random_state 一起洗牌

Question

我对一起使用 random_state 和 shuffle 有点困惑。我想拆分数据而不洗牌。在我看来，当我将 shuffle 设置为 False 时，我为 random_state 选择的数字是多少并不重要，我有相同的输出（random_state 42 或 2 的拆分相同， 7、17 等）。为什么？

X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.25,random_state=42,shuffle=False )

但是如果 shuffle 为 True，我对不同的 random_state 有不同的输出（拆分），这是有道理的。

X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.25,random_state=42)

Answer 1

如果您将 shuffle 设置为 False，train_test_split 只会按原始顺序读入您的数据。因此参数random_state被完全忽略。

示例：

X = [k for k in range(0, 50)] # create array with numbers ranging from 0 to 49
y = X # just for testing
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.25, random_state=42, shuffle=False)

print(X_train) // prints [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36]

一旦您将 shuffle 设置为 True，random_state 就会用作随机数生成器的种子。结果，您的数据集被随机分为训练集和测试集。

random_state=42 的示例：

X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.25, random_state=42, shuffle=True)

print(X_train) // prints [8, 3, 6, 41, 46, 47, 15, 9, 16, 24, 34, 31, 0, 44, 27, 33, 5, 29, 11, 36, 1, 21, 2, 43, 35, 23, 40, 10, 22, 18, 49, 20, 7, 42, 14, 28, 38]

random_state=44 的示例：

X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.25, random_state=44, shuffle=True)

print(X_train) // prints [13, 11, 2, 12, 34, 41, 30, 16, 39, 28, 24, 8, 18, 9, 4, 10, 0, 19, 21, 29, 14, 1, 48, 38, 7, 43, 25, 22, 23, 42, 46, 49, 32, 3, 45, 35, 20]

random_state 一起洗牌

random_state and shuffle together

python

shuffle

scikit-learn