iterate over list of dicts to find minimum values per unique value - python-3.8

Question

I am looking for the minimum of 'diff' per unique 'expiry' in the following list of dicts. Guessing itertools.groupby can help but really lost how to apply that.
import datetime as dt
data = [ {'expiry': dt.datetime(2020,6,26), 'strike': 138.0, 'diff': 0.305},
{'expiry': dt.datetime(2020,6,26), 'strike': 138.5, 'diff': 0.188},
{'expiry': dt.datetime(2020,6,26), 'strike': 139.0, 'diff': 0.688},
{'expiry': dt.datetime(2020,7,24), 'strike': 137.0, 'diff': 0.805},
{'expiry': dt.datetime(2020,7,24), 'strike': 137.5, 'diff': 0.305},
{'expiry': dt.datetime(2020,7,24), 'strike': 138.0, 'diff': 0.203},
{'expiry': dt.datetime(2020,7,24), 'strike': 138.5, 'diff': 0.703}]
desired output:
[{'expiry': dt.datetime(2020,6,26), 'strike': 138.5, 'diff': 0.188},
{'expiry': dt.datetime(2020,7,24), 'strike': 138.0, 'diff': 0.203}]
Looking for an efficient and fast way to solve this (without using pandas).

Answer 1

m = dict()
for d in data:
m.update({
d['expiry']: min(
d['diff'], m[d['expiry']] if d['expiry'] in m else d['diff']
)
})
for exp, min_of_diff in m.items():
print('{},{}'.format(exp,min_of_diff))
or if you want access to the whole dict:
m = dict()
for d in data:
if d['expiry'] not in m:
m.update({
d['expiry']: d
})
continue
# otherwise, get the lesser
smallest_diff = m[d['expiry']]['diff']
if d['diff'] < smallest_diff:
m.update({
d['expiry']: d
})
for exp, min_dict in m.items():
print('{},{}'.format(exp,min_dict))
I haven't tested these solutions.
But that should give you the idea.