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Find row number of a certain index in a ndarray

Time:09-24

I need to find the row and column number of a specific index/item in a ndarray. I found nothing to do this simultaneously, so I tried to achieve this separately. In the shown code I only had at least an idea how to do this for the rows.

For this code the expected output should be 0. My idea was to iterate over the number of entries with for i in range(0, len(array)):. For i = 0 the wanted item is not found obviously. So if row != None: is not fulfilled so the loop should start over. For i = 1 the condition should be fulfilled and the loop should stop there and give the correct row number. But the output is ValueError: 2 is not in list. I found solution to avoid this error but not in the combination with a ndarray.

It feels like I just can't get behind the logic of the combination of the for loop and the if statement. I hope someone can help me with my problem and maybe enlighten me on how this works exactly.

array = [[0,1],
         [2,3]]

row = None
item = 2

for i in range(0, len(array)):
    row = array[i].index(item)
    if row != None:
        break
    else:
        continue
print(row)

CodePudding user response:

If you want a pure python solution:

target = 0
for row, lst in enumerate(array):     # loop over rows
    for col, value in enumerate(lst): # loop over values
        if value == target:
            print(row, col)
            # uncomment below if you want to stop after the first match
            # break

output: 0 0

However, I would recommend to use n this is worth learning if you plan to do real ndarray operations:

array = [[0,1],
         [2,3]]

# convert to numpy array
import numpy as np
a = np.array(array)

target = 0
# identify the coordinate for all matching values
row, col = np.nonzero(a==target)

output (there will be as many values as there are matches):

# row
array([0])

# col
array([0])

First value:

first_match = row[0], col[0]

output: (0, 0)

CodePudding user response:

To find the row and column of a specific number in a 2d array, loop through the array and use the index function

arr = [[0, 1], [2, 3]]

def find(ele):
     for i in arr:
         if ele in i:
             row = arr.index(i)
             col = i.index(ele)
             print("Row: ", row, "Column: ", col)

find(1) # returns "Row: 0 Column: 1"
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