访问列表字典中列表的最小元素

Question

The keras model.fit(..) returns a callback history object from which we can retrieve a history of model's metrics logs for all epochs ie:

hist = model.fit(..)
off_history = hist.history

So that:

pprint(off_history)
{'accuracy': [0.5884908437728882,
              0.6860442757606506,
              0.7116397619247437,
              0.7207977771759033,
              0.7276809811592102],
 'loss': [1.5214883089065552,
          0.9418172836303711,
          0.8826016187667847,
          0.8565425276756287,
          0.8384354114532471],
 'val_accuracy': [0.569063663482666,
                  0.6021720170974731,
                  0.6034634709358215,
                  0.6066920757293701,
                  0.607513964176178],
 'val_loss': [1.3138036727905273,
              1.2316004037857056,
              1.201686143875122,
              1.1939105987548828,
              1.1881093978881836]}

I am interested in retrieving the minimum val_loss for all model fit. I know that I can do:

off_history['val_loss'][-1]

But since I'm not sure if keras actually returns a sorted list for accuracy, loss, val_accuracy... all the time, I cannot rely on this operation.

What then is the best way to get the min val_loss in all cases of model.fit() ?

Answer 1

Keras fit() returns a history object of training logs. So if you set:

hist = model.fit()

This history object contains all epoch logs of model training:

type(history.epoch) #epochs list [0,1,2,,...n_epochs]
list

So the hist.history dictionary you refers to, contains the corresponding metrics values for each epoch in history.epoch in the list above.

To answer your question, keras fit() returns those metrics in the order of epochs run. 1,2,3,4,5....

A way around your goal would be to use numpy 's min function, like this:

validation_loss = np.min(hist.history['val_loss'])

to retrieve minimum validation loss across all epochs of keras fit()

Answer 2

If you need the exact output then you can try using for loop

for k,v in off_history.items():
    if k == 'val_loss':
        print(v[-1])

This gives:

1.1881093978881836