blckbird - 9 months ago 47

Python Question

I created a numpy array with n values from 0 to 2pi. Now, I want to generate n test data points deviating from sin(x) normally distributed.

So i figured I need to do something like this:

`t = sin(x) + noise`

`noise = np.random.randn(mean, std)`

However, I do not know how I can calculate the noise when my mean is sin(x) (and not a constant).

Answer

The arguments to `numpy.random.randn`

are not the mean and standard deviation. For that, you want `numpy.random.normal`

. Its signature is

```
normal(loc=0.0, scale=1.0, size=None)
```

To add noise to your sin function, simply use a mean of 0 in the call of `normal()`

. The mean corresponds to the `loc`

argument (i.e. "location"), which by default is 0. So, given that `x`

is something like `np.linspace(0, 2*np.pi, n)`

, you can do this:

```
t = np.sin(x) + np.random.normal(scale=std, size=n)
```

You *could* use `numpy.random.randn`

, but you have to scale it by `std`

, because `randn`

returns samples from the standard normal distribution, with mean 0 and standard deviation 1. To use `randn`

, you would write:

```
t = np.sin(x) + std * np.random.randn(n)
```

Source (Stackoverflow)