Convert Strings,list of strings into a python dict or dictionary

"""
Strings ==> Dicts

"""

#problem1
#Input: s = "{'muffin' : 'lolz', 'foo' : 'kitty'}" --> a string
#output:  {'muffin' : 'lolz', 'foo' : 'kitty'} --> dictionary

>>> s
"{'muffin' : 'lolz', 'foo' : 'kitty'}"
>>> from ast import literal_eval
>>> literal_eval(s)
{'muffin': 'lolz', 'foo': 'kitty'}

>>> s
"{'muffin' : 'lolz', 'foo' : 'kitty'}"
>>> import json

>>> json_acceptable_string = s.replace("'", "\"")
>>> d = json.loads(json_acceptable_string)
>>> d
{'muffin': 'lolz', 'foo': 'kitty'}

"""
NOTE that if you have single quotes as a part of your keys or values this will fail due to improper character replacement

"""

#problem:2
#Input: mystring="a=0 b=1 c=3"
#output: {'a': 0, 'b': 1, 'c': 3}

"""
It's easy to convert list into a dict.

"""

In [1]: mystring = "a=0 b=1 c=3"

In [2]: mylist1=mystring.split() #using split for a string generates a list 

In [3]: mylist1
Out[3]: ['a=0', 'b=1', 'c=3']

In [4]: mylist2=[]

In [5]: for i in mylist1: #for every element in list1 i'm caling split at '='
   ...:     mylist2.append(i.split('='))
   ...:     

In [6]: mylist2
Out[6]: [['a', '0'], ['b', '1'], ['c', '3']]

In [7]: dict(mylist2)
Out[7]: {'a': '0', 'b': '1', 'c': '3'} #it worked but values in dictonary are strings not ints

In [8]: mylist2
Out[8]: [['a', '0'], ['b', '1'], ['c', '3']]

#convert the value item into an int  i.e '0'->0, '1'->1,'3'->3  ; mylist2 has 3 lists 
#So for every list in mylist2 i want to change the first element into a int
#mylist2[0][1] is '0'
#mylist2[1][1] is '1'

In [9]: for lists in mylist2: 
   ...:     lists[1]=int(lists[1])
   ...:     

In [10]: mylist2
Out[10]: [['a', 0], ['b', 1], ['c', 3]]


#we can use a single line answer for this

In [1]: mystring = "a=0 b=1 c=3"

In [2]: dict( (n,int(v)) for n,v in (i.split('=') for i in mystring.split() ) )
Out[2]: {'a': 0, 'b': 1, 'c': 3}


#using eval to solve 
#try to avoid using eval.
#eval() interprets a string as code.
>>> a='2*3'
>>> eval(a)
6
mystring = "a=0 b=1 c=3"


In [3]: mydict=eval('dict(%s)'%mystring.replace(' ',','))

In [4]: mydict
Out[4]: {'a': 0, 'b': 1, 'c': 3}

"""
This one took  me a while to understand.

try this 
"""
In [27]: dict(a=0,b=2,c=3)
Out[27]: {'a': 0, 'b': 2, 'c': 3}
#After trying this i came to understand

In [25]: 'dict(%s)' % mystring.replace(' ',',')
Out[25]: 'dict(a=0,b=1,c=3)'

# Invoking eval on the above line gives us desired dictionary





#problem 3
#Input:list_with_strings=["name","Ajay Kumar","age",25,"place","India"]

#Output:{'age': 25, 'name': 'Ajay Kumar', 'place': 'India'} 

In [1]: list_with_strings=["name","Ajay Kumar","age",25,"place","India"]

In [2]: dict(zip(*[iter(list_with_strings)]*2))
Out[2]: {'age': 25, 'name': 'Ajay Kumar', 'place': 'India'}

"""
func(*a) is the same as func(a[0], a[1], a[2], a[3] ... a[n]) if ahad n arguments
* is an argument unpacking
More @ http://stackoverflow.com/questions/287085/what-do-args-and-kwargs-mean/287582#287582
"""
In [37]: list_with_strings=["name","Ajay Kumar","age",25,"place","India"] 

In [38]: l = [iter(list_with_strings)]*2

In [39]: l
Out[39]: [, ]

In [40]: dict(zip(l[0], l[1]))
Out[40]: {'age': 25, 'name': 'Ajay Kumar', 'place': 'India'}


In [41]: def foo(a,b,c,d):
   ....:     print a,b,c,d
   ....:     

In [42]: l=[0,1] 

In [43]: d={"d":3,"c":2}

In [44]: foo(*l,**d) #for arguments we use * and keyword arguments we use **
0 1 2 3

#Easy way to understand this

n [3]: my_iterable=iter(list_with_strings) #iter keyword makes it iterable 

In [4]: dict(zip(my_iterable,my_iterable))
Out[4]: {'age': 25, 'name': 'Ajay Kumar', 'place': 'India'}

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