I have a = {'foo': 2, 'bar': 3, 'baz': 5 }
Is there anyway I can get a = {'foo': 0.2, 'bar': 0.3, 'baz': 0.5 }
in one line? Need to divide each value by total value... I just can't get it done.. :(
Thank you so much!
Sum the values, then use a dictionary comprehension to produce a new dictionary with the normalised values:
total = sum(a.itervalues(), 0.0)
a = {k: v / total for k, v in a.iteritems()}
You can squeeze that into a one-liner, but it won't be as readable:
a = {k: v / total for total in (sum(a.itervalues(), 0.0),) for k, v in a.iteritems()}
I gave sum()
with a floating point starting value to prevent the /
operator from using floor division in Python 2, which would happen if total
and v
would both be integers.
In Python 3, drop the iter*
prefixes:
a = {k: v / total for total in (sum(a.values()),) for k, v in a.items()}
Note that you do not want to use {k: v / sum(a.values()) for k, v in a.items()}
here; the value expression is executed for each iteration in the comprehension loop, recalculating the sum()
again and again. The sum()
loops over all N items in the dictionary, so you end up with a quadratic O(N^2) solution rather than a O(N) solution to your problem.
I did it using a function
a = {'foo': 2, 'bar': 3, 'baz': 5}
def func1(my_diction):
total = 0
for i in my_diction:
total = total + my_diction[i]
for j in my_diction:
my_diction[j] = (float)(my_diction[j])/total
return my_diction
print (func1(a))
def div_d(my_dict):
sum_p = sum(my_dict.values())
for i in my_dict:
my_dict[i] = float(my_dict[i]/sum_p)
return my_dict
ps i am totally new to programming but this is the best i could think off.
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