Python implements a way to create a dictionary without having to invoke it too much _python

This article illustrates the Python implementation method of creating a dictionary without having to invoke it too much. Share to everyone for your reference. The implementation methods are as follows:

1. Using the Itertools module

Import itertools
the_key = [' ab ', ', ']
the_vale = [' aaaa ', ' dddddddd ', ' 22222222222 ']
d = dict ( Itertools.izip (The_key,the_vale))
print D

2. Add parameter

Dict = dict (red = 1,bule = 2,yellow = 3)
Print Dict

The result is: {' Yellow ': 3, ' Bule ': 2, ' Red ': 1}

3. Use the built-in zip function
Zip ([iterable,...]) Returns a list of

The_key = [' ab ', ', ']
the_vale = [' aaaa ', ' dddddddd ', ' 22222222222 ']
dict2 = dict (Zip (The_key,the_vale))
Print type (Zip (the_key,the_vale))
print Dict2

Results:

<type ' list ' >
{: ' 22222222222 ', ' ab ': ' AAAA ', ': ' DDDDDDDD '}

The Fromkeys function of 4.dict
Each key created has the same value

Fromkeys (Seq[,value])
Create A new dictionary with the keys from SEQ and values set to value.

The_key = [' ab ', ', ']
the_vale = 0
D = Dict.fromkeys (The_key,the_vale)
Print 

Result: {33:0, ' ab ': 0, ' 22 ': 0}

Import string
Count_by_letter = Dict.fromkeys (string.ascii_lowercase,0)
print Count_by_letter

Results:

{' A ': 0, ' C ': 0, ' B ': 0, ' E ': 0, ' d ': 0, ' G: ' 0, ' F ': 0, ' I ': 0, ' h ': 0, ' K ': 0, ' J ': 0, ' m ': 0, ' l ': 0, ' O ': 0, ' n ': 0, ' Q ': 0, ' P ': 0, ' s ': 0, ' R ': 0, ' U ': 0, ' t ': 0, ' W ': 0, ' ' V ': 0, ' y ': 0, ' X ': 0, ' Z ': 0}

I hope this article will help you with your study of Python programming.