argmin & argmax

max() tells you the highest score was 103. But the next question is always "in which match?" or "by which player?". For that you need the position of the maximum, not its value. That's what argmax (and its partner argmin) return.

Definition

argmax returns the index (position) of the largest value, and argmin returns the index of the smallest value. If the extreme value appears more than once, the index of the first occurrence is returned.

• "arg" — short for argument: the input position that produces the max, as in maths ("arg max").
• Index, not value — argmax gives 6; you then use scores[6] to get the value 103.
• First occurrence — if two matches are both ducks, argmin returns the earlier match.

Syntax

np.argmax(a, axis=None, keepdims=False)     a.argmax(axis=None)
np.argmin(a, axis=None, keepdims=False)     a.argmin(axis=None)
np.unravel_index(flat_index, shape)          # flat index -> (row, col)

Example — best and worst match (1D)

import numpy as np

virat = np.array([72, 45, 88, 31, 64, 0, 103, 52, 19, 77])

best = virat.argmax()
worst = virat.argmin()
print("Best innings  : index", best, "-> match", best + 1, "->", virat[best], "runs")
print("Worst innings : index", worst, "-> match", worst + 1, "->", virat[worst], "runs")

Best innings  : index 6 -> match 7 -> 103 runs
Worst innings : index 5 -> match 6 -> 0 runs

How it works: indexes start at 0, so index 6 is match 7. Adding + 1 turns an index into a human "match number". Using the index to look back into the array (virat[best]) gives the value.

Example — on 2D data with axis

import numpy as np

players = ["Virat", "Rohit", "Hardik", "Shubman"]
scores = np.array([
    [72, 45, 88, 31, 64],
    [56,  0, 74, 88, 32],
    [45, 32,  0, 67, 18],
    [33, 61, 45,  0, 77],
])

print("Best match of each player (axis=1):", scores.argmax(axis=1))
top = scores.argmax(axis=0)
print("Top scorer row per match (axis=0)  :", top)
print("Top scorer names:", [players[i] for i in top])

Best match of each player (axis=1): [2 3 3 4]
Top scorer row per match (axis=0)  : [0 3 0 1 3]
Top scorer names: ['Virat', 'Shubman', 'Virat', 'Rohit', 'Shubman']

How it works: argmax(axis=1) looks along each row and returns a column index (which match). argmax(axis=0) looks down each column and returns a row index (which player), which we map to names.

Example — no axis on 2D: a flat index

Without axis, NumPy flattens the 2D array first, so the answer is a single flat index. Convert it back to (row, column) with np.unravel_index:

import numpy as np

scores = np.array([
    [72, 45, 88, 31, 64],
    [56,  0, 74, 88, 32],
    [45, 32,  0, 67, 18],
    [33, 61, 45,  0, 77],
])

flat = scores.argmax()
print("flat index:", flat)
row, col = np.unravel_index(flat, scores.shape)
print("row:", int(row), "col:", int(col), "value:", scores[row, col])

flat index: 2
row: 0 col: 2 value: 88

How it works: the maximum 88 appears twice (Virat match 3 and Rohit match 4). In the flattened array Virat's 88 is at position 2, which comes first, so argmax returns 2 → row 0, column 2.

Common mistakes

• ❌ Expecting argmax to return the value. ✅ It returns a position; use a[a.argmax()] or a.max() for the value.
• ❌ Using scores.argmax() on a 2D array and treating it as a row number. ✅ It is a flat index — use np.unravel_index or pass axis.
• ❌ Forgetting it is 0-based ("match 6" when it is really match 7). ✅ Add 1 for human numbering.
• ❌ Assuming ties return all positions. ✅ Only the first is returned; use np.flatnonzero(a == a.max()) or np.where(a == a.max()) to get all of them.
• ❌ Arrays with NaN: argmax returns the position of the NaN. ✅ Use np.nanargmax (see NaN-safe functions).

Interview tip

Q: How do you find the row and column of the maximum value in a 2D array? A: np.unravel_index(a.argmax(), a.shape). argmax() without axis returns an index into the flattened array, and unravel_index converts it back to a (row, col) tuple.