Question

任何人都可以在这个python 3.2代码中找到语法错误吗？

import math
import numpy

vinf = 10
vfr = 4
x = numpy.linspace(-10,10,50)
y = numpy.linspace(-10,10,50)
a = numpy.zeros(shape=(len(x),len(y)))
b = numpy.zeros(shape=(len(x),len(y)))
u = numpy.zeros(shape=(len(x),len(y)))
v = numpy.zeros(shape=(len(x),len(y)))
p = numpy.zeros(shape=(len(x),len(y)))
s = numpy.zeros(shape=(len(x),len(y)))


for i in range (0,len(x)):
    for j in range (0,len(y)):
        a[i][j]=x[i]
        b[i][j]=y[j]


for i in range (0,len(x)):
    for j in range (0,len(y)):

        v[i][j] = (vfr/(2*math.pi))*math.pow((b[i][j]+.0047)/(a[i][j]+.05)+(b[i][j]+.0517)/(a[i][j]+.2084)+(b[i][j]-.0271)/(a[i][j]+.2084)+(b[i][j]+.0583)/(a[i][j]+.525)+(b[i][j]-.0228)/(a[i][j]+.525)+(b[i][j]+.0242)/(a[i][j]+.8417)+(b[i][j]-.0087)/(a[i][j]+.8417),-2)*((b[i][j]+.0047)/(math.pow(a[i][j]+.05,2)*(1+math.pow((b[i][j]+.0047)/(a[i][j]+.05),2)))+(b[i][j]+.0517)/(math.pow(a[i][j]+.2084,2)*(1+math.pow((b[i][j]+.0517)/(a[i][j]+.2084),2)))+(b[i][j]-.0271)/(math.pow(a[i][j]+.2084,2)*(1+math.pow((b[i][j]-.0271)/(a[i][j]+.2084),2)))+(b[i][j]+.0583)/(math.pow(a[i][j]+.525,2)*(1+math.pow((b[i][j]+.0583)/(a[i][j]+.525),2)))+(b[i][j]-.0228)/(math.pow(a[i][j]+.525,2)*(1+math.pow((b[i][j]-.0228)/(a[i][j]+.525),2)))+(b[i][j]+.0242)/(math.pow(a[i][j]+.8417,2)*(1+math.pow((b[i][j]+.0242)/(a[i][j]+.8417),2)))+(b[i][j]-.0087)/(math.pow(a[i][j]+.8417,2)*(1+math.pow((b[i][j]+.0087)/(a[i][j]+.8417),2))))
        u[i][j] = vinf+(vfr/(2*math.pi))*math.pow((b[i][j]+.0047)/(a[i][j]+.05)+(b[i][j]+.0517)/(a[i][j]+.2084)+(b[i][j]-.0271)/(a[i][j]+.2084)+(b[i][j]+.0583)/(a[i][j]+.525)+(b[i][j]-.0228)/(a[i][j]+.525)+(b[i][j]+.0242)/(a[i][j]+.8417)+(b[i][j]-.0087)/(a[i][j]+.8417),-2)*(1/((a[i][j]+.05)*(1+math.pow((b[i][j]+.0047)/(a[i][j]+.05),2)))+1/((a[i][j]+.2084)*(1+math.pow((b[i][j]+.0517)/(a[i][j]+.2084),2)))+1/((a[i][j]+.2084)*(1+math.pow((b[i][j]-.0271)/(a[i][j]+.2084),2)))+1/((a[i][j]+.525)*(1+math.pow((b[i][j]+.0583)/(a[i][j]+.525),2)))+1/((a[i][j]+.525)*(1+math.pow((b[i][j]-.0228)/(a[i][j]+.525),2)))+1/(a[i][j]+.8417)*(1+math.pow((b[i][j]+.0242)/(a[i][j]+.8417),2)))+1/((a[i][j]+.8417)*(1+math.pow((b[i][j]+.0087)/(a[i][j]+.8417),2))))
        p[i][j] = 1-math.pow(math.sqrt(math.pow(u[i][j],2)+math.pow(v[i][j],2)),2)

file = open("data.dat","w")
file.write('title = "Driven Cavity"\r\n')
file.write('variables = "x", "y", "u", "v", "p"\r\n')
file.write('ZONE T="All"\r\n')
file.write(' I={}'.format(len(x)))
file.write(' J={}'.format(len(x)))
file.write(' K=1')
file.write(' ZONETYPE=Ordered\r\n')
file.write(' DATAPACKING=POINT\r\n')

for i in range (0,len(x)):
    for j in range (0,len(y)):
        file.write('{}\t'.format(a[i][j]))
        file.write('{}\t'.format(b[i][j]))
        file.write('{}\t'.format(u[i][j]))
        file.write('{}\t'.format(v[i][j]))
        file.write('{}\t\n'.format(p[i][j]))

file.close()

错误表明它位于u [i] [j] =行的末尾它可能是一个括号或我无法看到的东西。从昨晚5点开始，我一直在努力，现在是第二天下午3点40分。

Answer 1

行尾的

额外)，用以下内容替换该行：

u[i][j] = vinf+(vfr/(2*math.pi))*math.pow((b[i][j]+.0047)/(a[i][j]+.05)+(b[i][j]+.0517)/(a[i][j]+.2084)+(b[i][j]-.0271)/(a[i][j]+.2084)+(b[i][j]+.0583)/(a[i][j]+.525)+(b[i][j]-.0228)/(a[i][j]+.525)+(b[i][j]+.0242)/(a[i][j]+.8417)+(b[i][j]-.0087)/(a[i][j]+.8417),-2)*(1/((a[i][j]+.05)*(1+math.pow((b[i][j]+.0047)/(a[i][j]+.05),2)))+1/((a[i][j]+.2084)*(1+math.pow((b[i][j]+.0517)/(a[i][j]+.2084),2)))+1/((a[i][j]+.2084)*(1+math.pow((b[i][j]-.0271)/(a[i][j]+.2084),2)))+1/((a[i][j]+.525)*(1+math.pow((b[i][j]+.0583)/(a[i][j]+.525),2)))+1/((a[i][j]+.525)*(1+math.pow((b[i][j]-.0228)/(a[i][j]+.525),2)))+1/(a[i][j]+.8417)*(1+math.pow((b[i][j]+.0242)/(a[i][j]+.8417),2)))+1/((a[i][j]+.8417)*(1+math.pow((b[i][j]+.0087)/(a[i][j]+.8417),2)))

Answer 2

将该行分成几个较小的操作，以便隔离错误。尝试从u[i][j] = 0开始，然后连续添加整行的小部分，直到错误发生，然后你就会知道在哪里寻找它。

Answer 3

将其粘贴到vim并突出显示最后一个字符表明，至少还有一个额外的关闭。但是，我建议将其分成若干行，以便可以手动调试。出于可读性原因，任何语言中的长线都不是一个好主意。

python 3.2语法错误：语法无效

3 个答案: