This seems fundamental (though I do come from a MySQL background).
a = [(1,f,e),(7,r,None),(2,s,f),(32,None,q)]
b = [(32,dd), (1,pp)]
If I do this in MySQL (LEFT JOIN):
SELECT a.*, b.* FROM a LEFT JOIN b ON a[0] = b[0]
I get:
[(1,f,e,1,pp),(7,r,None,None,None),(2,s,f,None,None),(32,None,q,32,dd)]
How does one accomplish this pythonically?
(Maybe I'm not searching for the right term... but I don't (think) I want to append or merge or concat...)
You can solve it by making a dictionary out of the second input list and then looking up into it:
>>> a = [(1,'f','e'),(7,'r',None),(2,'s','f'),(32,None,'q')]
>>> b = [(32,'dd'), (1,'pp')]
>>>
>>> b_dict = {item[0]: item for item in b}
>>> [item + b_dict.get(item[0], (None, None)) for item in a]
[
(32, None, 'q', 32, 'dd'),
(1, 'f', 'e', 1, 'pp'),
(2, 's', 'f', None, None),
(7, 'r', None, None, None)
]
Since we are iterating over a to form a resulting list, and looking up values of the second list, this would act as a "LEFT JOIN" - returning all the items from the left "table" even if they are not present in the right "table".
You could choose pandas as a solution. pandas is a python module related with data process, it has MySQL interface and could mock database operation(like filter, join, groupby) in its DataFrame, please check here for detail.
import pandas as pd
# origin data
#a = [(1,f,e),(7,r,None),(2,s,f),(32,None,q)]
#b = [(32,dd), (1,pp)]
# new data
a = [{'a1':1,'a2':'f', 'a3':'e'}, {'a1':2, 'a2':'r', 'a3':None}]
b = [{'b1':32, 'b2':'dd'}, {'b1':1, 'b2':'pp'}]
pd_a = pd.DataFrame(a)
pd_b = pd.DataFrame(b)
result = pd.merge(pd_a, pd_b, left_on='a1', right_on='b1', how='left')
print result
output as below:
a1 a2 a3 b1 b2
0 1 f e 1 pp
1 2 r None NaN NaN
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