Question

我想将所有SkinThickness的零值分配给每个患者的平均值处于Age的特定范围内。

因此，我将数据帧按Age分组，以获取每个年龄段的SkinThickness的平均值。

为了将SkinThickness列中的每个零值分配给根据年龄分组计算出的相应平均值。

ageSkinMean = df_clean.groupby("Age_Class")["SkinThickness"].mean()
>>> ageSkinMean

Age_Class
21-22 years     82.163399
23-25 years    103.171429
26-30 years     91.170254
31-38 years     80.133028
39-47 years     73.685851
48-58 years     89.130233
60+ years       40.899160
Name: Insulin, dtype: float64

目前，我正在运行的代码太少了……使用iterrows()

花费的时间太长

start = time.time()
for i, val in df_clean[df_clean.SkinThickness == 0].iterrows():
    if val[7] < 22:
        df_clean.loc[i, "SkinThickness"] = ageSkinMean[0]
    elif val[7] < 25:
        df_clean.loc[i, "SkinThickness"] = ageSkinMean[1]
    elif val[7] < 30:
        df_clean.loc[i, "SkinThickness"] = ageSkinMean[2]
    elif val[7] < 38:
        df_clean.loc[i, "SkinThickness"] = ageSkinMean[3]
    elif val[7] < 47:
        df_clean.loc[i, "SkinThickness"] = ageSkinMean[4]
    elif val[7] < 58:
        df_clean.loc[i, "SkinThickness"] = ageSkinMean[5]
    else:
        df_clean.loc[i, "SkinThickness"] = ageSkinMean[6]
print(time.time() - start)

我想知道是否有任何 pandas 优化以使这样的代码块运行更快

Answer 1

您可以使用pandas转换功能将SkinThickness 0值替换为平均值

    age_skin_thickness_mean = df_clean.groupby('Age_Class')['SkinThickness'].mean()

    def replace_with_mean_thickness(row):
       row['SkinThickness'] = age_skin_thickness_mean[row['Age_Class']]
       return row

    df_clean.loc[df_clean['SkinThickness'] == 0] = df_clean.loc[df_clean['SkinThickness'] == 0].transform(replace_with_mean_thickness, axis=1)

df_clean中所有SkinThickness == 0的行现在将具有等于其年龄组平均值的SkinThickness。

如何在熊猫中将价值分配给多价值

1 个答案: