iloc 函数在 iris 数据集中做了啥？

Posted 2023-03-11

【中文标题】iloc 函数在 iris 数据集中做了啥？

【英文标题】：what does the function iloc do in the iris dataset?iloc 函数在 iris 数据集中做了什么？

【发布时间】：2021-08-14 07:49:09

【问题描述】：

谁能解释一下这段代码的粗体部分。我已经阅读了 pandas 和 sklearn 的文档，但仍然有点难以理解它。我想为自己的数据修改此内容，并希望对此有更多了解。

X = df.iloc[0:100, **[0,1]**].values
plt.scatter(**X[:50, 0], X[:50, 1]**,alpha=0.5, c='b', edgecolors='none', label='setosa %2s'%(y[0]))
plt.scatter(**X[50:100, 0], X[50:100, 1]**,alpha=0.5, c='r', edgecolors='none', label='versicolor %2s'%(y[50]))

完整代码如下

%matplotlib inline
import matplotlib.pyplot as plt
import pandas as pd
import numpy as np
#from sklearn import cross_validation
from sklearn.model_selection import train_test_split
from sklearn import preprocessing
from mlclass2 import simplemetrics, plot_decision_2d_lda

df = pd.read_csv('https://archive.ics.uci.edu/ml/'
        'machine-learning-databases/iris/iris.data', header=None)
X = df.iloc[0:100, **[0,1]**].values
y = df.iloc[0:100, 4].values
y = np.where(y == 'Iris-setosa', 0, 1)

X_train, X_test, y_train, y_test = train_test_split(
    X, y, test_size=0.2, random_state=5)

stdscaler = preprocessing.StandardScaler().fit(X_train)
X_scaled  = stdscaler.transform(X)
X_train_scaled = stdscaler.transform(X_train)
X_test_scaled  = stdscaler.transform(X_test)

# plot data
plt.scatter(X[:50, 0], X[:50, 1],alpha=0.5, c='b', edgecolors='none', label='setosa %2s'%(y[0]))
plt.scatter(X[50:100, 0], X[50:100, 1],alpha=0.5, c='r', edgecolors='none', label='versicolor %2s'%(y[50]))
plt.xlabel('sepal length [cm]')
plt.ylabel('petal length [cm]')
plt.legend(loc='lower right')
plt.show()

【参考方案1】：

.values 只返回删除了轴标签的数据框的值。

.iloc 使用基于整数位置的索引。

代码的.iloc 部分是说，我们的自变量只需要第 0 列和第 1 列的前 100 行，而因变量只需要第 4 行的前 100 行。如果这部分仍然令人困惑，我建议您查看切片符号。简单地说，.iloc 上的切片符号简化为 .iloc[start:stop]。

原始数据框：

import matplotlib.pyplot as plt
import pandas as pd
import numpy as np
from sklearn.model_selection import train_test_split
from sklearn import preprocessing


df = pd.read_csv('https://archive.ics.uci.edu/ml/'
        'machine-learning-databases/iris/iris.data', header=None)
X = df.iloc[0:100, [0,1]].values
y = df.iloc[0:100, 4].values
y = np.where(y == 'Iris-setosa', 0, 1)

X_train, X_test, y_train, y_test = train_test_split(
    X, y, test_size=0.2, random_state=5)

stdscaler = preprocessing.StandardScaler().fit(X_train)
X_scaled  = stdscaler.transform(X)
X_train_scaled = stdscaler.transform(X_train)
X_test_scaled  = stdscaler.transform(X_test)

print(df)

输出：

       0    1    2    3               4
0    5.1  3.5  1.4  0.2     Iris-setosa
1    4.9  3.0  1.4  0.2     Iris-setosa
2    4.7  3.2  1.3  0.2     Iris-setosa
3    4.6  3.1  1.5  0.2     Iris-setosa
4    5.0  3.6  1.4  0.2     Iris-setosa
..   ...  ...  ...  ...             ...
145  6.7  3.0  5.2  2.3  Iris-virginica
146  6.3  2.5  5.0  1.9  Iris-virginica
147  6.5  3.0  5.2  2.0  Iris-virginica
148  6.2  3.4  5.4  2.3  Iris-virginica
149  5.9  3.0  5.1  1.8  Iris-virginica

[150 rows x 5 columns]

iloc[0:100, [0,1]].values - 看看我们在这里如何只返回第 0 列和第 1 列？从索引值 0 开始，到 100 结束，[start:stop]。我们只选择第 0 列和第 1 列，因为 [0,1] 要清楚。

[[5.1 3.5]
 [4.9 3. ]
 [4.7 3.2]
 [4.6 3.1]
 [5.  3.6]
 [5.4 3.9]
 [4.6 3.4]
 [5.  3.4]
 [4.4 2.9]
 [4.9 3.1]
 [5.4 3.7]
 [4.8 3.4]
 [4.8 3. ]
 [4.3 3. ]
 [5.8 4. ]
 [5.7 4.4]
 [5.4 3.9]
 [5.1 3.5]

df.iloc[0:100, 4].values - 与上面相同，但只选择第 4 列。

['Iris-setosa' 'Iris-setosa' 'Iris-setosa' 'Iris-setosa' 'Iris-setosa'
 'Iris-setosa' 'Iris-setosa' 'Iris-setosa' 'Iris-setosa' 'Iris-setosa'
 'Iris-setosa' 'Iris-setosa' 'Iris-setosa' 'Iris-setosa' 'Iris-setosa'
 'Iris-setosa' 'Iris-setosa' 'Iris-setosa' 'Iris-setosa' 'Iris-setosa'
 'Iris-setosa' 'Iris-setosa' 'Iris-setosa' 'Iris-setosa' 'Iris-setosa'
 'Iris-setosa' 'Iris-setosa' 'Iris-setosa' 'Iris-setosa' 'Iris-setosa']