Python Functional Programming
This requires proper look at it, after all functional programming in Python also made reference.
python in functional programming support:
The filter function corresponds to the function of the filter. A Boolean function call bool_functo iterate through each element in seq; so returns a bool_seqreturn value of true elements in a sequence.
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>>>a = [1,2,3,4,5,6,7] >>>b = filter(lambda x: x > 5, a) >>>print b >>>[6,7] |
function map each item is a sequence of function sequentially executed, the following is a sequence for each item multiplied by 2:
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>>> a = map(lambda x:x*2,[1,2,3]) >>> list(a) [2, 4, 6] |
Iterative reduce function call is a function of each sequence item, following a factorial 3:
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>>> reduce(lambda x,y:x*y,range(1,4)) 6
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python replication, deep and shallow copy copy difference (https://home.cnblogs.com/u/xueli/ from: Li m bloggers )
In python, the object assignment is actually a reference to an object. When you create an object and assign it to another variable of time, python and no copy of the object, but only a copy of this object reference
There are three general methods,
alist=[1,2,3,["a","b"]]
(1) direct assignment, refers to the default shallow copy of the object passed it, changed the original list, the assigned b will do the same change
>>> b=alist
>>> print b
[1, 2, 3, ['a', 'b']]
>>> alist.append(5)
>>> print alist;print b
[1, 2, 3, ['a', 'b'], 5]
[1, 2, 3, ['a', 'b'], 5]
(2) copy a shallow copy, no copies of the child object, the original data is changed, the child object will change
>>> import copy
>>> c=copy.copy(alist)
>>> print alist;print c
[1, 2, 3, ['a', 'b']]
[1, 2, 3, ['a', 'b']]
>>> alist.append(5)
>>> print alist;print c
[1, 2, 3, ['a', 'b'], 5]
[1, 2, 3, ['a', 'b']]
Alist >>> [. 3]
[ 'A', 'B']
>>> alist [. 3] .append ( 'CCCC')
>>> Print alist; Print C
[. 1, 2,. 3, [ 'A', 'B', 'CCCC'],. 5]
[. 1, 2,. 3, [ 'a', 'B', 'CCCC']] inside the sub-object is changed
(3) a deep copy, copy contains objects from inside the object, so changing the original object change will not cause a deep copy in any sub-elements
>>> import copy
>>> d=copy.deepcopy(alist)
>>> print alist;print d
[1, 2, 3, ['a', 'b']]
[1, 2, 3, ['a', 'b']]始终没有改变
>>> alist.append(5)
>>> print alist;print d
[1, 2, 3, ['a', 'b'], 5]
[1, 2, 3, ['a', 'b']]始终没有改变
>>> alist[3]
['a', 'b']
>>> alist[3].append("ccccc")
>>> print alist;print d
[1, 2, 3, ['a', 'b', 'ccccc'], 5]
[1, 2, 3, ['a', 'b']] 始终没有改变
/******************************************************************************************************************/
之前见过一道面试题目，还挺有意思，输出测试返回true 还是false
考察点：1，不可变原组 和 可变列表的深拷贝，浅拷贝区别
在python中，对象赋值实际上是对象的引用。当创建一个对象，然后把它赋给另一个变量的时候，python并没有拷贝这个对象，而只是拷贝了这个对象的引用
一般有三种方法，
alist=[1,2,3,["a","b"]]
（1）直接赋值,默认浅拷贝传递对象的引用而已,原始列表改变，被赋值的b也会做相同的改变
>>> b=alist
>>> print b
[1, 2, 3, ['a', 'b']]
>>> alist.append(5)
>>> print alist;print b
[1, 2, 3, ['a', 'b'], 5]
[1, 2, 3, ['a', 'b'], 5]
（2）copy浅拷贝，没有拷贝子对象，所以原始数据改变，子对象会改变
>>> import copy
>>> c=copy.copy(alist)
>>> print alist;print c
[1, 2, 3, ['a', 'b']]
[1, 2, 3, ['a', 'b']]
>>> alist.append(5)
>>> print alist;print c
[1, 2, 3, ['a', 'b'], 5]
[1, 2, 3, ['a', 'b']]
>>> alist[3]
['a', 'b']
>>> alist[3].append('cccc')
>>> print alist;print c
[1, 2, 3, ['a', 'b', 'cccc'], 5]
[1, 2, 3, ['a', 'b', 'cccc']] 里面的子对象被改变了
（3）深拷贝，包含对象里面的自对象的拷贝，所以原始对象的改变不会造成深拷贝里任何子元素的改变
>>> import copy
>>> d=copy.deepcopy(alist)
>>> print alist;print d
[1, 2, 3, ['a', 'b']]
[1, 2, 3, ['a', 'b']]始终没有改变
>>> alist.append(5)
>>> print alist;print d
[1, 2, 3, ['a', 'b'], 5]
[1, 2, 3, ['a', 'b']]始终没有改变
>>> alist[3]
['a', 'b']
>>> alist[3].append("ccccc")
>>> print alist;print d
[1, 2, 3, ['a', 'b', 'ccccc'], 5]
[1, 2, 3, ['a', 'b']] 始终没有改变
/******************************************************************************************************************/
之前见过一道面试题目，还挺有意思，输出测试返回true 还是false
考察点：1，不可变原组 和 可变列表的深拷贝，浅拷贝区别