python - 如何在 Python 中压缩一堆函数？

Question

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!

Answer 1

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