I have been trying to select a particular set of columns from a dataset for all the rows. I tried something like below.
train_features = train_df.loc[,[0,4,5
You can access the column values via the the underlying numpy array
Consider the dataframe df
df = pd.DataFrame(np.random.randint(10, size=(5, 20)))
df
You can slice the underlying array
slc = [0,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18]
df.values[:, slc]
array([[1, 3, 9, 8, 3, 2, 1, 6, 6, 0, 3, 9, 8, 5, 9, 9],
[8, 0, 2, 3, 7, 8, 9, 2, 7, 2, 1, 3, 2, 5, 4, 9],
[1, 1, 9, 3, 5, 8, 8, 8, 8, 4, 8, 0, 5, 4, 9, 0],
[6, 3, 1, 8, 0, 3, 7, 9, 9, 0, 9, 7, 6, 1, 4, 8],
[3, 2, 3, 3, 9, 8, 3, 8, 3, 4, 1, 6, 4, 1, 6, 4]])
Or you can reconstruct a new dataframe from this slice
pd.DataFrame(df.values[:, slc], df.index, df.columns[slc])
This is not as clean and intuitive as
df.iloc[:, slc]
You could also use slc
to slice the df.columns
object and pass that to df.loc
df.loc[:, df.columns[slc]]
If need select by positions use iloc:
train_features = train_df.iloc[:, [0,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18]]
print (train_features)
age default housing loan equities contact duration campaign pdays \
0 56 1 1 1 1 0 261 1 999
1 37 1 0 1 1 0 226 1 999
2 56 1 1 0 1 0 307 1 999
previous poutcome emp.var.rate cons.price.idx cons.conf.idx euribor3m \
0 0 2 1.1 93.994 -36.4 3.299552
1 0 2 1.1 93.994 -36.4 0.743751
2 0 2 1.1 93.994 -36.4 1.282652
nr.employed
0 5191
1 5191
2 5191
Another solution is drop unnecessary columns:
cols= ['job','marital','education','y']
train_features = train_df.drop(cols, axis=1)
print (train_features)
age default housing loan equities contact duration campaign pdays \
0 56 1 1 1 1 0 261 1 999
1 37 1 0 1 1 0 226 1 999
2 56 1 1 0 1 0 307 1 999
previous poutcome emp.var.rate cons.price.idx cons.conf.idx euribor3m \
0 0 2 1.1 93.994 -36.4 3.299552
1 0 2 1.1 93.994 -36.4 0.743751
2 0 2 1.1 93.994 -36.4 1.282652
nr.employed
0 5191
1 5191
2 5191