This question already has answers here:
Is there any pythonic way to combine two dicts (adding values for keys that appear in both)?
(22 answers)
Closed 5 years ago.
How to add the values of two dictionary ?
Ex :
a = {'a':10,'b':11,'c':20}
b = {'a':1,'b':1,'c':1}
result must be
c = {'a':11,'b':12,'c':21}
You can easily add two dictionaries by using Counter class of collections library for ex:
from collections import Counter
a = {'a':10,'b':11,'c':20}
b = {'a':1,'b':1,'c':1}
a = Counter(a)
b = Counter(b)
c = dict(a + b)
print c
OUTPUT
{'c': 21, 'b': 12, 'a': 11}
Next some please show some effort..
a = {'a':10,'b':11,'c':20}
b = {'a':1,'b':1,'c':1}
c = {k: a[k] + b[k] for k in a}
print(c) # {'c': 21, 'b': 12, 'a': 11}
The above works fine if we assume that a and b have the same keys.
If that is not the case, you can try the following:
a = {'a': 10, 'b': 11, 'c': 20, 'h': 5}
b = {'a': 1, 'b': 1, 'c': 1, 'd': 12}
all_keys = set(a.keys()) # in Python 3 it can be simplified as `all_keys = set(a)`
all_keys.update(b.keys()) # in Python 3 it can be simplified as `all_keys.update(b)`
c = {k: a.get(k, 0) + b.get(k, 0) for k in all_keys}
print(c) # {'c': 21, 'h': 5, 'a': 11, 'b': 12, 'd': 12}
Notice that i am using get on both dictionaries to skip the check on the existence of the key.
Related
This question already has answers here:
How do I create a nested dict from a dict by splitting the keys on a delimiter in python?
(4 answers)
Closed 1 year ago.
I want to write a method which takes a dict in the form of {a: 1, b-c: 2, b-d: 3, e-f-g: 4} and returns {a: 1, b: {c: 2, d: 3}, e: {f: {g: 4}}}. Basically split the keys containing - into subdicts. I've tried to do it iteratively and recursively but I got stuck. Any tips?
You can use collections.defaultdict with recursion:
from collections import defaultdict
d = {'a': 1, 'b.c': 2, 'b.d': 3, 'e.f.g': 4}
def group(d):
nd = defaultdict(list)
for [a, *b], c in d:
nd[a].append([b, c])
return {a:b[0][-1] if not any(j for j, _ in b) else group(b) for a, b in nd.items()}
result = group([[a.split('.'), b] for a, b in d.items()])
Output:
{'a': 1, 'b': {'c': 2, 'd': 3}, 'e': {'f': {'g': 4}}}
I have three dictionaries:
X = {'a':2, 'b':3,'e':4}
Y = {'c':3, 'b':4,'a':5, 'd':7}
Z = {'c':8, 'b':7,'a':9, 'e':10,'f':10}
I want to add elements of X and Y if they are present in both dicts and then subtract them from z i.e. Z-X+Y
How can I do that ?
expected result:
res = {'a':2,'b':0,'c':5,'d':7,'e':6,'f':10}
What I tried:
from collections import Counter
xy = Counter(X) + Counter(Y)
res = Counter(Z) - xy
which return:
Counter({'c': 5, 'a': 2, 'e': 6, 'f': 10})
as you can see b and d are missing from my attempt
Your expected result is actually an operation of symmetric difference in terms of sets, but since collections.Counter doesn't support such an operation, you can emulate it with:
xy = Counter(X) + Counter(Y)
z = Counter(Z)
res = z - xy | xy - z
res becomes:
Counter({'f': 10, 'd': 7, 'e': 6, 'c': 5, 'a': 2})
But if you do want keys with value of 0, which Counter would hide from its output, you would have to iterate through a union of the keys of the 3 dicts:
{k: res.get(k, 0) for k in {*X, *Y, *Z}}
This returns:
{'a': 2, 'd': 7, 'e': 6, 'b': 0, 'f': 10, 'c': 5}
How can I iterate over only X number of dictionary items? I can do it using the following bulky way, but I am sure Python allows a more elegant way.
d = {'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5}
x = 0
for key in d:
if x == 3:
break
print key
x += 1
If you want a random sample of X values from a dictionary you can use random.sample on the dictionary's keys:
from random import sample
d = {'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5}
X = 3
for key in sample(d, X):
print key, d[key]
And get output for example:
e 5
c 3
b 2
This question already has answers here:
How do I count the occurrences of a list item?
(29 answers)
Closed 4 years ago.
I have 5 set of lists and some of strings in those lists are repetitive
now! I wanna know the number of repetition! for example the word "A" is in all of my lists by "B" is just in "3" or "C" is in 4 of them.
How can I sort this problem, by using remove() I faced to wrong answer
Thank you in advance!
Take a look at Counter
from collections import Counter
a = ['a','a','b','b']
b = ['b','b','c','d']
c = a+b
cnt = Counter()
for x in c:
cnt[x] +=1
print(cnt)
Counter({'a': 2, 'b': 4, 'c': 1, 'd': 1})
The above will get you counts of each but it seems like you're more concerned at a list level.
from collections import defaultdict
f = defaultdict(list)
a = ['a','a','b','b']
b = ['b','b','c','d']
c = ['e','f','g','h']
d = a + b + c
for i in d:
f[i] = 0
if i in b:
f[i] += 1
if i in c:
f[i] +=1
if i in a:
f[i] +=1
print (f)
defaultdict(list,
{'a': 1, 'b': 2, 'c': 1, 'd': 1, 'e': 1, 'f': 1, 'g': 1, 'h': 1})
A = {0:{a:1, b:7}, 1:{a:5,b:5}, 2:{a:4,b:6}}
I want to attach an item guess to each sub dictionary based on the value b accounting of all b's in each sub dictionary.
Saying, in Dictionary A:
0-b-7 percentage of b: 7/(7+5+6)
1-b-5 percentage of b: 5/(7+5+6)
2-b-6 percentage of b: 1 - 7/(7+5+6) - 5/(7+5+6)
The desired Dictionary should be like
A = {0:{a:1, b:7, 'guess': 7/(7+5+6)},
1:{a:5,b:5, 'guess': 5/(7+5+6)},
2:{a:4,b:6, 'guess': 1 - 7/(7+5+6) - 5/(7+5+6)}}
I don't know how to incorporate the other two b's for a specific subdictionary.
One approach is to precompute the sum of all bs and then use it to add a new key-value pair to your dictionary.
b_total = float(sum(A[k]['b'] for k in A))
for k in A:
A[k]['guess'] = A[k]['b'] / b_total
#{0: {'a': 1, 'b': 7, 'guess': 0.3888888888888889},
# 1: {'a': 5, 'b': 5, 'guess': 0.2777777777777778},
# 2: {'a': 4, 'b': 6, 'guess': 0.3333333333333333}}
A = {0:{"a":1, "b":7}, 1:{"a":5,"b":5}, 2:{"a":4,"b":6}}
char = "b"
denominator = 0
# =========================
# First Calculate the sum
# =========================
for key in A:
inner_map = A[key]
denominator += inner_map[char]
# ========================================
# Now insert the new key to the inner_map
# ========================================
for key in A:
inner_map = A[key]
inner_map["guess"] = inner_map[char]/denominator
print(A)
Output:
{0: {'a': 1, 'b': 7, 'guess': 0.3888888888888889}, 1: {'a': 5, 'b': 5, 'guess': 0.2777777777777778}, 2: {'a': 4, 'b': 6, 'guess': 0.3333333333333333}}
Try this:
def add_calc(my_dict):
total_guesses = sum(map(lambda x: my_dict.get(x).get('b'), my_dict))
for item in my_dict.itervalues():
item.update({'guess': 1.0 * item.get('b') / total_guesses})
return my_dict
d = add_calc(A)
{0: {'a': 1, 'b': 7, 'guess': 0.3888888888888889},
1: {'a': 5, 'b': 5, 'guess': 0.2777777777777778},
2: {'a': 4, 'b': 6, 'guess': 0.3333333333333333}}
I'm on Python 2 btw, you didn't specify version
You can use dictionary unpacking:
A = {0:{'a':1, 'b':7}, 1:{'a':5, 'b':5}, 2:{'a':4, 'b':6}}
results = {a:{**b, **{'guess':b['b']/float(sum(c['b'] for _, c in A.items()))}} for a, b in A.items()}
Output:
{0: {'guess': 0.3888888888888889, 'b': 7, 'a': 1}, 1: {'guess': 0.2777777777777778, 'b': 5, 'a': 5}, 2: {'guess': 0.3333333333333333, 'b': 6, 'a': 4}}