samplelist1 = [('A', 'Pass', 20), ('A', 'Failed', 5), ('B', 'Pass', 10), ('B', 'Failed', 5) ]
samplelist2 = [('A', 'Pass', 2), ('A', 'Failed', 1), ('B', 'Failed', 2)]
# Expected Output: result = [('A', 'Pass', 18), ('A', 'Failed', 4), ('B', 'Pass', 10), ('B', 'Failed', 3)
Can I Still use this code ??? Using this code as my reference
from collections import defaultdict
d = defaultdict(int)
for letter, status, value in samplelist:
d[(letter, status)] += value
res = [key + (val,) for key, val in d.items()] # convert to required format
print(res)
Your current code won't work, because it uses samplelist, but your data only contains samplelist1 and samplelist2. I'm assuming this was a error and you forgot to include the loop where you iterate the sample lists.
In any case, I don't think you need a defaultdict. Just traverse samplelist1, set the values in a dictionary, then traverse samplelist2, and deduct the values from the dictionary.
samplelist1 = [('A', 'Pass', 20), ('A', 'Failed', 5), ('B', 'Pass', 10), ('B', 'Failed', 5) ]
samplelist2 = [('A', 'Pass', 2), ('A', 'Failed', 1), ('B', 'Failed', 2)]
d = {}
for x, y, z in samplelist1:
d[x, y] = z
for x, y, z in samplelist2:
d[x, y] -= z
result = [(k1, k2, v) for (k1, k2), v in d.items()]
print(result)
If we really want to use a defaultdict, we could modify your approach to first collect the items in a list, then minus the first item from the second item(if more than one element exists):
from collections import defaultdict
samplelist1 = [('A', 'Pass', 20), ('A', 'Failed', 5), ('B', 'Pass', 10), ('B', 'Failed', 5) ]
samplelist2 = [('A', 'Pass', 2), ('A', 'Failed', 1), ('B', 'Failed', 2)]
d = defaultdict(list)
for samplelist in (samplelist1, samplelist2):
for x, y, z in samplelist:
d[x, y].append(z)
result = [(k1, k2, v[0] - v[1] if len(v) > 1 else v[0]) for (k1, k2), v in d.items()]
print(result)
Output:
[('A', 'Pass', 18), ('A', 'Failed', 4), ('B', 'Pass', 10), ('B', 'Failed', 3)]