erfinv | The inverse error function
kandi X-RAY | erfinv Summary
kandi X-RAY | erfinv Summary
Implementation of the inverse error function based on the rational approximation of percentage points of normal distribution available from erfinv(x) = --------- ppnd | ------- | sqrt(2) \ 2 /. The code has been tested on an x86_64 machine with Intel Core i7, the tests provided in this repository might not pass for different hardware configuration due to difference in floating point operations. golang's math library uses the same implementation for erfinv.
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erfinv Examples and Code Snippets
def erfinv(x, name="erfinv"):
"""The inverse function for erf, the error function.
Args:
x: `Tensor` of type `float32`, `float64`.
name: Python string. A name for the operation (default="erfinv").
Returns:
x: `Tensor` with `dtype=
def _ErfinvGrad(op, grad):
"""Returns grad * sqrt(pi) / 2 * exp(erfinv(x)**2)."""
root_pi_over_two = constant_op.constant(np.sqrt(np.pi) / 2, dtype=grad.dtype)
with ops.control_dependencies([grad]):
return grad * root_pi_over_two * math_ops
Community Discussions
Trending Discussions on erfinv
QUESTION
I'm trying to optimize a custom model (no fancy ML whatsoever) that has 13 parameters, 12 of which I know to be normally distributed. I've gotten decent results using the hyperopt
library:
ANSWER
Answered 2021-Nov-11 at 22:46You can cheat optuna
by using uniform distribution and transforming it into normal distribution. To do that one of the method is inversed error function implemented in scipy
.
Function takes uniform distribution from in range <-1, 1> and converts it to standard normal distribution
QUESTION
I am trying to translate or convert this observable notebook example into plain javascript and d3 in order for me to use it in a website.
I tried to follow along this answer right here where something similar was achieved. But it didn't work for me, I get an error saying "Unexpected value NaN parsing cy attribute."
Here's what I have done until now:
...ANSWER
Answered 2021-Aug-13 at 01:14Here's a working example:
QUESTION
I am using a custom log gamma distribution in Scipy and I want to use it with a variety of predefined Scipy Distributions. My issue is that it seems Custom Distributions need to be a handled a bit different than predefined distributions. Specifically, it seems I need to initialize the custom distribution before it can be used. See below:
...ANSWER
Answered 2021-Apr-12 at 14:26You are not initializing your distribution; it's just how SciPy's distribution infrastructure is built. You are confused between an instance of the rv_continuous
class and the rv_frozen
class.
When you call your distribution class (custom_lgamma
in your case), SciPy generates documentation and code for the user interface which creates the distributions we see and use from scipy.stats
(e.g. norm
, gamma
) (this is what happens when you do custom_lgamma()
). This instance contains methods such as pdf
, cdf
, etc that are called with the shape parameters (just like other scipy.stats
distributions).
Calling this instance again (e.g. norm()
) will create a frozen distribution (an instance of the rv_frozen
class) whose shape parameters cannot be changed (which is what would happen if you call your custom_lgamma
instance again or do custom_lgamma()()
). This instance has the pdf
method that just accepts the quantile parameter.
Here is the convention that SciPy uses:
QUESTION
I have redefined the lognormal distribution using custom scipy class. I have simulated this distribution and I am trying to recover the original parameters I have specified, however, the fit method is returning different parameters.
...ANSWER
Answered 2021-Feb-09 at 20:19I've corrected code with proper _ppf, and it seems to produce proper fits for mu and sigma
Code, Python 3.9 Windows 10 x64
QUESTION
The goal is to say: "These values lie within a band of 95 % of values around the mean in a normal distribution."
Now, I am trying to convert percentage to z-score, so then I can get the precise range of values. Something like would be enough.
So I need something like
...ANSWER
Answered 2020-Dec-15 at 16:24I say you were "close enough".
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