This question is similar to this one and this one, and I am afraid we have not had a satisfactory answer yet.
Using take()
and skip()
requires knowing the dataset size. What if I don't know that, or don't want to find out?
Using shard()
only gives 1 / num_shards
of dataset. What if I want the rest?
I try to present a better solution below, tested on TensorFlow 2 only. Assuming you already have a shuffled dataset, you can then use filter()
to split it into two:
import tensorflow as tf
all = tf.data.Dataset.from_tensor_slices(list(range(1, 21))) \
.shuffle(10, reshuffle_each_iteration=False)
test_dataset = all.enumerate() \
.filter(lambda x,y: x % 4 == 0) \
.map(lambda x,y: y)
train_dataset = all.enumerate() \
.filter(lambda x,y: x % 4 != 0) \
.map(lambda x,y: y)
for i in test_dataset:
print(i)
print()
for i in train_dataset:
print(i)
The parameter reshuffle_each_iteration=False
is important. It makes sure the original dataset is shuffled once and no more. Otherwise, the two resulting sets may have some overlaps.
Use enumerate()
to add an index.
Use filter(lambda x,y: x % 4 == 0)
to take 1 sample out of 4. Likewise, x % 4 != 0
takes 3 out of 4.
Use map(lambda x,y: y)
to strip the index and recover the original sample.
This example achieves a 75/25 split.
x % 5 == 0
and x % 5 != 0
gives a 80/20 split.
If you really want a 70/30 split, x % 10 < 3
and x % 10 >= 3
should do.
UPDATE:
As of TensorFlow 2.0.0, above code may result in some warnings due to AutoGraph's limitations. To eliminate those warnings, declare all lambda functions separately:
def is_test(x, y):
return x % 4 == 0
def is_train(x, y):
return not is_test(x, y)
recover = lambda x,y: y
test_dataset = all.enumerate() \
.filter(is_test) \
.map(recover)
train_dataset = all.enumerate() \
.filter(is_train) \
.map(recover)
This gives no warning on my machine. And making is_train()
to be not is_test()
is definitely a good practice.