I have the following code:
我有以下代码：
l = 2
h = 1
p = 2
q = -2
x = Symbol('x')
f = Piecewise ( (0, x < 0),
(p, 0 <= x <= l/3),
(h/l * x - h, l/3 < x < 2*l/3),
(q, 2*l/3 <= x <= l),
(0, x > l)
)
which results into a Error:
这会导致错误：
TypeError: cannot determine truth value of Relational
What is the proper way to define this function? I need sympy for this because I plan to integrate it later.
定义此功能的正确方法是什么？我需要同情，因为我打算稍后整合它。
1 个解决方案
#1
2
SymPy's Piecewise class does not support chained inequalities like 0 < x < 1. It is possible to use And(0 < x, x < 1) instead, but in your example this is unnecessary. Recall that the conditions are evaluated from left to right; the first that evaluates to True wins. So you don't need 0 <= x in the second condition, or l/3 < x in the third, etc. Your function should be defined by
SymPy的Piecewise类不支持像0
f = Piecewise ( (0, x < 0),
(p, x <= l/3),
(h/l * x - h, x < 2*l/3),
(q, x <= l),
(0, True)
)
#1
2
SymPy's Piecewise class does not support chained inequalities like 0 < x < 1. It is possible to use And(0 < x, x < 1) instead, but in your example this is unnecessary. Recall that the conditions are evaluated from left to right; the first that evaluates to True wins. So you don't need 0 <= x in the second condition, or l/3 < x in the third, etc. Your function should be defined by
SymPy的Piecewise类不支持像0
f = Piecewise ( (0, x < 0),
(p, x <= l/3),
(h/l * x - h, x < 2*l/3),
(q, x <= l),
(0, True)
)