python - 在数据框中使用 z 变换

Question

我使用 RetailRocket 作为我的数据集。我为每个事件分配了一个值，view = 1，addtocart = 2，transaction = 3。现在我想使用 z 变换来规范化这些值。不幸的是我得到了一个错误。我的错误在哪里？

这是我的 z 变换代码:

df = df.sample(frac=1, random_state=42)
x = df[["visitorid", "itemid"]].values
#y = df["code"].values
y = df["code"].apply(lambda x: (x - x.mean()) / x.std()).values
# Assuming training on 90% of the data and validating on 10%.
train_indices = int(0.9 * df.shape[0])
x_train, x_val, y_train, y_val = (
    x[:train_indices],
    x[train_indices:],
    y[:train_indices],
    y[train_indices:],
)
print(y)

我用 numpy 找到了这个 z 变换的公式:

X = (X - X.mean()) / X.std()

错误:

---------------------------------------------------------------------------
AttributeError                            Traceback (most recent call last)
<ipython-input-7-2712d78bf2a4> in <module>()
      2 x = df[["visitorid", "itemid"]].values
      3 #y = df["code"].values
----> 4 y = df["code"].apply(lambda x: (x - x.mean()) / x.std()).values
      5 # Assuming training on 90% of the data and validating on 10%.
      6 train_indices = int(0.9 * df.shape[0])

1 frames
pandas/_libs/lib.pyx in pandas._libs.lib.map_infer()

<ipython-input-7-2712d78bf2a4> in <lambda>(x)
      2 x = df[["visitorid", "itemid"]].values
      3 #y = df["code"].values
----> 4 y = df["code"].apply(lambda x: (x - x.mean()) / x.std()).values
      5 # Assuming training on 90% of the data and validating on 10%.
      6 train_indices = int(0.9 * df.shape[0])

AttributeError: 'int' object has no attribute 'mean'

Answer 1

由于您使用apply(lambda x: ...)，x 将只是一个值。当您尝试对该单个值使用 x.mean() 时，将会出现错误。

您要做的是在整个列上使用 mean 和 std。使用 apply，可以按如下方式完成:

col = 'code'
df['z_score'] = df[col].apply(lambda x: (x - df[col].mean()) / df[col].std())

但是，不使用 apply 会更快:

df['z_score'] = (df[col] - df[col].mean())/df[col].std()