
     [84, 76, 44, 9, 2]]
a_ind = [p[x] for p in a for x in range(len(p))] # Construct 1d array as index for a to search efficiently.


def find(x):
    return a_ind.index(x) // 5 + 1, a_ind.index(x) % 5 + 1 # Here 5 is the number of columns


print(find(98), find(58), find(40))

#Output
(1, 1) (3, 5) (3, 4)

edited Jun 12, 2020 at 6:37

answered Jun 12, 2020 at 6:31

Abhishek J

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user10067135 · Accepted Answer · 2020-06-12 07:00:40Z

1

You can use numpy.where function.

r = np.random.randint(1,10, size=(5,5))

# array([[6, 5, 3, 1, 8],
#        [3, 9, 7, 5, 6],
#        [6, 2, 5, 5, 8],
#        [1, 5, 1, 1, 1],
#        [1, 6, 5, 8, 6]])

s = np.where(r == 8)

# the first array is row indices, second is column indices
# (array([0, 2, 4], dtype=int64), array([4, 4, 3], dtype=int64))

s = np.array(np.where(r == 8)).T

# transpose to get 2d array of indices
# array([[0, 4],
#        [2, 4],
#        [4, 3]], dtype=int64)

answered Jun 12, 2020 at 7:00

user10067135

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