我的 Coursera 问题答案

【问题标题】：Coursera issue for me我的 Coursera 问题
【发布时间】：2021-10-16 16:50:56
【问题描述】：

请帮帮我，我在做一个在线课程，有人问我这个问题

fractional_part 函数将分子除以分母，只返回小数部分（0 到 1 之间的数字）。完成函数的主体，使其返回正确的数字。注意：由于除以 0 会产生错误，因此如果分母为 0，则函数应返回 0 而不是尝试除法。

这是我的代码：

def fractional_part(numerator, denominator):
# Operate with numerator and denominator to 
# keep just the fractional part of the quotient
   if denominator == 0 or numerator == 0:
      print("0")
   else:
        fraction = numerator / denominator
        while fraction > 1:
           fraction1 = fraction - 1
   return fraction1

print(fractional_part(5, 5)) # Should be 0
print(fractional_part(5, 4)) # Should be 0.25
print(fractional_part(5, 3)) # Should be 0.66...
print(fractional_part(5, 2)) # Should be 0.5
print(fractional_part(5, 0)) # Should be 0
print(fractional_part(0, 5)) # Should be 0

这是我的输出：

UnboundLocalError：在赋值之前引用了局部变量“fraction1”

【问题讨论】：

fraction1 在while fraction > 1: 中定义。这意味着如果 fraction 为 1 或更小，则不会定义 fraction1
在语句的 else 部分，在 while 循环中。如果分数小于 1，fraction1 = fraction - 1 将被忽略并跳转到 return 语句。在 return 语句中，它不知道 fraction1 是什么，因为您从未使用过它/从未将其分配给某物
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标签： python php python-3.x function

【解决方案1】：

你必须在你的代码中定义fraction1。

您只是在while 循环内创建fraction1 变量，并且while 循环仅在fraction > 1 时运行。

如果fraction <= 1 怎么办？您没有在代码中的任何位置创建任何名为fraction1 的变量。

最后，您的函数返回未定义的 fraction1。

您还必须在循环内将 fraction 减 1，否则循环将无限期地继续。

def fractional_part(numerator, denominator):
# Operate with numerator and denominator to 
# keep just the fractional part of the quotient
   if denominator == 0 or numerator == 0:
       return 0
   else:
       fraction = numerator / denominator
       fraction1 = 0
       while fraction > 1:
           fraction1 = fraction - 1
           fraction -= 1
       return fraction1

【讨论】：

【解决方案2】：

如果将fraction1 更改为fraction 将解决fraction1 未定义的问题。那么你需要在第一个if之后添加fraction = 0来解决如果denominator或numerator是0

def fractional_part(numerator, denominator):
# Operate with numerator and denominator to 
# keep just the fractional part of the quotient
   if denominator == 0 or numerator == 0:
      fraction = 0 #to solve if denominator or numerator is 0
      print("0")
   else:
        fraction = numerator / denominator
        while fraction > 1:
           fraction = fraction - 1
   return fraction

【讨论】：

【解决方案3】：

def fractional_part(numerator, denominator):
    # Operate with numerator and denominator to
    # keep just the fractional part of the quotient
    if denominator == 0:
        return 0 #raise Exception('Not possible')
    elif numerator == 0:
        return 0
    # check if result of division is already smaller than 1
    if numerator < denominator:
       return numerator / denominator

    remainder  = numerator % denominator
    return remainder / denominator

print(fractional_part(23, 5))
print(fractional_part(5, 4))

输出

0.6
0.25

【讨论】：

ẁhile 方法给出近似值：fractional_part(23, 5) -> 0.59999999... 而不是 0.6

【解决方案4】：

要解决这个问题，您需要删除 Floor division，所以我们使用这个 Python 算术运算符 //

def fractional_part(numerator, denominator):
      if denominator == 0 or numerator == 0:
         return 0

      else:
          fraction = ((numerator / denominator)-(numerator // denominator))
          return fraction
    

print(fractional_part(5, 5)) # Should be 0
print(fractional_part(5, 4)) # Should be 0.25
print(fractional_part(5, 3)) # Should be 0.66...
print(fractional_part(5, 2)) # Should be 0.5
print(fractional_part(5, 0)) # Should be 0
print(fractional_part(0, 5)) # Should be 0

输出

0.0
0.25
0.6666666666666667
0.5
0
0

【讨论】：

【解决方案5】：

def fractional_part(numerator, denominator):
    if numerator==0 or denominator==0:
        return 0
    else:
        result=numerator/denominator
        while result>=1:
            result-=1
        return result   

print(fractional_part(5, 5)) # Should be 0
print(fractional_part(5, 4)) # Should be 0.25
print(fractional_part(5, 3)) # Should be 0.66...
print(fractional_part(5, 2)) # Should be 0.5
print(fractional_part(5, 0)) # Should be 0
print(fractional_part(0, 5)) # Should be 0

【讨论】：

您的答案可以通过额外的支持信息得到改进。请edit 添加更多详细信息，例如引用或文档，以便其他人可以确认您的答案是正确的。你可以找到更多关于如何写好答案的信息in the help center。

【解决方案6】：

def fractional_part(numerator, denominator):
    if numerator == denominator or denominator ==0 or numerator == 0:
        return 0
    else:
        result=numerator/denominator
        if result > 1 : 
           result1=result-int(result)
           return result1   
        return result              
    print(fractional_part(5, 5)) # Should be 0
    print(fractional_part(5, 4)) # Should be 0.25
    print(fractional_part(5, 3)) # Should be 0.66...
    print(fractional_part(5, 2)) # Should be 0.5
    print(fractional_part(5, 0)) # Should be 0
    print(fractional_part(0, 5)) # Should be 0

【讨论】：

【解决方案7】：

def fractional_part(numerator, denominator):
    if denominator == 0 or numerator == 0:
       return 0
    else:
      result = numerator / denominator
      if result >= 2:
            result1 = result - int(2)
            return result1
      elif result >= 1:
            result2 = result - int(1)
            return result2

print(fractional_part(5, 5)) # Should be 0
print(fractional_part(5, 4)) # Should be 0.25
print(fractional_part(5, 3)) # Should be 0.66...
print(fractional_part(5, 2)) # Should be 0.5
print(fractional_part(5, 0)) # Should be 0
print(fractional_part(0, 5)) # Should be 0

【讨论】：