python之如何在 Python 中压缩一堆函数

所以我正在尝试制作一个转换单位的程序，这样我就可以避免在我的物理课上使用因子标签方法。我已经弄清楚了它的通用代码，但到最后我将定义 35 个函数。下面是我到目前为止的代码，我希望能够在每个单元之间进行转换。在它们的正下方是使用这些单位的每种可能的转换。问题是我不想为每个可能的转换创建函数。到目前为止我已经完成的代码只适用于其中的几个单元，但可以在这个链接的小饰品中找到它:

             # km  &  m  &  cm  &  mm  &  M  &  ft  &  In  
#km to m   |  m to km  |  cm to km  |  mm to km  |  M to km  |  ft to km  |  In to km 
#km to cm  |  m to cm  |  cm to m   |  mm to m   |  M to m   |  ft to m   |  In to m 
#km to mm  |  m to mm  |  cm to mm  |  mm to cm  |  M to cm  |  ft to cm  |  In to cm 
#km to M   |  m to M   |  cm to M   |  mm to M   |  M to mm  |  ft to mm  |  In to mm 
#km to ft  |  m to ft  |  cm to ft  |  mm to ft  |  M to ft  |  ft to M   |  In to M 
#km to In  |  m to In  |  cm to In  |  mm to In  |  M to In  |  ft to In  |  In to ft 
 
 
 
def km_to_M_conv(): 
  km=float(input("How many km?")) 
  result = km * .621371192 
  sentence = '{} km is equal to {} M.'.format(km, result) 
  print sentence 
 
def M_to_km_conv(): 
  M=float(input("How many m?")) 
  result = M * 1.60934 
  sentence = '{} M is equal to {} km.'.format(M, result) 
  print sentence 
 
def km_to_m_conv(): 
  km=float(input("How many km?")) 
  result = km * 1000 
  sentence = '{} km is equal to {} m.'.format(km, result) 
  print sentence 
 
def mm_to_cm_conv(): 
  mm=float(input("How many mm?")) 
  result = mm * .1 
  sentence = '{} mm is equal to {} cm.'.format(mm, result) 
  print sentence 
 
def cm_to_mm_conv(): 
  cm=float(input("How many cm?")) 
  result = cm * 10 
  sentence = '{} cm  is equal to {} mm.'.format(cm, result) 
  print sentence 
 
welcome=input("What would you like to convert?") 
if welcome == ("mm to cm"): 
  mm_to_cm_conv() 
if welcome == ("cm to mm"): 
  cm_to_mm_conv() 
if welcome == ("km to M"): 
  km_to_M_conv() 
if welcome == ("M to km"): 
  M_to_km_conv() 

我对 Python 还是很陌生，所以请多多包涵。谢谢!

请您参考如下方法:

提取通用代码:

def converter(source, target, factor): 
  qty = float(input("How many %s?")) 
  result = qty * factor 
  print '{} {} is equal to {} {}.'.format(qty, source, result, target) 
 
def km_to_miles(km): 
   converter('km', 'miles', 0.621371) 

更新:

如果你想变得非常聪明，你可以像这样动态生成函数:

CONVERSIONS = ( 
    ('km', 'miles', 0.621371), 
    ('metres', 'feet', 3.28084), 
    ('litres', 'gallons', 0.264172), 
) 
 
for from, to, factor in CONVERSIONS: 
    func_name = '%s_to_%s' % (from, to) 
    func = lambda x: x * factor 
    globals()[func_name] = func 
 
    reverse_func_name = '%s_to_%s' % (to, from) 
    reverse_func = lambda x: x * (1 / factor) 
    globals()[reverse_func_name] = reverse_func