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Immutable object trap

Idea: generate 1 million random strings, merge strings using non-object or mutable objects, respectively, and compare the time spent.

The code that generates a random string:

Import random import time

Big = [chr (I) for i in range (65mem91)] small = [chr (I) for i in range (97123)]

Def randNStr (n):...: result = []...: for i in range (n):...: strlen = random.randint (3Magne9)...: result.append ('.join (random.sample (small+big,strlen)...: return result

Use the immutable object method: combine+s is then assigned to combine, this statement: because combine, s are immutable objects, the result is a new object is generated, objects continue to pile up.

Def immfn (n):...: combine=''...: res = randNStr (n): beg = time.time (): print ('beg time% s'%str (beg))...: for s in res:...: combine= combine + s...: # print (combine). ..: end = time.time ().: print ('end time% s'%str (end)).: print (' elispe time:% s'%str (end-beg))

In order to avoid repeatedly generating new objects, using mutable objects, the objects that Python solves are mutable and immutable. It is mentioned that generally iterated objects are mutable.

Def mutfn (n):...: combine=''...: res = randNStr (n): beg = time.time (): print ('beg time% s'%str (beg))...: combine='' .join (res).: # print (' print result:%s'%combine)..: End = time.time ().: print ('end time% s'%str (end)).: print (' elispe time:% s'%str (end-beg))

Str.join (iter) is used here to accept iterable objects and return only one new object. Compare the calculation time of the above two methods, from 100000, step 100000, termination to 1 million, and compare and analyze the time of merging strings.

We can see that the method in which variable objects generate only one new object has more obvious advantages than immutable objects in the case of larger amount of data, showing an expanded trumpet. When the data is 900000, the time cost savings are as high as (0.025) / 0.025 = 6.6times. When the amount of data increases to 10 million or more, the advantage is more obvious.

Mutable object trap

BAT interview question 26: a Python blank question

The title is as follows:

Def f (xPowerl = []):...: for i in range (x):...: l.append (iTuni)...: print (l)

F (2) f (3)

The result is:

In [2]: F (2) [0,1]

In [3]: F (3) [0,1,0,1,4]

For the reason, the variable object is used as the function parameter and is set to the default value, and when the function is called again, the original value of the function parameter will be indexed again.

To avoid this problem, the default value of a mutable object is generally constant: None

Def f (XMagneNone):...: if l is None:...: l = [].: for i in range (x):...: l.append (iTuni)...: print (l)...:

When called continuously in this way, the output will be as expected.

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