From: Stan West <stan.west@nr...>  20120606 18:16:17

> From: Eric Firing [mailto:efiring@...] > Sent: Wednesday, June 06, 2012 13:41 > To: matplotlibusers@... > Subject: Re: [Matplotlibusers] 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 oneelement 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 10000element 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') 