[英]How to iterate through two pandas columns and create a new column
I am trying to create a new column by concatenating two columns with certain conditions.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
The problem is with master['work_action'].append(a + " " + b)
问题出在
master['work_action'].append(a + " " + b)
If I change my code to this: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)
I get exactly what I want in a list.我在列表中得到了我想要的东西。 But I want it in a pandas column.
但我希望它在 pandas 列中。 How do I create a new pandas column with the conditions above?
如何在上述条件下创建新的 pandas 列?
If performance is important, I would use numpy
's select
: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']
)
Which results in:结果是:
repair_location work_service work_action
0 Field 1 Field 1
1 Depot 2 Depot 2
2 Other 3 Other
Append method is for insert values at the end. Append 方法用于在末尾插入值。 You are trying to concatenate two strings values.
您正在尝试连接两个字符串值。 Use
apply
method: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)
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