Drop multiple columns in python - pandas

Question

I have 1000 columns in my dataframe, and I want to drop 3rd, 5th, 7th, 9th....999th columns, but need to keep the 1st column. I am not sure if df.drop function can deal with this issue. any ideas? The following code drops all odd columns include 1st column. but I need to keep the 1st column
col_to_drop = df.columns[np.array([i for i in range(df.shape[1]) if i%2!= 1 and i != 1])]
df.drop(col_to_drop, axis=1)

Answer 1

Try this:
df = pd.DataFrame(np.arange(100).reshape(10,-1), columns=[*'ABCDEFGHIJ'])
df.drop(df.columns[np.r_[2,4,5,6,9]], axis=1)
# or df.drop(df.columns[np.array([2,4,5,6,9])], axis=1)
Output:
A B D H I
0 0 1 3 7 8
1 10 11 13 17 18
2 20 21 23 27 28
3 30 31 33 37 38
4 40 41 43 47 48
5 50 51 53 57 58
6 60 61 63 67 68
7 70 71 73 77 78
8 80 81 83 87 88
9 90 91 93 97 98
Update for 1,000 columns.
Drops every odd colum except 1. -> 3,5,7,9...999
col_to_drop = df.columns[np.array([i for i in range(df.shape[1]) if i%2 != 0 and i != 0])]
df.drop(col_to_drop, axis=1)

Answer 2

I think you can try like below:
>>> df
A B C D E
0 A1 B1 C1 D1 E1
1 A2 B2 C2 D2 E2
2 A3 B3 C3 D3 E3
3 A4 B4 C4 D4 E4
4 A5 B5 C5 D5 E5
Drop unwanted columns:
>>> df.drop(ex.columns[[0, 4, 2]], axis = 1, inplace = True)
Result:
>>> df
B D
0 B1 D1
1 B2 D2
2 B3 D3
3 B4 D4
4 B5 D5
Another way around:
>>> df
A B C D E F G H I J K L M N O P
0 24 27 61 40 12 20 52 95 47 23 92 97 86 73 89 84
1 88 33 39 38 87 7 38 31 77 0 66 41 30 34 80 4
2 8 45 68 99 21 66 6 54 28 48 29 18 12 56 30 3
3 22 79 37 1 32 5 53 13 0 57 54 97 71 86 84 12
4 94 19 8 99 91 91 77 90 56 50 41 4 70 36 56 58
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..
95 70 22 73 15 41 50 81 38 31 78 94 72 62 56 78 17
96 13 45 40 83 25 24 73 30 67 85 98 63 32 1 70 97
97 24 58 74 64 0 43 36 27 17 76 51 36 85 39 15 77
98 29 3 73 57 39 70 45 53 83 98 92 51 49 45 85 71
99 74 22 77 87 4 16 69 77 39 88 71 73 39 8 63 34
[100 rows x 16 columns]
As you want to drop all the odd columns hence other way around is to opt all the even column then while using .loc indexer as follows.
>>> df.loc[:, ::2]
A C E G I K M O
0 24 61 12 52 47 92 86 89
1 88 39 87 38 77 66 30 80
2 8 68 21 6 28 29 12 30
3 22 37 32 53 0 54 71 84
4 94 8 91 77 56 41 70 56
.. .. .. .. .. .. .. .. ..
95 70 73 41 81 31 94 62 78
96 13 40 25 73 67 98 32 70
97 24 74 0 36 17 51 85 15
98 29 73 39 45 83 92 49 85
99 74 77 4 69 39 71 39 63
[100 rows x 8 columns]
OR
>>> df[df.columns[::2]]
A C E G I K M O
0 24 61 12 52 47 92 86 89
1 88 39 87 38 77 66 30 80
2 8 68 21 6 28 29 12 30
3 22 37 32 53 0 54 71 84
4 94 8 91 77 56 41 70 56
.. .. .. .. .. .. .. .. ..
95 70 73 41 81 31 94 62 78
96 13 40 25 73 67 98 32 70
97 24 74 0 36 17 51 85 15
98 29 73 39 45 83 92 49 85
99 74 77 4 69 39 71 39 63
[100 rows x 8 columns]

Answer 3

A little change from scott Boston's answer works perfectly for this case
col_to_drop = df.columns[np.array([i for i in range(df.shape[1]) if i%2 != 1 and i != 0])]
df.drop(col_to_drop, axis=1)