Question

我有一个定义积分的函数。

稍后通过将所述函数乘以0.5 / pi来定义阵列A [i，j]。

我收到错误：

TypeError：*：＆＃39; float＆＃39;不支持的操作数类型和＆＃39; NoneType＆＃39;

显然这是因为我将一些没有类型的东西乘以一个数字，而python无法解释。

你们所建议的是最容易解决这个错误的方法吗？

我在下面包含了我的代码（不包括所有代码，因此sum变量可能看起来未定义）

def integral_normal(p_i, p_j):
"""Evaluates the contribution of a panel at the center-point of another,
in the normal direction.

Arguments
---------
p_i -- panel on which the contribution is calculated.
p_j -- panel from which the contribution is calculated.

Returns
-------
Integral over the panel of the influence at a control-point.
"""
def func(s):
    return ( (+(p_i.xc-(p_j.xa-sin(p_j.beta)*s))*cos(p_i.beta)
              +(p_i.yc-(p_j.ya+cos(p_j.beta)*s))*sin(p_i.beta))
            /((p_i.xc-(p_j.xa-sin(p_j.beta)*s))**2
              +(p_i.yc-(p_j.ya+cos(p_j.beta)*s))**2) )
    return integrate.quad(lambda s:func(s), 0., p_j.length)[0]

# computes the source influence matrix
A = np.empty((N_panels, N_panels), dtype=float)
np.fill_diagonal(A, 0.5)

# use enumerate to access element panels individually from 0 to i, locates element of A to fill.
for i, p_i in enumerate(panels):
    for j, p_j in enumerate(panels):
        if i != j:
            A[i,j] = 0.5/pi*integral_normal(p_i, p_j)

# computes right hand side of linear system
b = - u_inf * np.cos([p.beta for p in panels])

# solves the linear system
sigma = np.linalg.solve(A, b)

for i, panel in enumerate(panels):
    panel.sigma = sigma[i]

这是错误：

TypeError                                 Traceback (most recent call last)
<ipython-input-101-b5c8b9263c2d> in <module>()
      26     for j, p_j in enumerate(panels):
      27         if i != j:
----> 28             A[i,j] = 0.5/pi*integral_normal(p_i, p_j)
      29 
      30 # computes right hand side of linear system

TypeError: unsupported operand type(s) for *: 'float' and 'NoneType'

Answer 1

假设这是integral_normal()的整个定义，您实际上永远不会从函数返回值。默认情况下，如果您没有return语句，python将返回None，这意味着您对第28行的integral_normal(p_i, p_j)的递归调用将始终返回None。这有很多话要说“你需要实际返回一个值。”

TypeError - 无法乘以＆＃39; float＆＃39; by＆＃39; NoneType＆＃39;

1 个答案: