python中的reduce内建函数使用方法指南 python中的reduce内建函数使用方法指南

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) 
#这种方式用lambda表示当做参数，因为没有提供reduce的第三个参数，所以第一次执行时x=1,y=2,第二次x=1+2,y=3,即列表的第三个元素 
 
 
#或者 
lst = [1,2,3,4,5] 
print reduce(lambda x,y:x+y,lst,0) 
#这种方式用lambda表示当做参数，因为指定了reduce的第三个参数为0，所以第一次执行时x=0,y=1,第二次x=0+1,y=2,即列表的第二个元素， 
假定指定reduce的第三个参数为100，那么第一次执行x=100,y仍然是遍历列表的元素，最后得到的结果为115 
 
 
 
#或者 
def add(x,y): 
  return x+y 
 
print reduce(add, lst) 
#与方式1相同，只不过把lambda表达式换成了自定义函数 
 
#或者 
def add(x,y): 
  return x+y 
 
print reduce(add, lst,0) 
#与方式2相同，只不过把lambda表达式换成了自定义函数 

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,{})  
#提供第三个参数，第一次，初始字典为空，作为statistics的第一个参数，然后遍历lst,作为第二个参数，然后将返回的字典集合作为下一次的第一个参数 
 
或者 
d = {} 
d.extend(lst) 
print reduce(statistics,d) 
#不提供第三个参数，但是要在保证集合的第一个元素是一个字典对象，作为statistics的第一个参数，遍历集合依次作为第二个参数