如何遍历两个 pandas 列并创建一个新列

Question

我正在尝试通过连接具有特定条件的两列来创建一个新列。

master['work_action'] = np.nan
for a,b in zip(master['repair_location'],master['work_service']):
    if a == 'Field':
        master['work_action'].append(a + " " + b)
    elif a == 'Depot':
        master['work_action'].append(a + " " + b)
    else:
        master['work_action'].append(a)

TypeError: cannot concatenate object of type '<class 'str'>'; only Series and DataFrame objs are valid

问题出在master['work_action'].append(a + " " + b)

如果我将代码更改为此：

test = []
for a,b in zip(master['repair_location'],master['work_service']):
    if a == 'Field':
        test.append(a + " " + b)
    elif a == 'Depot':
        test.append(a + " " + b)
    else:
        test.append(a)

我在列表中得到了我想要的东西。 但我希望它在 pandas 列中。 如何在上述条件下创建新的 pandas 列？

Answer 1

如果性能很重要，我会使用numpy的select ：

master = pd.DataFrame(
    {
        'repair_location': ['Field', 'Depot', 'Other'],
        'work_service':[1, 2, 3]
    }
)

master['work_action'] = np.select(
    condlist= [
        master['repair_location'] == 'Field',
        master['repair_location'] == 'Depot'
    ],
    choicelist= [
        master['repair_location'] + ' ' + master['work_service'].astype(str),
        master['repair_location'] + ' ' + master['work_service'].astype(str)
    ],
    default= master['repair_location']
)

结果是：

  repair_location  work_service work_action
0           Field             1     Field 1
1           Depot             2     Depot 2
2           Other             3       Other

Answer 2

Append 方法用于在末尾插入值。 您正在尝试连接两个字符串值。 使用apply方法：

def fun(a,b):
    if a == 'Field':
        return a + " " + b
    elif a == 'Depot':
        return a + " " + b
    else:
       return a

master['work_action'] = master.apply(lambda x: fun(x['repair_location'], x['work_service']), axis=1)