Micro- and macro-averages (for whatever metric) will compute slightly different things, and thus their interpretation differs. A macro-average will compute the metric independently for each class and then take the average (hence treating all classes equally), whereas a micro-average will aggregate the contributions of all classes to compute the average metric. In a multi-class classification setup, micro-average is preferable if you suspect there might be class imbalance (i.e you may have many more examples of one class than of other classes).
To illustrate why, take for example precision $Pr=\frac{TP}{(TP+FP)}$. Let's imagine you have a One-vs-All (there is only one correct class output per example) multi-class classification system with four classes and the following numbers when tested:
* Class A: 1 TP and 1 FP
* Class B: 10 TP and 90 FP
* Class C: 1 TP and 1 FP
* Class D: 1 TP and 1 FP
You can see easily that $Pr_A = Pr_C = Pr_D = 0.5$, whereas $Pr_B=0.1$.
* A macro-average will then compute: $Pr=\frac{0.5+0.1+0.5+0.5}{4}=0.4$
* A micro-average will compute: $Pr=\frac{1+10+1+1}{2+100+2+2}=0.123$
These are quite different values for precision. Intuitively, in the macro-average the "good" precision (0.5) of classes A, C and D is contributing to maintain a "decent" overall precision (0.4). While this is technically true (across classes, the average precision is 0.4), it is a bit misleading, since a large number of examples are not properly classified. 964
These examples predominantly correspond to class B, so they only contribute 1/4 towards the average in spite of constituting 94.3% of your test data. The micro-average will adequately capture this class imbalance, and bring the overall precision average down to 0.123 (more in line with the precision of the dominating class B (0.1)).
For computational reasons, it may sometimes be more convenient to compute class averages and then macro-average them. If class imbalance is known to be an issue, there are several ways around it. One is to report not only the macro-average, but also its standard deviation (for 3 or more classes). Another is to compute a weighted macro-average, in which each class contribution to the average is weighted by the relative number of examples available for it. In the above scenario, we obtain:
$Pr_{macro-mean}={0.25·0.5+0.25·0.1+0.25·0.5+0.25·0.5}=0.4$ $Pr_{macro-stdev}=0.173$
$Pr_{macro-weighted}={0.0189·0.5+0.943·0.1+0.0189·0.5+0.0189·0.5}={0.009+0.094+0.009+0.009}=0.123$
The large standard deviation (0.173) already tells us that the 0.4 average does not stem from a uniform precision among classes, but it might be just easier to compute the weighted macro-average, which in essence is another way of computing the micro-average. 915