0
for i in range(0,9):
    num = training_data[training_data[i]==row[i]&training_data[9]==1].groupby([i,9]).size()
    print num
    count_equals.append(num)

DataFrame

      0  1  2   3  4   5  6  7  8  9

314   1  1  1   1  1   1  2  1  1  1 
431   5  1  1   3  4   1  3  2  1  1
260  10  5  8  10  3  10  5  1  3 -1
91    3  1  1   2  2   1  1  1  1  1
337   1  1  1   1  2   1  3  1  1  1

I need counts into a list else groupby without second condition works. if row =[1,1,1,1,1,1,1,1] then the count_equals list should be [2,4,4,2,4,1,3,4]

Error:
Traceback (most recent call last):
File "naive.py", line 46, in
num = training_data[training_data[i]==row[i]&training_data[9]==1].groupby([i,9]).size()
File "/usr/local/lib/python2.7/dist-packages/pandas/core/generic.py", line 917, in nonzero .format(self.class.name))
ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

marc_s
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vishnu
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1 Answers1

0

In training_data[i]==row[i]&training_data[9]==1, the operator & has higher precedence than ==. Surround the relational expressions with parentheses before doing the AND:

(training_data[i]==row[i]) & (training_data[9]==1)
DYZ
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