Understanding Logits and Class Scores

A classifier rarely answers with a class name. For each input it produces a row of numbers, one number for each possible class. Each number describes how strongly the model leans toward that class. A row might say that the first class is the most likely, but before anyone can count how many predictions were right, that row has to be turned into one decision. This article follows the step that makes that decision: picking the position of the largest value in each row.

The function torch.argmax with dim=1 returns the index of the highest value in each row. Each row of scores becomes a single integer, and that integer is the class label for that row. The dim argument decides the direction of the search. With dim=1 the function scans each row from left to right and reports where its largest value sits.

Read the output row by row. Row 0 has its largest score, 2.0, at position 0, so its label is 0. Row 1 has its largest score, 3.0, at position 1, so its label is 1. Row 2 has its largest score, 1.5, at position 1, so its label is 1. The third row shows that two different samples can share the same label, which is normal for a classifier.

Changing dim from 1 to 0 changes the question entirely. With dim=0 the function scans down each column instead of across each row, and it returns, for every column, the row position of its largest value. That second result answers a different question. It does not give one label per sample, so using the wrong dimension is one of the most common ways to get label predictions that look plausible but mean something else.

The same argmax operation applies to probabilities and to the raw outputs of the network. Computing softmax first and then calling argmax is a valid path. It is not required, however, because the position of the largest value is the same whether you look at the logits or at the probabilities computed from them. Softmax changes the numbers, but it keeps their order within a row, so the winning position does not move.

Once you have predicted labels, the next question is whether they match the true labels. For a small set you could compare the two tensors by eye. The equality operator makes this exact. The == operator compares two tensors element by element and returns a tensor of booleans, one for each position. True means the prediction at that position matched the true label.

The count of correct predictions becomes accuracy by dividing it by the total number of examples and multiplying by 100 to express the result as a percentage. The rule is the same for any dataset.

The source's own training set has five examples, and all five predictions are correct. Five correct out of five gives an accuracy of 100 percent. If a comparison had produced three matches out of four positions, the accuracy would be three quarters, or 75 percent.

When an accuracy number surprises you, the mistake almost always sits in one of three places: the dimension passed to argmax, the tensor passed into argmax, or the true labels it is compared against. Work through the chain in that order, because each step depends on the one before it.