Nav*_*ian 3 python python-2.7 python-3.x pandas
我一直在尝试从数据集中为所有行选择一组特定的列。我尝试了以下类似的方法。
train_features = train_df.loc[,[0,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18]]
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我想提一下,所有行都包含在内,但只需要编号的列即可。有没有更好的方法来解决这个问题。
样本数据:
age job marital education default housing loan equities contact duration campaign pdays previous poutcome emp.var.rate cons.price.idx cons.conf.idx euribor3m nr.employed y
56 housemaid married basic.4y 1 1 1 1 0 261 1 999 0 2 1.1 93.994 -36.4 3.299552287 5191 1
37 services married high.school 1 0 1 1 0 226 1 999 0 2 1.1 93.994 -36.4 0.743751247 5191 1
56 services married high.school 1 1 0 1 0 307 1 999 0 2 1.1 93.994 -36.4 1.28265179 5191 1
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我试图忽略数据集中的工作,婚姻,教育和y栏。y列是目标变量。
如果需要按位置选择,请使用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
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另一个解决方案是drop不必要的列:
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
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您可以通过底层 numpy 数组访问列值
考虑数据框df
df = pd.DataFrame(np.random.randint(10, size=(5, 20)))
df
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您可以对底层数组进行切片
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]])
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或者您可以从此切片重建一个新的数据帧
pd.DataFrame(df.values[:, slc], df.index, df.columns[slc])
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这不像
df.iloc[:, slc]
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您还可以使用对对象slc进行切片df.columns并将其传递给df.loc
df.loc[:, df.columns[slc]]
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