converting list of string to list of integers in python

Question

I have a dictionary like :

a={
   'sig2/Bit 2': ['0', '0', '1'], 
   'sig1/Bit 1': ['1', '0', '1']
  }

I want to convert it like :

a={
   'sig2/Bit 2': [0, 0, 1], 
   'sig1/Bit 1': [1, 0, 1]
  }

Please help me out.

Answer 1

I will not provide you with the complete answer but rather guidance on how to do it. Here is an example:

>>> d = ['0', '0', '1']
>>> print [int(i) for i in d]
[0,0,1]

Hopefully that should be enough to guide you to figure this out.
Hint: iterate through the dictionary and do this for each key/value pair. Good Luck

Answer 2

for i in a.keys():
    a[i] = [int(j) for j in a[i]]

Answer 3

May be there are better ways, but here is one.

for key in a:
    a[key] = [int(x) for x in a[key]];

Answer 4

Try this out

In [54]: dict([(k, [int(x) for x in v]) for k,v in a.items()])
Out[54]: {'sig1/Bit 1': [1, 0, 1], 'sig2/Bit 2': [0, 0, 1]}

Answer 5

for key, strNumbers in a.iteritems():
  a[key] = [int(n) for n in strNumbers]

Answer 6

You can try this ,

>>> a={
   'sig2/Bit 2': ['0', '0', '1'], 
   'sig1/Bit 1': ['1', '0', '1']
  }

>>> for i in a.keys():
        a[i]=map(int,a[i])

Output:

>>> print a
{'sig2/Bit 2': [0, 0, 1], 'sig1/Bit 1': [1, 0, 1]}

Answer 7

It is as easy as:

a = {
    'sig2/Bit 2': ['0', '0', '1'], 
    'sig1/Bit 1': ['1', '0', '1']
}

b = dict((x, map(int, y)) for x, y in a.iteritems())

UPDATE

For last line in code above you could use dict comprehension:

b = {x: map(int, y) for x, y in a.iteritems()}