Python Create and Access dictionaries

Source: Internet
Author: User

>>> Dict1 = {' A ': ' 1 ', ' B ': ' 2 ', ' C ': ' 3 ', ' d ': ' 4 '}

>>> print (' A's value is: ', dict1[' a '])

The value of a is: 1


>>> Dict4 = dict (' I ' = ' happy ', you = ' sadness ')

Syntaxerror:keyword can ' t be an expression


>>> dict4[' you '] = ' change sadness '

>>> Dict4

{' I ': ' Happy ', ' you ': ' Change Sadness '}


>>> dict4[' he '] = ' Add one '

>>> Dict4

{' I ': ' Happy ', ' you ': ' Change sadness ', ' he ': ' Add one '}


Fromkeys (S[,v]) creates and returns a new dictionary

>>> Dict1 = {}

>>> Dict1.fromkeys ((+))

{1:none, 2:none, 3:none}


>>> Dict1.fromkeys ((+), ' num ')

{1: ' num ', 2: ' Num ', 3: ' Num '}


>>> Dict1.fromkeys ((1,3), ' number ') # If you want to change the value of the key with Fromkeys, the new dictionary will not be created successfully

{1: ' Number ', 3: ' Number '}


Ways to access a dictionary

Keys ()

VALUES ()

Items ()


>>> dict2 = {}

>>> Dict2 = Dict2.fromkeys (range (2), ' the ' only ')

>>> Dict2

{0: ' the ' ", 1: ' The '"}

>>> for Eachkey in Dict2. keys (): #values as ' an '

Print (Eachkey)

0

1


>>> for Eachitem in Dict2. items ():

Print (Eachitem)

(0, ' a few ')

(1, ' a few ')


Get () Find Method

>>> Dict2 = Dict2.fromkeys (range (10), ' the ' only ')

>>> Dict2

{0: ' A few ', 1: ' The ', ' 2: ' The ', ' 3: ' The ', ' 4: ' The ', ' 5: ' The ', ' 6: ' The ', ' 7

>>> dict2.get (5, ' none ') #如果有则返回值, if not, return ' none '

' A few '

>>> Dict2.get (10, ' none ')

No


>>> 5 in Dict2 #成员操作符判定

True

>>> in Dict2

False


Clean () Empty A dictionary

>>> Dict2.fromkeys (Range (1), ' happy ')

{0: ' Happy '}

>>> Dict2.clear ()

>>> Dict2

{}


Copy ()

>>> a = {1: ' One ', 2: ' One ', 3: ' Three ', 4: ' Four '}

>>> A

{1: ' One ', 2: ' Both ', 3: ' Three ', 4: ' Four '}

>>> B = a.copy ()

>>> C = A

>>> b

{1: ' One ', 2: ' Both ', 3: ' Three ', 4: ' Four '}

>>> C

{1: ' One ', 2: ' Both ', 3: ' Three ', 4: ' Four '}

>>>

>>> c[5] = ' five '

>>> C

{1: ' One ', 2: ' Both ', 3: ' Three ', 4: ' Four ', 5: ' Five '}

>>> A

{1: ' One ', 2: ' Both ', 3: ' Three ', 4: ' Four ', 5: ' Five '}

>>> b

{1: ' One ', 2: ' Both ', 3: ' Three ', 4: ' Four '}


Pop ()

Popitem () Random Popup

>>> A.pop (2)

' Both '

>>> A

{1: ' One ', 3: ' Three ', 4: ' Four ', 5: ' Five '}


>>> A.popitem ()

(5, ' five ')

>>> A

{1: ' One ', 3: ' Three ', 4: ' Four '}


>>> A.setdefault(7, ' seven ')

' Seven '

>>> A

{1: ' One ', 3: ' Three ', 4: ' Four ', 7: ' Seven '}

>>> B = {7: ' SEVEN '}

>>> A

{1: ' One ', 3: ' Three ', 4: ' Four ', 7: ' Seven '}

>>> A.update (b)

>>> A

{1: ' One ', 3: ' Three ', 4: ' Four ', 7: ' SEVEN '}








Python Create and Access dictionaries

Related Article

Contact Us

The content source of this page is from Internet, which doesn't represent Alibaba Cloud's opinion; products and services mentioned on that page don't have any relationship with Alibaba Cloud. If the content of the page makes you feel confusing, please write us an email, we will handle the problem within 5 days after receiving your email.

If you find any instances of plagiarism from the community, please send an email to: info-contact@alibabacloud.com and provide relevant evidence. A staff member will contact you within 5 working days.

A Free Trial That Lets You Build Big!

Start building with 50+ products and up to 12 months usage for Elastic Compute Service

  • Sales Support

    1 on 1 presale consultation

  • After-Sales Support

    24/7 Technical Support 6 Free Tickets per Quarter Faster Response

  • Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.