Apply function of two arguments cumulatively to the items of iterable, from left to right, so as to reduce the iterable to a single value. For example, reduce(lambda x, y: x+y, [1, 2, 3, 4, 5]) calculates ((((1+2)+3)+4)+5). The left argument, x, is the accumulated value and the right argument, y, is the update value from the iterable. If the optional initializer is present, it is placed before the items of the iterable in the calculation, and serves as a default when the iterable is empty. If initializer is not given and iterable contains only one item, the first item is returned. Roughly equivalent to：


          

            
def reduce(function, iterable, initializer=None): 
  it = iter(iterable) 
  if initializer is None: 
    try: 
      initializer = next(it) 
    except StopIteration: 
      raise TypeError('reduce() of empty sequence with no initial value') 
  accum_value = initializer 
  for x in iterable: 
    accum_value = function(accum_value, x) 
  return accum_value 

          

            
sum(lst) 
          

            
def customer_sum(lst): 
  result = 0 
  for x in lst: 
    result+=x 
  return result 
 
#或者 
def customer_sum(lst): 
  result = 0 
  while lst: 
      temp = lst.pop(0) 
      result+=temp 
  return result 
 
if __name__=="__main__": 
  lst = [1,2,3,4,5] 
  print customer_sum(lst) 

          

            
def add(lst,result): 
  if lst: 
    temp = lst.pop(0) 
    temp+=result 
    return add(lst,temp) 
  else: 
    return result 
 
if __name__=="__main__": 
  lst = [1,2,3,4,5] 
  print add(lst,0) 

          

            
lst = [1,2,3,4,5] 
print reduce(lambda x,y:x+y,lst) 
#這種方式用lambda表示當做參數，因為沒有提供reduce的第三個參數，所以第一次執行時x=1,y=2,第二次x=1+2,y=3,即列表的第三個元素 
 
 
#或者 
lst = [1,2,3,4,5] 
print reduce(lambda x,y:x+y,lst,0) 
#這種方式用lambda表示當做參數，因為指定了reduce的第三個參數為0，所以第一次執行時x=0,y=1,第二次x=0+1,y=2,即列表的第二個元素， 
假定指定reduce的第三個參數為100，那么第一次執行x=100,y仍然是遍歷列表的元素，最后得到的結果為115 
 
 
 
#或者 
def add(x,y): 
  return x+y 
 
print reduce(add, lst) 
#與方式1相同，只不過把lambda表達式換成了自定義函數 
 
#或者 
def add(x,y): 
  return x+y 
 
print reduce(add, lst,0) 
#與方式2相同，只不過把lambda表達式換成了自定義函數 
          

            
def statistics(lst): 
  dic = {} 
  for k in lst: 
    if not k in dic: 
      dic[k] = 1 
    else: 
      dic[k] +=1 
  return dic 
 
lst = [1,1,2,3,2,3,3,5,6,7,7,6,5,5,5] 
print(statistics(lst)) 

          

            
def statistics2(lst): 
  m = set(lst) 
  dic = {} 
  for x in m: 
    dic[x] = lst.count(x) 
 
  return dic 
 
lst = [1,1,2,3,2,3,3,5,6,7,7,6,5,5,5] 
print statistics2(lst) 

          

            
def statistics(dic,k): 
  if not k in dic: 
    dic[k] = 1 
  else: 
    dic[k] +=1 
  return dic 
 
lst = [1,1,2,3,2,3,3,5,6,7,7,6,5,5,5] 
print reduce(statistics,lst,{})  
#提供第三個參數，第一次，初始字典為空，作為statistics的第一個參數，然后遍歷lst,作為第二個參數，然后將返回的字典集合作為下一次的第一個參數 
 
或者 
d = {} 
d.extend(lst) 
print reduce(statistics,d) 
#不提供第三個參數，但是要在保證集合的第一個元素是一個字典對象，作為statistics的第一個參數，遍歷集合依次作為第二個參數