Reputation: 133
I have the following dictionary:
d= {
'2018-01-01': [[1.0], [2.0], [3.0]],
'2018-01-02': [[4.0], [5.0], [6.0]],
'2018-01-03': [[7.0], [8.0], [9.0]],
'2018-01-04': [[10.0], [11.0], [12.0]],
'2018-01-31': [[13.0, 14.0], [15.0, 16.0], [17.0]]
}
My aim is to count the number of values (floats in this case), assuming some of these 'cells' might contain more than one value (e.g.: values for '2018-01-31'
). In other words, I need:
result=17
So far, I was able to count the number of 'cells' (sublists), but I can't find out how can I count several values inside a single 'cell'. I tried:
cols=len(d[[k for k in d.keys() if d[k]==max(d.values(),key=len)][0]])
cells = [[1 for j in range(cols)] for i in range(len(d))]
result = len([item for sublist in cells for item in sublist])
print(str(result))
Output:
15
Thank you very much in advance.
Upvotes: 2
Views: 270
Reputation: 164793
You do not need to create a flattened list for this task. This is one way:
from itertools import chain
res = sum(1 for _ in chain.from_iterable(map(chain.from_iterable, d.values())))
We deal with nested lists by applying chain.from_iterable
twice and counting the results after this process.
Upvotes: 1
Reputation: 403020
Option 1
100% flattening + len
>>> len([k for i in d.values() for j in i for k in j])
17
Option 2
sum
+ len
+ partial flattening
Use a sum
with a generator comprehension. 3-list form:
>>> sum(len([*a, *b, *c]) for (a, b, c) in d.values())
17
For a the generic solution (with more/less than 3 sublists), use itertools.chain
:
>>> from itertools import chain
>>> sum(len(list(chain.from_iterable(v))) for v in d.values())
17
Upvotes: 2
Reputation: 18950
>>> len([c for a in d.values() for b in a for c in b])
17
you should really try these in an interactive interpreter, start from the simplest [x for x in d.values()]
and work your way up.
Upvotes: 2