> From: Eric Firing [mailto:efir...@hawaii.edu]
> Sent: Wednesday, June 06, 2012 13:41
> To: matplotlib-users@lists.sourceforge.net
> Subject: Re: [Matplotlib-users] scatter plot with constant x
>
> On 06/06/2012 06:42 AM, Ethan Gutmann wrote:
> >> ...
> >> No, but you can do this:
> >>
> >> plt.plot([3] * 4, [60, 80, 120, 180], ...)
> >
> > This started from a simple enough question, but it got me
> thinking about what the fastest way to do this is (in case
> you have HUGE arrays, or many loops over them).
[...]
> Since we end up needing float64 anyway:
>
> In [3]: %timeit l=np.empty(10000,dtype=np.float64); l.fill(3)
> 100000 loops, best of 3: 14.1 us per loop
>
> In [4]: %timeit l=np.zeros(10000,dtype=np.float64);l[:]=3
> 10000 loops, best of 3: 26.6 us per loop
>
> Eric
Numpy's as_strided came to mind; it can make a large array that's really a
view of a one-element array:
In [1]: as_strided = np.lib.stride_tricks.as_strided
In [2]: s = as_strided(np.array([3], dtype=np.float64), shape=(10000,),
...: strides=(0,))
In [3]: s[0] = 4
In [4]: s[9999] # all elements share data
Out[4]: 4.0
It's somewhat slower to create the base array and the view than to create and
fill a 10000-element array:
In [5]: %timeit l = np.empty(10000, dtype=np.float64); l.fill(3)
100000 loops, best of 3: 10.1 us per loop
In [6]: %timeit s = as_strided(np.array([3], dtype=np.float64),
shape=(10000,), strides=(0,)) # line broken for email
10000 loops, best of 3: 21.6 us per loop
However, once created, its contents may be changed much more quickly:
In [7]: l = np.empty(10000, dtype=np.float64)
In [8]: %timeit l.fill(3)
100000 loops, best of 3: 7.71 us per loop
In [9]: %timeit s[0] = 3
10000000 loops, best of 3: 116 ns per loop
Numpy's broadcast_arrays uses as_strided under the hood. Code could look
like:
x, y = np.broadcast_arrays(3, [60, 80, 120, 180])
plt.plot(x, y, '+')
x[0] = 21 # new x for all samples
plt.plot(x, y, 'x')
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