I know this is dumb, but I need the equivalent of np.sqrt(2.0*np.pi) in my loss function. How can I get it? Statements like this give error: 'float object has no attribute dtype':
pi = np.pi
def myLoss(...):
k = K.sqrt(2.0*pi)
...
Even K.sqrt(2.0*3.14159) is disallowed.
Use it like this:
k = K.sqrt(tf.constant([2.0*np.pi]))
Since, it accepts an object which has dtype. One option is a Tensor.
Another option is to not using keras backend, but using numpy:
k = np.sqrt(2.0*np.pi)
Related
I'm trying to use the shape of an incoming tensor to form the output, sort of like this:
import tensorflow.keras.backend as K
def myFunc(x):
sz = tf.shape(x)[1]
# .. other stuff
z = K.repeat_elements(y, sz, axis=1)
This results in TypeError: Tensor object cannot be interpreted as integer.
How do I get around this?
If you know are that the dimension of x is known in advance, you can use x.shape[1] instead of tf.shape(x)[1], which will return an integer.
But I would advise to use tf.repeat instead of tf.keras.backend.repeat_elements. tf.repeat will work regardless the usage of tf.shape(x) or x.shape.
I want to change the variable_scope by the value of some tensors. For an easy example, I defined a very simple code like this:
import tensorflow as tf
def calculate_variable(scope):
with tf.variable_scope(scope or type(self).__name__, reuse=tf.AUTO_REUSE):
w = tf.get_variable('ww', shape=[5], initializer=tf.truncated_normal_initializer(mean=0.0, stddev=0.1))
return w
w = calculate_variable('in_first')
w1 = calculate_variable('in_second')
The function is very simple. It just returns value defined in a certain variable scope. Now, 'w' and 'w1' would have different values.
What I want to do is to select this variable scope by some condition of tensors. Assuming I have two tensors x, y, if their value is same, I want to get value from the function above with certain variable scope.
x = tf.constant(3)
y = tf.constant(3)
condi = tf.cond(tf.equal(x, y), lambda: 'in_first', lambda: 'in_second')
w_cond = calculate_variable(condi)
I tried many other methods and searched the Internet. However, whenever I want to determine variable_scope from condition of tensors in a similar way to this example, it shows an error.
TypeError: Using a `tf.Tensor` as a Python `bool` is not allowed. Use `if t is not None:` instead of `if t:` to test if a tensor is defined, and use TensorFlow ops such as tf.cond to execute subgraphs conditioned on the value of a tensor.
Is there any good workaround?
The way you stated it, this is not possible. variable_scope class explicitly checks that name_or_scope argument is either a string or a VariableScope instance:
...
if not isinstance(self._name_or_scope,
(VariableScope,) + six.string_types):
raise TypeError("VariableScope: name_or_scope must be a string or "
"VariableScope.")
It does not accept a Tensor. This is reasonable, because variable scope is part of graph definition, one can't define variables dynamically.
The closest supported expression is this:
x = tf.constant(3)
y = tf.constant(3)
w_cond = tf.cond(tf.equal(x, y),
lambda: calculate_variable('in_first'),
lambda: calculate_variable('in_second'))
... where you can select any of the two variables at runtime.
Can I just assign values to certain entries in a tensor? I got this problems when I compute the cross correlation matrix of a NxP feature matrix feats, where N is observations and P is dimension. Some columns are constant so the standard deviation is zero, and I don't want to devide by std for those constant column. Here is what I did:
fmean, fvar = tf.nn.moments(feats, axes = [0], keep_dims = False)
fstd = tf.sqrt(fvar)
feats = feats - fmean
sel = (fstd != 0)
feats[:, sel] = feats[:, sel]/ fstd[sel]
corr = tf.matmul(tf.transpose(feats), feats)
However, I got this error: TypeError: 'Tensor' object does not support item assignment. Is there any workaround for such issue?
You can make your feats a tf.Variable and use tf.scatter_update to update locations selectively.
It's a bit awkward in that scatter_update needs a list of linear indices to update, so you'd need to convert your [:, sel] implicit 2D specification into explicit list of 1D indices. There's example of constructing 1D indices from 2D here
There's some work in simplifying this kind of use-case in issue #206
I have tensor that has shape (?, 3), looks like this [x, y, z] and I need to create function that take argmax of it, creates new vector and assign values with respect to dimension and argmax.
Example:
f(y):
v = tf.variable(tf.zeros(y.get_shape()))
index = tf.argmax(y)
v[index] = 1.0
return v
Unfortunately this doesn't work and I can't figure out how can one do it.
Are you sure that you want to create and assign to a tf.Variable here? It would probably be simpler to use the tf.one_hot() op (available from version 0.8 onwards) to build the result functionally, as you wouldn't have to worry about initialization, etc. For example, you could do the following:
def f(y):
index = tf.argmax(y, 1)
return tf.one_hot(index, tf.shape(y)[1], 1.0, 0.0)
I have a pandas dataframe and passing df[list_of_columns] as X and df[[single_column]] as Y to a Random Forest regressor.
What does the following warnning mean and what should be done to resolve it?
DataConversionWarning: A column-vector y was passed when a 1d array was expected. Please change the shape of y to (n_samples, ), for example using ravel(). probas = cfr.fit(trainset_X, trainset_Y).predict(testset_X)
Simply check the shape of your Y variable, it should be a one-dimensional object, and you are probably passing something with more (possibly trivial) dimensions. Reshape it to the form of list/1d array.
You can use df.single_column.values or df['single_column'].values to get the underlying numpy array of your series (which, in this case, should also have the correct 1D-shape as mentioned by lejlot).
Actually the warning tells you exactly what is the problem:
You pass a 2d array which happened to be in the form (X, 1), but the method expects a 1d array and has to be in the form (X, ).
Moreover the warning tells you what to do to transform to the form you need: y.values.ravel().
Use Y = df[[single_column]].values.ravel() solves DataConversionWarning for me.