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Convert String representation of a List to a Dictionary Python

How can I convert a str representation of the list, such as the below string into a dictionary?
a = '[100:0.345,123:0.34,145:0.86]'
Expected output :
{100:0.345,123:0.34,145:0.86}
First tried to convert the string into a list using ast.literal_eval. But it's showing an error as : invalid syntax

It's showing as invalid syntax because it has the wrong brackets, so you could do
ast.literal_eval(a.replace("[","{").replace("]", "}"))
Or alternatively parse the string yourself in a dictionary comprehension
{x.split(":")[0]: x.split(":")[1] for x in a[1:-1].split(",")}
and if as mentioned there are [ or ] elsewhere in your string the following may be more robust
ast.literal_eval("{" + a[1:-1] +"}")

I would simply try
eval(a.replace('[', '{').replace(']', '}'))

To convert to a dict:
Code:
data = '[100:0.345,123:0.34,145:0.86]'
new_data = dict(y.split(':') for y in (x.strip().strip('[').strip(']')
for x in data.split(',')))
print(new_data)
Or if you need numbers not strings:
new_data = dict((map(float, y.split(':'))) for y in (
x.strip().strip('[').strip(']') for x in data.split(',')))
Results:
{'100': '0.345', '123': '0.34', '145': '0.86'}
{145.0: 0.86, 123.0: 0.34, 100.0: 0.345}

Translate brackets to braces, literal_eval.
rebracket = {91: 123, 93: 125}
a = '[100:0.345,123:0.34,145:0.86]'
literal_eval(a.translate(rebracket))

Given a string representation of a dict:
a = '[100:0.345,123:0.34,145:0.86]'
Ignore the containing braces [..], and break up the elements on commas:
a = a[1:-1].split(",")
For each element, separate the key and value:
d1 = [x.split(":") for x in a]
Reconstitute the parsed data as a dict:
d2 = { int(k[0]) : float(k[1]) for k in d1}

How to replace text between parentheses in Python?

I have a dictionary containing the following key-value pairs: d={'Alice':'x','Bob':'y','Chloe':'z'}
I want to replace the lower case variables(values) by the constants(keys) in any given string.
For example, if my string is:
A(x)B(y)C(x,z)
how do I replace the characters in order to get a resultant string of :
A(Alice)B(Bob)C(Alice,Chloe)
Should I use regular expressions?

re.sub() solution with replacement function:
import re
d = {'Alice':'x','Bob':'y','Chloe':'z'}
flipped = dict(zip(d.values(), d.keys()))
s = 'A(x)B(y)C(x,z)'
result = re.sub(r'\([^()]+\)', lambda m: '({})'.format(','.join(flipped.get(k,'')
for k in m.group().strip('()').split(','))), s)
print(result)
The output:
A(Alice)B(Bob)C(Alice,Chloe)
Extended version:
import re
def repl(m):
val = m.group().strip('()')
d = {'Alice':'x','Bob':'y','Chloe':'z'}
flipped = dict(zip(d.values(), d.keys()))
if ',' in val:
return '({})'.format(','.join(flipped.get(k,'') for k in val.split(',')))
else:
return '({})'.format(flipped.get(val,''))
s = 'A(x)B(y)C(x,z)'
result = re.sub(r'\([^()]+\)', repl, s)
print(result)
Bonus approach for particular input case A(x)B(y)C(Alice,z):
...
s = 'A(x)B(y)C(Alice,z)'
result = re.sub(r'\([^()]+\)', lambda m: '({})'.format(','.join(flipped.get(k,'') or k
for k in m.group().strip('()').split(','))), s)
print(result)

I assume you want to replace the values in a string with the respective keys of the dictionary. If my assumption is correct you can try this without using regex.
First the swap the keys and values using dictionary comprehension.
my_dict = {'Alice':'x','Bob':'y','Chloe':'z'}
my_dict = { y:x for x,y in my_dict.iteritems()}
Then using list_comprehension, you replace the values
str_ = 'A(x)B(y)C(x,z)'
output = ''.join([i if i not in my_dict.keys() else my_dict[i] for i in str_])

Hope this is what you need ;)
Code
import re
d={'Alice':'x','Bob':'y','Chloe':'z'}
keys = d.keys()
values = d.values()
s = "A(x)B(y)C(x,z)"
for i in range(0, len(d.keys())):
rx = r"" + re.escape(values[i])
s = re.sub(rx, keys[i], s)
print s
Output
A(Alice)B(Bob)C(Alice,Chloe)

Also you could use the replace method in python like this:
d={'x':'Alice','y':'Bob','z':'Chloe'}
str = "A(x)B(y)C(x,z)"
for key in d:
str = str.replace(key,d[key])
print (str)
But yeah you should swipe your dictionary values like Kishore suggested.

This is the way that I would do it:
import re
def sub_args(text, tosub):
ops = '|'.join(tosub.keys())
for argstr, _ in re.findall(r'(\(([%s]+?,?)+\))' % ops, text):
args = argstr[1:-1].split(',')
args = [tosub[a] for a in args]
subbed = '(%s)' % ','.join(map(str, args))
text = re.sub(re.escape(argstr), subbed, text)
return text
text = 'A(x)B(y)C(x,z)'
tosub = {
'x': 'Alice',
'y': 'Bob',
'z': 'Chloe'
}
print(sub_args(text, tosub))
Basically you just use the regex pattern to find all of the argument groups and substitute in the proper values--the nice thing about this approach is that you don't have to worry about subbing where you don't want to (for example, if you had a string like 'Fn(F,n)'). You can also have multi-character keys, like 'F(arg1,arg2)'.

python distinguish number and string solution

I'm new to python and trying to solve the distinguish between number and string
For example :
Input: 111aa111aa
Output : Number: 111111 , String : aaaa

Here is your answer
for numbers
import re
x = '111aa111aa'
num = ''.join(re.findall(r'[\d]+',x))
for alphabets
import re
x = '111aa111aa'
alphabets = ''.join(re.findall(r'[a-zA-Z]', x))

You can use in-built functions as isdigit() and isalpha()
>>> x = '111aa111aa'
>>> number = ''.join([i for i in x if i.isdigit()])
'111111'
>>> string = ''.join([i for i in x if i.isalpha()])
'aaaa'
Or You can use regex here :
>>> x = '111aa111aa'
>>> import re
>>> numbers = ''.join(re.findall(r'\d+', x))
'111111'
>>> string = ''.join(re.findall(r'[a-zA-Z]', x))
'aaaa'

>>> my_string = '111aa111aa'
>>> ''.join(filter(str.isdigit, my_string))
'111111'
>>> ''.join(filter(str.isalpha, my_string))
'aaaa'

Try with isalpha for strings and isdigit for numbers,
In [45]: a = '111aa111aa'
In [47]: ''.join([i for i in a if i.isalpha()])
Out[47]: 'aaaa'
In [48]: ''.join([i for i in a if i.isdigit()])
Out[48]: '111111'
OR
In [18]: strings,numbers = filter(str.isalpha,a),filter(str.isdigit,a)
In [19]: print strings,numbers
aaaa 111111

As you mentioned you are new to Python, most of the presented approaches using str.join with list comprehensions or functional styles are quite sufficient. Alternatively, I present some options using dictionaries that can help organize data, starting from basic to intermediate examples with arguably increasing intricacy.
Basic Alternative
# Dictionary
d = {"Number":"", "String":""}
for char in s:
if char.isdigit():
d["Number"] += char
elif char.isalpha():
d["String"] += char
d
# {'Number': '111111', 'String': 'aaaa'}
d["Number"] # access by key
# '111111'
import collections as ct
# Default Dictionary
dd = ct.defaultdict(str)
for char in s:
if char.isdigit():
dd["Number"] += char
elif char.isalpha():
dd["String"] += char
dd

Perform mathematical expression written as a string using variables values from dictionary (python)

Let's say I have this operation as a string variable:
formula = '{a} + {b}'
And I have a dictionary such as
data = {'a': 3, 'b': 4}
Is there such a functionality in some library where:
evaluate(operation = formula, variables = data)
gives:
7

If you are using Python3 you can do something like this with string formatting:
>>> import ast
>>> data = {'a': 3, 'b': 4}
>>> formula = '{a} + {b}'
>>> res_string = formula.format(a=data['a'], b=data['b'])
>>> res = ast.literal_eval(res_string)
>>> print(res)
7
Or even better as pointed by Steven in the comments:
res_string = formula.format(**data)
Or if you are using Python3.6 you can even do this with the cool f-string:
>>> f"{sum(data.values())}"
'7'

Although not recomended, you can use eval(). Check out:
>>> data = {'a': 3, 'b': 4}
>>> eval('{a} + {b}'.format(**data))
>>> 7
eval() will try to execute the given string as python code.
For more information about python format you can take a look at the really nice pyformat site.

First you need to parse your string, then you need to have a proper dictionary in order to map the founded operators to their equivalent functions, which you can use operator module for this aim:
In [54]: from operator import add
In [55]: operators = {'+': add} # this is an example, if you are dealing with more operations you need to add them to this dictionary
In [56]: def evaluate(formula, data):
a, op, b = re.match(r'{(\w)} (\W) {(\w)}', formula).groups()
op = operators[op]
a, b = data[a], data[b]
return op(a, b)
....:
In [57]: evaluate(formula, data)
Out[57]: 7

How to encode categorical values in Python

Given a vocabulary ["NY", "LA", "GA"],
how can one encode it in such a way that it becomes:
"NY" = 100
"LA" = 010
"GA" = 001
So if I do a lookup on "NY GA", I get 101

you can use numpy.in1d:
>>> xs = np.array(["NY", "LA", "GA"])
>>> ''.join('1' if f else '0' for f in np.in1d(xs, 'NY GA'.split(' ')))
'101'
or:
>>> ''.join(np.where(np.in1d(xs, 'NY GA'.split(' ')), '1', '0'))
'101'

vocab = ["NY", "LA", "GA"]
categorystring = '0'*len(vocab)
selectedVocabs = 'NY GA'
for sel in selectedVocabs.split():
categorystring = list(categorystring)
categorystring[vocab.index(sel)] = '1'
categorystring = ''.join(categorystring)
This is the end result of my won testing, turns out Python doesn't support string item assignment, somehow i thought it did.
Personally i think behzad's solution is better, numpy does a better job and is faster.

Or you can
vocabulary = ["NY","LA","GA"]
i=pow(10,len(vocabulary)-1)
dictVocab = dict()
for word in vocabulary:
dictVocab[word] = i
i /= 10
yourStr = "NY LA"
result = 0
for word in yourStr.split():
result += dictVocab[word]

Another solution using numpy. It looks like you're tyring to binary encode a dictionary, so the code below feels natural to me.
import numpy as np
def to_binary_representation(your_str="NY LA"):
xs = np.array(["NY", "LA", "GA"])
ys = 2**np.arange(3)[::-1]
lookup_table = dict(zip(xs,ys))
return bin(np.sum([lookup_table[k] for k in your_str.split()]))
It's also not needed to do it in numpy, but it is probably faster in case you have large arrays to work on. np.sum can be replaced by the builtin sum then and the xs and ys can be transformed to non-numpy equivalents.

To create a lookup dictionary, reverse the vocabulary, enumerate it, and take the power of 2:
>>> vocabulary = ["NY", "LA", "GA"]
d = dict((word, 2 ** i) for i, word in enumerate(reversed(vocabulary)))
>>> d
{'NY': 4, 'GA': 1, 'LA': 2}
To query the dictionary:
>>> query = "NY GA"
>>> sum(code for word, code in d.iteritems() if word in query.split())
5
If you want it formatted to binary:
>>> '{0:b}'.format(5)
'101'
edit: if you want a 'one liner':
>>> '{0:b}'.format(
sum(2 ** i
for i, word in enumerate(reversed(vocabulary))
if word in query.split()))
'101'
edit2: if you want padding, for example with six 'bits':
>>> '{0:06b}'.format(5)
'000101'