Slicing multiple ranges of columns in Pandas, by list of names

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I think you need numpy.r_ for concanecate positions of columns, then use iloc for selecting:

print (df.iloc[:, np.r_[1:3, 6:len(df.columns)]])

and for second approach subset by list:

print (df[years_month])

Sample:

df = pd.DataFrame({'2000-1':[1,3,5],
                   '2000-2':[5,3,6],
                   '2000-3':[7,8,9],
                   '2000-4':[1,3,5],
                   '2000-5':[5,3,6],
                   '2000-6':[7,8,9],
                   '2000-7':[1,3,5],
                   '2000-8':[5,3,6],
                   '2000-9':[7,4,3],
                   'A':[1,2,3],
                   'B':[4,5,6],
                   'C':[7,8,9]})

print (df)
   2000-1  2000-2  2000-3  2000-4  2000-5  2000-6  2000-7  2000-8  2000-9  A  \
0       1       5       7       1       5       7       1       5       7  1   
1       3       3       8       3       3       8       3       3       4  2   
2       5       6       9       5       6       9       5       6       3  3   

   B  C  
0  4  7  
1  5  8  
2  6  9  

print (df.iloc[:, np.r_[1:3, 6:len(df.columns)]])
   2000-2  2000-3  2000-7  2000-8  2000-9  A  B  C
0       5       7       1       5       7  1  4  7
1       3       8       3       3       4  2  5  8
2       6       9       5       6       3  3  6  9

You can also sum of ranges (cast to list in python 3 is necessary):

rng = list(range(1,3)) + list(range(6, len(df.columns)))
print (rng)
[1, 2, 6, 7, 8, 9, 10, 11]

print (df.iloc[:, rng])
   2000-2  2000-3  2000-7  2000-8  2000-9  A  B  C
0       5       7       1       5       7  1  4  7
1       3       8       3       3       4  2  5  8
2       6       9       5       6       3  3  6  9

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