如何使用matplotlib绘制此步进函数？

Question

How can I plot the following function on matplotlib:

For all intervals t in [n,n+1] , f(t)=1 for n even and f(t)=-1 for n odd. So this is basically a step function with f(t)=1 from 0 to 1, f(t)=-1 from 1 to 2, f(t)=1 from 2 to 3, f(t)=-1 from 3 to 4, and so on.

This is my code so far:

t = arange(0,12)

def f(t):
    if t%2 == 0:
        for t in range(t,t+1):
            f = 1
        if t%2 != 0:
            for t in range(t,t+1):
                f = -1

This is the process of this code:

1. Define t to be in the range 0 to 12.
2. Define the function f(t) .
3. Use the statement, if t is an even integer, so it will consider t=0,2,4,6,8,10,12 .
4. Use the for loop to allow us to define f=1 for each of these integers.
5. Repeat for odd values of t.

Can you see anything fundamentally wrong with this code? Am I complicating things?

When I try to plot using

matplotlib.pyplot.plot(t,f,'b-')
matplotlib.pyplot.show()

I get a ValueError saying "x and y must have same first dimension".

What is going wrong here?

Answer 1

You could use numpy.repeat to double up elements in your array t , and build the (-1,1) pattern with 1 - 2 * (t%2) :

t = np.arange(13)
f = np.repeat(1 - 2 * (t%2), 2)[:-1]
t = np.repeat(t, 2)[1:]

In [6]: t
Out[6]: 
array([ 0,  1,  1,  2,  2,  3,  3,  4,  4,  5,  5,  6,  6,  7,  7,  8,  8,
        9,  9, 10, 10, 11, 11, 12, 12])

In [7]: f
Out[7]: 
array([ 1,  1, -1, -1,  1,  1, -1, -1,  1,  1, -1, -1,  1,  1, -1, -1,  1,
        1, -1, -1,  1,  1, -1, -1, 1])

Perhaps even easier is:

In  [8]: n = 12
In  [9]: t = np.repeat(np.arange(n+1), 2)[1:-1]
In [10]: f = np.array([1,1,-1,-1]*(n//2))