Python中DataFrame多条件自定义函数应用报错排查与优化

1. Cannot convert type '<class 'list'>' to date. #

创建德国假日对象时，holidays.DE(years=[])传入了空列表，holidays库无法识别空年份列表，导致后续判断date in german_holidays时无法正确转换日期类型，抛出类型转换错误。需要指定数据对应的年份（比如你的数据是2019年，应改为holidays.DE(years=2019)）。

2. The truth value of a Series is ambiguous... #

• 日期类型问题：如果df['date']列是字符串类型而非datetime64类型，在进行date + pd.Timedelta(days=1)或日期判断时会隐式转换失败，导致返回Series而非单个布尔值，触发模糊真值错误。
• weekday调用错误：date.weekday是方法而非属性，正确写法应为date.weekday()，直接使用属性会拿到方法对象，无法和数字（5、6等）做比较，导致逻辑判断异常。

先解决上述错误，调整代码如下：

import pandas as pd
import holidays

# 1. 正确初始化德国假日，指定年份
german_holidays = holidays.DE(years=2019)
# 2. 确保date列是datetime类型
df['date'] = pd.to_datetime(df['date'])

def eighteen(date):
    if date in german_holidays:
        return 120
    elif (date + pd.Timedelta(days=1)) in german_holidays:
        return 1320
    elif date.weekday() == 5:  # 改为调用weekday()方法
        return 1080
    elif date.weekday() == 6:
        return 120
    else:
        return 1440

def fifteenone(date):
    if date in german_holidays:
        return 120
    elif (date + pd.Timedelta(days=1)) in german_holidays:
        return 1320
    elif date.weekday() == 4:
        return 1320
    elif date.weekday() == 5:
        return 0
    elif date.weekday() == 6:
        return 120
    else:
        return 1440

def fifteentwo(date):
    if date in german_holidays:
        return 120
    elif (date + pd.Timedelta(days=1)) in german_holidays:
        return 1320
    elif date.weekday() == 4:
        return 1320
    elif date.weekday() == 5:
        return 0
    elif date.weekday() == 6:
        return 0
    elif date.weekday() == 0:
        return 1080
    else:
        return 1440

def twelve(date):
    if date in german_holidays:
        return 0
    elif date.weekday() == 5:
        return 0
    elif date.weekday() == 6:
        return 0
    else:
        return 960

def WT(shift, date):
    if shift == "18-sh":
        return eighteen(date)
    elif shift == "15.1-sh":
        return fifteenone(date)
    elif shift == "15.2-sh":
        return fifteentwo(date)
    elif shift == "12-sh":
        return twelve(date)

df['WT'] = df.apply(lambda row: WT(row['shift'], row['date']), axis=1)

apply是逐行遍历，数据量大时效率极低，建议用向量化操作代替，步骤如下：

1. 预处理衍生列：

df['date'] = pd.to_datetime(df['date'])
# 标记当日是否为德国假日
df['is_holiday'] = df['date'].isin(german_holidays)
# 标记次日是否为德国假日
df['next_day_holiday'] = df['date'].shift(-1).isin(german_holidays)
# 获取星期几（0=周一，6=周日）
df['weekday'] = df['date'].dt.weekday

2. 针对不同班次定义规则，用np.select批量计算：

import numpy as np

# 定义18-sh的规则和结果
rules_18sh = [
    df['is_holiday'],
    df['next_day_holiday'],
    df['weekday'] == 5,
    df['weekday'] == 6
]
results_18sh = [120, 1320, 1080, 120]

# 定义15.1-sh的规则和结果
rules_151sh = [
    df['is_holiday'],
    df['next_day_holiday'],
    df['weekday'] == 4,
    df['weekday'] == 5,
    df['weekday'] == 6
]
results_151sh = [120, 1320, 1320, 0, 120]

# 定义15.2-sh的规则和结果
rules_152sh = [
    df['is_holiday'],
    df['next_day_holiday'],
    df['weekday'] == 4,
    df['weekday'] == 5,
    df['weekday'] == 6,
    df['weekday'] == 0
]
results_152sh = [120, 1320, 1320, 0, 0, 1080]

# 定义12-sh的规则和结果
rules_12sh = [
    df['is_holiday'],
    df['weekday'] == 5,
    df['weekday'] == 6
]
results_12sh = [0, 0, 0]

# 按班次批量计算WT
df['WT'] = np.where(df['shift'] == '18-sh',
                   np.select(rules_18sh, results_18sh, default=1440),
                   np.where(df['shift'] == '15.1-sh',
                            np.select(rules_151sh, results_151sh, default=1440),
                            np.where(df['shift'] == '15.2-sh',
                                     np.select(rules_152sh, results_152sh, default=1440),
                                     np.select(rules_12sh, results_12sh, default=960))))

这种方式完全利用pandas和numpy的向量化运算，避免逐行遍历，效率提升显著，且规则更清晰易维护。

内容的提问来源于stack exchange，提问作者user20216792