选择最大的奇数python

Question

我试图用Python编写一个简单的程序来计算x,y,z值中最大的奇数.如何为用户提供选择x,y和z值的选项？

所以程序会询问x,y和z是什么,然后说"x,y,z是最大的奇数"或者数字都是偶数.

我到目前为止的内容如下.这至少是一个不错的开始吗？

  # This program exmamines variables x, y, and z 
  # and prints the largest odd number among them

  if x%2 !== 0 and x > y and y > z:
      print 'x is the largest odd among x, y, and z'
  elif y%2 !== 0 and y > z and z > x:
     print 'y is the largest odd among x, y, and z'
  elif z%2 !== 0 and z > y and y > x:
     print 'z is the largest odd among x, y, and z'
  elif x%2 == 0 or y%2 == 0 or z%2 == 0:
     print 'even'

有了thkang帖子,我现在有:

  # This program exmamines variables x, y, and z 
  # and prints the largest odd number among them

  if x%2 !== 0:
    if y%2 !== 0:
      if z%2 !== 0:
        if x > y and x > z: #x is the biggest odd
        elif y > z and y > x: #y is the biggest odd
        elif z > x and z > y: #z is the biggest odd

      else: #z is even
        if x > y: #x is the biggest odd
        else: #y is the biggest odd

    else: #y is even
      if z%2 != 0: #z is odd
        if x > z: #x is the biggest odd
        else: #z is the biggest odd
      else: #y,z are even and x is the biggest odd

  else: #x is even
    if y%2 != 0 and z%2 != 0; #y,z is odd
      if y > z: #y is the biggest odd
      else: #z is the biggest odd
    else: #x and y is even
      if z%2 != 0: #z is the biggest odd

Answer 1

途径

避免使用if-stmts以找到最大值.使用python内置max.使用任一发生器或filter仅查找奇数.

使用这样的内置函数更安全/更可靠,因为组合它们更简单,代码经过了充分测试,代码主要在C中执行(而不是多字节代码指令).

码

def find_largest_odd(*args):
    return max(arg for arg in args if arg & 1)

要么:

def find_largest_odd(*args):
    return max(filter(lambda x: x & 1, args))

测试

>>> def find_largest_odd(*args):
...     return max(arg for arg in args if arg & 1)
... 
>>> print find_largest_odd(1, 3, 5, 7)
7
>>> print find_largest_odd(1, 2, 4, 6)
1

和:

>>> def find_largest_odd(*args):
...     return max(filter(lambda x: x & 1, args))
>>> print find_largest_odd(1, 3, 5, 7)
7
>>> print find_largest_odd(1, 2, 4, 6)
1

如果您传递一个空序列或仅提供偶数,您将获得ValueError:

>>> find_largest_odd(0)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 2, in find_largest_odd
ValueError: max() arg is an empty sequence

参考

• 过滤
• 最大
• 发电机

Answer 2

如果有人正在寻找使用条件语句的简单解决方案

x,y,z = 4,1,6
largest = None
if x%2:
 largest = x
if y%2:
 if y > largest:
  largest = y
if z%2:
 if z > largest:
  largest = z
if largest:
 print "largest number is" largest
else
 print "there are no odd numbers"

Answer 3

最好先过滤数字，然后对它们进行排序。

numbers = [x, y, z]

sorted_odd_nums = sorted((x for x in enumerate(numbers) if x[1]%2), 
                         key = lambda x:x[1], 
                         reverse=True)

if not sorted_odd_nums:
   # all numbers were even and filtered out.
elif sorted_odd_nums[0][0] == 0:
   # x is the biggest odd number
elif sorted_odd_nums[0][0] == 1:
   # y is the biggest odd number
elif sorted_odd_nums[0][0] == 2:
   # z is the biggest odd number

它能做什么：

enumerate(numbers)返回一个(index, item)对的序列。由于原始列表是，所以即使在过滤和排序之后，[x, y, z]我们也可以跟踪x, 。yz

(x for x in enumerate(numbers) if x[1]%2)如果给定元组中的第二项不是偶数，则过滤上述枚举。

sort( ... , key=lambda x:x[1], reverse=True)使用第二个索引项目的值（原始数字）按降序对过滤项目进行排序。

用户输入

要从用户那里读取，最简单的方法是使用raw_input(py2) / input(py3k)。

number = int(raw_input('enter a number: '))

仅使用 if 语句

你必须嵌套 if 语句。喜欢：

if x%2: # x is odd
  if y%2: # y is odd
    if z%2: #z is odd
      if x>y and x>z: #x is the biggest odd number
      elif y>z and y>x: #y is the biggest odd number
      elif z>x and z>y: #z is the biggest odd number

    else: #z is even
      if x>y: #x is the biggest odd number
      else: #y is the biggest odd number
  else: #y is even
    if z%2: #z is odd
...