Pandas基于两列条件生成新列出错，求问题排查与优化方案

错误原因 #

你代码里的逻辑判断用了&，这是Python中的按位与运算符，而非逻辑与运算符。而且&的优先级高于==，导致条件被错误解析。例如row['T1']==1 & row['T2']==1会被执行成row['T1'] == (1 & row['T2']) == 1，完全偏离了原本的判断逻辑，最终导致结果错误。

修正后的代码 #

有两种修正方式：

方式1：替换为逻辑与运算符and #

def merge(row):
    if row['T1'] == 1 and row['T2'] == 1:
        val = '1_1'
    elif row['T1'] == 1 and row['T2'] == 2:
        val = '1_2'  
    elif row['T1'] == 1 and row['T2'] == 3:
        val = '1_3'
    elif row['T1'] == 1 and row['T2'] == 4:
        val = '1_4'
    elif row['T1'] == 2 and row['T2'] == 1:
        val = '2_1'
    elif row['T1'] == 2 and row['T2'] == 2:
        val = '2_2'
    elif row['T1'] == 2 and row['T2'] == 3:
        val = '2_3'
    else:
        val = '2_4'
    return val

df.loc[:, 'new'] = df.apply(merge, axis=1)

方式2：保留&但给每个条件加括号 #

如果习惯使用pandas的按位与操作，需要给每个==判断加上括号，确保优先级正确：

def merge(row):
    if (row['T1'] == 1) & (row['T2'] == 1):
        val = '1_1'
    elif (row['T1'] == 1) & (row['T2'] == 2):
        val = '1_2'  
    elif (row['T1'] == 1) & (row['T2'] == 3):
        val = '1_3'
    elif (row['T1'] == 1) & (row['T2'] == 4):
        val = '1_4'
    elif (row['T1'] == 2) & (row['T2'] == 1):
        val = '2_1'
    elif (row['T1'] == 2) & (row['T2'] == 2):
        val = '2_2'
    elif (row['T1'] == 2) & (row['T2'] == 3):
        val = '2_3'
    else:
        val = '2_4'
    return val

df.loc[:, 'new'] = df.apply(merge, axis=1)

更优实现方法 #

从你的需求来看，本质是将T1和T2的值用下划线拼接成字符串，完全不需要写大量条件判断。直接使用pandas的向量化字符串操作，效率远高于逐行处理的apply：

df['new'] = df['T1'].astype(str) + '_' + df['T2'].astype(str)

如果偏好lambda表达式写法，也可以用：

df['new'] = df.apply(lambda x: f"{x['T1']}_{x['T2']}", axis=1)

前者是整列批量处理，速度更快，适合大数据场景。

内容的提问来源于stack exchange，提问作者Kajette