Reputation: 145
I have a dictionary like below and I want to store the values meaning 1, 1 in a list.
sc_dict=[('n', {'rh': 1}), ('n', {'rhe': 1}), ('nc', {'rhex': 1})]
I want an array [1,1,1]
.
This is my code:
dict_part = [sc[1] for sc in sc_dict]
print(dict_part[1])
L1=[year for (title, year) in (sorted(dict_part.items(), key=lambda t: t[0]))]
print(L1)
Upvotes: 3
Views: 75
Reputation: 115
I tried the simple way like below:
sc_dict=[('n', {'rh': 1}), ('n', {'rhe': 1}), ('nc', {'rhex': 1})]
l = []
for i in range(len(sc_dict)):
l.extend(sc_dict[i][1].values())
print l
The output l
would be [1, 1, 1]
Upvotes: 0
Reputation: 164843
You can use next
to retrieve the first value of your dictionary as part of a list comprehension.
This works since your dictionaries have length 1.
sc_dict=[('n', {'rh': 1}), ('n', {'rhe': 1}), ('nc', {'rhex': 1})]
res = [next(iter(i[1].values())) for i in sc_dict]
# [1, 1, 1]
Upvotes: 0
Reputation: 18950
>>> [v for t1, t2 in sc_dict for k, v in t2.items()]
[1, 1, 1]
t1
and t2
being respectively the first and second item of each tuple, and k
, v
the key-value pairs in the dict t2
.
Upvotes: 3
Reputation: 71471
You can use unpacking:
sc_dict=[('n', {'rh': 1}), ('n', {'rhe': 1}), ('nc', {'rhex': 1})]
new_data = [list(b.values())[0] for _, b in sc_dict]
Output:
[1, 1, 1]
It can become slightly cleaner with one additional step:
d = [(a, b.items()) for a, b in sc_dict]
new_data = [i for _, [(c, i)] in d]
Upvotes: 1