[JavaScript] python之函数式编程与函数闭包

In [1]: def a(x):
...:   if x > 20:
...:     return True
...:   else:
...:     return False
...:
In [2]: l1 = [1,2,3,4,20,21,25,40,50,32,47]
In [3]: filter(a,l1)
Out[3]: [21, 25, 40, 50, 32, 47]

In [1]: l1 = [0,1,2,3,4,5,6]
In [2]: l2 = ['Sun','Mon','Tue','Wed','Thu','Fri','Sat']
In [3]: map(None,l1,l2)
Out[3]:
[(0, 'Sun'),
(1, 'Mon'),
(2, 'Tue'),
(3, 'Wed'),
(4, 'Thu'),
(5, 'Fri'),
(6, 'Sat')]

In [4]: l1 = [0,1,2,3,4,5,6]
In [5]: l2 = ['Sun','Mon','Tue','Wed','Thu','Fri','Sat']
In [6]: def a(x,y):      #这里的函数参数个数要与map函数中序列的个数一致
...:   return x2,y2
...:
In [7]: map(a,l1,l2)
Out[7]:
[(0, 'SunSun'),
(2, 'MonMon'),
(4, 'TueTue'),
(6, 'WedWed'),
(8, 'ThuThu'),
(10, 'FriFri'),
(12, 'SatSat')]

In [9]: l1 = [0,1,2,3,4,5,6]
In [10]: def a(x,y):
...:   return x + y
...:
In [11]: reduce(a,l1)    #返回所有参数之和
Out[11]: 21
In [12]: reduce(a,l1,10)   #返回所有参数+初始值之和
Out[12]: 31

In [1]: def func1(x):      #外层函数
...:   def func2(y):    #内层函数
...:     return y ** x
...:   return func2
...:
In [2]: f4 = func1(4)
In [3]: type(f4)
Out[3]: function
In [4]: f4(2)
Out[4]: 16
In [5]: f4(3)
Out[5]: 81

In [6]: def startPos(m,n):    #象棋起始位置
...:   def newPos(x,y):    #象棋新位置
...:     return "The old position is (%d,%d),and the new position is (%d,%d)."% (m,n,m+x,n+y)
...:   return newPos
...:
In [7]: action = startPos(10,10)
In [8]: action(1,2)
Out[8]: 'The old position is (10,10),and the new position is (11,12).'
In [9]: action = startPos(11,12)
In [10]: action(3,-2)
Out[10]: 'The old position is (11,12),and the new position is (14,10).'