How I can convert a Python Tuple into Dictionary

Question

For converting t = [('a', 1), ('b', 2), ('c', 3), ('a', 4)] to dict, I want the output to be {'a':5,'b':2,'c':3} instead of {'a':4,'b':2,'c':3} . Is there a way to add up the value of the same key item?

I am using this line to convert the tuple:

dict((x,y) for x, y in t)

Answer 1

My approach: use get with default value 0

t = [('a', 1), ('b', 2), ('c', 3), ('a', 4)]
d = dict()
for x,y in t:
    d[x] = d.get(x, 0) + y;
print(d)

Answer 2

Use itertools.groupby :

from itertools import groupby

t = [('a', 1), ('b', 2), ('c', 3), ('a', 4)]

f = lambda x: x[0]
print({k: sum(x[1] for x in g) for k, g in groupby(sorted(t, key=f), key=f)})

# {'a': 5, 'b': 2, 'c': 3}

Or collections.defaultdict :

from collections import defaultdict

t = [('a', 1), ('b', 2), ('c', 3), ('a', 4)]

d = defaultdict(int)
for x in t:
    d[x[0]] += x[1]

print(d)

# {'a': 5, 'b': 2, 'c': 3}

Answer 3

from collections import Counter

t = [('a', 1), ('b', 2), ('c', 3), ('a', 4)]

h = Counter()
for k, v in t:
    h[k] += v

h
Counter({'a': 5, 'c': 3, 'b': 2})

Answer 4

You can write a function that checks whether the key is already present in the dictionary or not. Something like this would work:

d = dict()
for x, y in t:
    if x in d:
        d[x]+=y
    else:
        d[x] = y

And that would give you the proper output:

{'a': 5, 'b': 2, 'c': 3}

Answer 5

This is almost verbatim the example from the collections.defaultdict

from collections import defaultdict

t = [('a', 1), ('b', 2), ('c', 3), ('a', 4)]
char_to_int = defaultdict(int)

for k, v in t:
    char_to_int[k] += v

Verified in the interpreter

In [1]: from collections import defaultdict
   ...:
   ...: t = [('a', 1), ('b', 2), ('c', 3), ('a', 4)]
   ...: char_to_int = defaultdict(int)
   ...:
   ...: for k, v in t:
   ...:     char_to_int[k] += v
   ...:

In [2]: char_to_int
Out[2]: defaultdict(int, {'a': 5, 'b': 2, 'c': 3})

Answer 6

Method #1:

Here is the fastest method:

t = [('a', 1), ('b', 2), ('c', 3), ('a', 4)]

out = {}
for x, y in t:
    if x in out:out[x] += y
    else:out[x] = y

Output:

 {'a': 5, 'b': 2, 'c': 3} 

Method #2:

This is the simplest method:

t = [('a', 1), ('b', 2), ('c', 3), ('a', 4)]

d = dict.fromkeys(dict(t), 0)
for x, y in t: d[x] += y

Output:

 {'a': 5, 'b': 2, 'c': 3} 

---

Proof of my fastest method:

I used timeit to calculate that how much time does it take by each method answered by users. The fastest was mine ( Method #1 ):

from timeit import timeit

from itertools import groupby #for method 1
from collections import defaultdict #for method 2 and 5
from collections import Counter  #for method 3

t = [('a', 1), ('b', 2), ('c', 3), ('a', 4)]



###############################
#########  METHODS  ###########
###############################


def method1():     #By Austin
    f = lambda x: x[0]
    {k: sum(x[1] for x in g) for k, g in groupby(sorted(t, key=f), key=f)}

###############################


def method2():     #By Austin
    d = defaultdict(int)
    for x in t:
        d[x[0]] += x[1]

###############################


def method3():     #By Jean-Claude Arbaut
    h = Counter()
    for k, v in t:
        h[k] += v

###############################


def method4():     #By PHP
    d = dict()
    for x,y in t:
        d[x] = d.get(x, 0) + y;

###############################


def method5():     #By Skam
    char_to_int = defaultdict(int)

    for k, v in t:
        char_to_int[k] += v

###############################


def method6():     #By Aquiles Carattino
    d = dict()
    for x, y in t:
        if x in d:
            d[x]+=y
        else:
            d[x] = y

###############################


def method7():   #My method 2
    d = dict.fromkeys(dict(t), 0)
    for x, y in t: d[x] += y

###############################


def method8():   #My method 1
    out = {}
    for x, y in t:
        if x in out:out[x] += y
        else:out[x] = y



print(timeit(method1))
print(timeit(method2))
print(timeit(method3))
print(timeit(method4))
print(timeit(method5))
print(timeit(method6))
print(timeit(method7))
print(timeit(method8))

The output on my platform was:

 5.7539954 1.7488919999999997 4.2322991000000005 1.0104673999999996 1.3571206 0.8095216999999995 1.5506056999999984 0.6387891000000003 

So the fastest method is method 8 which I answered ( Method #1 ).

Answer 7

Try:

 >>> t = ((1, 'a'),(2, 'b')) >>> dict((y, x) for x, y in t) {'a': 1, 'b': 2}