使用if-statement定义函数
使用if-statement定义函数
提问于 2016-07-29 14:14:57
def roots4(a,b,c,d):
d = b * b - 4 * a * c
if a != 0 and d == 0:
roots4(a,b,c,d)
x = -b/ (2*a)
if a != 0 and d > 0:
roots4(a,b,c,d)
x1 = (-b + math.sqrt(d)) / 2.0 / a
x2 = (-b - math.sqrt(d)) / 2.0 / a
if a != 0 and d < 0:
roots4(a,b,c,d)
xre = (-b) / (2*a)
xim = (math.sqrt(d))/ (2*a)
print x1 = xre + ixim
strx1 = "x2 = %6.2f + i %6.2f" %(xre, xim)
print strx1 这是我的项目代码的一部分。我想要做的是定义roots4(a,b,c,d)。例如，在a != 0 and d == 0然后roots4(a,b,c,d)将转到通过求解方程x = -b/ (2*a)来查找x的情况下。以此类推。我不知道我做错了什么。有什么建议吗？
回答 3
Stack Overflow用户
回答已采纳
发布于 2016-07-29 15:49:48
正如评论和jimmy的回答中所指出的，不需要递归调用。
这里有一个如何纠正它的例子(以你喜欢的方式调整它)：
#!/usr/bin/env python2
# -*- coding: utf-8 -*-
import math
def roots4(a, b, c):
"""Prints the solutions of ax² + bx + c = 0, if a != 0"""
if a == 0:
print 'This function is meant to solve a 2d degree equation, '\
'not a 1st degree one.'
else:
d = b * b - 4 * a * c
solutions = []
if d == 0:
solutions = [str(-b / (2 * a))]
elif d > 0:
solutions = [str((-b + math.sqrt(d)) / 2.0 / a),
str((-b - math.sqrt(d)) / 2.0 / a)]
elif d < 0:
xre = str((-b) / (2 * a))
xim = str((math.sqrt(-d)) / (2 * a))
solutions = [xre + " + i" + xim,
xre + " - i" + xim]
print "\nEquation is: {}x² + {}x + {} = 0".format(a, b, c)
if len(solutions) == 1:
print "There's only one solution: " + solutions[0]
else:
print "Solutions are: " + " and ".join(solutions)
roots = [(0.0, 0.0, 0.0),
(0.0, 0.0, 1.0),
(0.0, 2.0, 4.0),
(1.0, 2.0, 1.0),
(1.0, -5.0, 6.0),
(1.0, 2.0, 3.0)]
for r in roots:
roots4(*r)输出：
$ ./test_script2.py
This function is meant to solve a 2d degree equation, not a 1st degree one.
This function is meant to solve a 2d degree equation, not a 1st degree one.
This function is meant to solve a 2d degree equation, not a 1st degree one.
Equation is: 1.0x² + 2.0x + 1.0 = 0
There's only one solution: -1.0
Equation is: 1.0x² + -5.0x + 6.0 = 0
Solutions are: 3.0 and 2.0
Equation is: 1.0x² + 2.0x + 3.0 = 0
Solutions are: -1.0 + i1.41421356237 and -1.0 - i1.41421356237Stack Overflow用户
发布于 2016-07-29 16:05:47
你很可能堆积了
...
roots4(a,b,c,d)
...这会导致无限循环。
首先，为什么你需要一个递归调用？d参数是做什么用的？
其次，什么是ixim？它应该是像xim * 1j一样的东西吗？你对print x1 = xre + ixim有什么期望？
如果你只想在d < 0的情况下打印，这是很好的
from math import sqrt
def roots4(a,b,c):
if a != 0.:
x_left = -b/(2*a)
d = b * b - 4 * a * c
if d == 0.:
x_right = 0.
elif d > 0.:
x_right = sqrt(d) / (2 * a)
else:
xim = sqrt(-d) / (2 * a)
strx1 = "x1 = %6.2f + i %6.2f" %(x_left, xim)
print strx1
strx2 = "x2 = %6.2f - i %6.2f" %(x_left, xim)
print strx2
x_right = xim * 1j
x1 = x_left + x_right
x2 = x_left - x_right
else:
raise ValueError("incorrect leading coefficient in given square equation")Stack Overflow用户
发布于 2016-07-29 15:31:18
好的，看起来你已经用你的递归函数做了一个无限循环。现在看来，你永远不会做任何计算，除了
d = b * b - 4 * a * c一遍又一遍。
考虑程序流。每次你到达
roots4(a,b,c,d)你最终会再次名列前茅。
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原文链接：
https://stackoverflow.com/questions/38651714
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