Question

我试图编写一个程序作为练习，用2个数学函数之一计算从a到b的积分值。我的函数集成应该有f作为集成的数学函数。

from math import *

def g(x):
        return float(x) * float(x) + 3

def h(x):
    return math.cos(float(x) * float(x))

def integrate(f, a, b, n):
    H = (abs(float(a) - float(b)))/float(n)
    ans = 0
    xWaarde = a - H/2
    print xWaarde
    for k in range(1, n+1):
        xWaarde = xWaarde + H
        ans = ans + f(xWaarde) * H
    return ans

print 'available functions:'
print 'g(x) = x^2+3'
while True:
    print 'h(x) = cos(x^2)'
    aIn = float(raw_input('integral from a = '))
    bIn = float(raw_input('to b = '))
    nIn = int(raw_input('Number of subintervals: '))
    while True:
        funcIn = raw_input('Which function do you want to use? (g or h): ')
        if funcIn == 'g':
            integrate(g,aIn,bIn,nIn)
            break
        elif funcIn == 'h':
            integrate(h,aIn,bIn,nIn)
            break
        else:
            print 'This function is not available'


    print 'The definite integral is', integrate(funcIn, aIn, bIn, nIn)
    doorg = raw_input('Do you want to continue? (y or n): ')
    if doorg == 'n':
        break
    else:
        print

完整的回溯如下：

Traceback (most recent call last):
  File "C:/Users/Nick van Stijn/Desktop/Python/Assignment 3.1.py", line 38, in <module>
    print 'The definite integral is', integrate(funcIn, aIn, bIn, nIn)
  File "C:/Users/Nick van Stijn/Desktop/Python/Assignment 3.1.py", line 16, in integrate
    ans = ans + f(xWaarde) * H
TypeError: 'str' object is not callable

编辑：已解决我在调用函数时犯了一个错误，我根本没有调用它。

Answer 1

问题是您使用正确的函数integrate或f来呼叫g，然后丢弃结果，而是拨打{{1} 再次用于打印，这次只传递函数integrate的名称。

相反，您应该将结果存储在变量中，例如：

funcIn

此外，您可以使用result = None while result is None: funcIn = raw_input('Which function do you want to use? (g or h): ') if funcIn == 'g': result = integrate(g,aIn,bIn,nIn) elif funcIn == 'h': result = integrate(h,aIn,bIn,nIn) else: print 'This function is not available' print 'The definite integral is', result将函数名称映射到实际函数，而不是使用可能大量的dict：

if/elif/else

Answer 2

您正在以字符串的形式使用函数的文本名称，而不是对函数对象本身的引用。虽然有一些hacky技术从字符串名称派生函数对象，但它们很难维护并且容易出错。由于在python函数中的对象与任何其他对象（所谓的“第一类”对象）一样，它们并没有真正命名，只有函数的引用才有名称。

这是一个很好的示例，其中字典派上用场，特别是如果您希望稍后添加更多功能。我们可以将文本键（用户输入的内容）映射到任何python对象，包括函数：

from math import *

def g(x):
        return float(x) * float(x) + 3

def h(x):
    return math.cos(float(x) * float(x))

# Store references to the functions in a dictionary
# with the keys as the text name (the names need not match)
funcs = {'g': g, 'h': h}        #  <<<<  ADDED

def integrate(f, a, b, n):
    H = (abs(float(a) - float(b)))/float(n)
    ans = 0
    xWaarde = a - H/2
    print xWaarde
    for k in range(1, n+1):
        xWaarde = xWaarde + H
        ans = ans + f(xWaarde) * H
    return ans

print 'available functions:'
print 'g(x) = x^2+3'
while True:
    print 'h(x) = cos(x^2)'
    aIn = float(raw_input('integral from a = '))
    bIn = float(raw_input('to b = '))
    nIn = int(raw_input('Number of subintervals: '))
    while True:
        funcIn = raw_input('Which function do you want to use? (g or h): ')

        # THIS CODE CHANGED - note the simplification
        # we just test for membership of the dictionary
        if funcIn in funcs:
            integrate(funcs[funcIn],aIn,bIn,nIn)
            break
        else:
            print 'This function is not available'

    # THIS CODE CHANGED (note first argument to integrate)
    print 'The definite integral is', integrate(funcs[funcIn], aIn, bIn, nIn)
    doorg = raw_input('Do you want to continue? (y or n): ')
    if doorg == 'n':
        break
    else:
        print

＆＃39; STR＆＃39; object不可调用，作为args

2 个答案: