Precision is defined as:

p = true positives / (true positives + false positives)

What is the value of precision if (true positives + false positives) = 0? Is it just undefined?

Same question for recall:

r = true positives / (true positives + false negatives)

In this case, what is the value of recall if (true positives + false negatives) = 0?

P.S. This question is very similar to the question What are correct values for precision and recall in edge cases?.

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    $\begingroup$ Heh, it is even answered in the duplicate; but let's call it a good duplicate. $\endgroup$
    – user88
    Commented Mar 8, 2011 at 19:51

3 Answers 3


The answers to the linked earlier question apply here too.

If (true positives + false negatives) = 0 then no positive cases in the input data, so any analysis of this case has no information, and so no conclusion about how positive cases are handled. You want N/A or something similar as the ratio result, avoiding a division by zero error

If (true positives + false positives) = 0 then all cases have been predicted to be negative: this is one end of the ROC curve. Again, you want to recognise and report this possibility while avoiding a division by zero error.

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    $\begingroup$ Thanks Henry for the answer. If I understand correctly, in the former case, you do not want to recognize and report the result, whereas in the latter case you do. Is that correct? $\endgroup$
    – khatchad
    Commented Mar 8, 2011 at 19:56
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    $\begingroup$ Yes: In the no-positives-in-input case, precision is meaningless; in the no-positives-predicted case, you want to report that the test has been set to be extremely negative. $\endgroup$
    – Henry
    Commented Mar 8, 2011 at 22:29
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    $\begingroup$ I like the added intuition of "no information" vs "information" in the two cases. I wonder why scikit-learn handles these with replacing div by zero with 1 or 0? $\endgroup$ Commented Jul 17, 2021 at 3:44

An interesting answer is offered here: https://github.com/dice-group/gerbil/wiki/Precision,-Recall-and-F1-measure

The authors of the module output different scores for precision and recall depending on whether true positives, false positives and false negatives are all 0. If they are, the outcome is ostensibly a good one.

In some rare cases, the calculation of Precision or Recall can cause a division by 0. Regarding the precision, this can happen if there are no results inside the answer of an annotator and, thus, the true as well as the false positives are 0. For these special cases, we have defined that if the true positives, false positives and false negatives are all 0, the precision, recall and F1-measure are 1. This might occur in cases in which the gold standard contains a document without any annotations and the annotator (correctly) returns no annotations. If true positives are 0 and one of the two other counters is larger than 0, the precision, recall and F1-measure are 0.

I'm not sure if this type of scoring would be useful in other situations outside of their special case, but it's worth some thought.

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    $\begingroup$ I think this makes sense for any application. Here "annotations" stand-in for "relevant cases". Nice find. I'd like to see how they handle the recall situation, so I'll click the link. $\endgroup$ Commented Jul 17, 2021 at 3:48

When evaluating a classifier at high thresholds, the precision might (often actually) not be 1 when recall is 0. It's usually N/A! I think there is something wrong about how people plot the P/R curve. Avoiding N/A samples is a bias in the sense that you avoid singularity samples.

I computed the average precision wrt to the average recall ignoring N/A samples and I never got a classifier starting at 1 for 0 recall for a shallow neural net in object detection. This was also true for curves computed with the tp,fp,fn numbers. It's quite easy to verify by paper and pencil with a single image. For example: I have a classifier that outputs for a single image:

preds=[.7 .6 .5 .1 .05]
truth=[n  y   n  n  y ]

By computing the confusion matrices with the various thresholds we have:

tp=[2 1 1 1 0 0]
fn=[0 1 1 1 2 2]
fp=[3 3 2 1 1 0]

then recall rec=[1 .5 .5 .5 0 0], and precision prec=[.4 .25 1/3 .5 0 NaN]. I don't see how it would make sense to replace a NaN or the precision(@recall==0) with 1. 1 should be an upper bound, not a value we replace precision(@recall==0) with.


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