当熊猫的一些行之间有NaN时，如何计算从出生日期开始的年龄?

Question

我想在我的pandas数据框中计算从出生日期开始的年龄。但是，列中有一些日期是NaN，由于格式不同，导致我出现了一些错误。这是我的代码:

dob = {'DOB': ['11/29/1986', 'NaN', '02/23/2006']}

# Creating dataframe
df33 = pd.DataFrame(data = dob)
# This function converts given date to age
def age(born):
born = datetime.strptime(born, "%m/%d/%Y").date()
today = date.today()
return today.year - born.year - ((today.month, 
today.day) < (born.month, 
born.day))

df33['Age'] = df33['DOB'].apply(age)
display(df33)

我可以知道我应该如何修改代码，以便它可以忽略NaN值并继续计算其他行的年龄吗?这些有NaN的行可以留下NaN。任何帮助或建议将非常感激!

Answer 1

您可以通过在age函数中添加异常处理来修改age函数以保持行不变。

import pandas as pd
from datetime import datetime, date
# added pd.NaT to posted data
dob = {'DOB': ['11/29/1986', 'NaN', pd.NaT, '02/23/2006']}
# Creating dataframe
df33 = pd.DataFrame(data = dob)
def age(born):
try:
born = datetime.strptime(born, "%m/%d/%Y").date()
today = date.today()
return today.year - born.year - ((today.month, 
today.day) < (born.month, 
born.day))
except (ValueError, TypeError):
return born    # leave unchanged
dob = {'DOB': ['11/29/1986', 'NaN', '02/23/2006']}
df33['Age'] = df33['DOB'].apply(age)
display(df33)

输出

DOB Age
0   11/29/1986  35
1   NaN        NaN
2   NaT        NaT
3   02/23/2006  16

Answer 2

这可以使用relativedelta完成，而无需使用单独的函数。

安装模块

pip install python-dateutil

import pandas as pd
from datetime import datetime
from dateutil.relativedelta import relativedelta
import numpy as np
dob = {'DOB': ['11/29/1986', np.nan, '02/23/2006']}
# Creating dataframe
df33 = pd.DataFrame(data=dob)
df33["DOB"] = pd.to_datetime(df33["DOB"])
df33["Age"] = df33.apply(lambda x: relativedelta(datetime.now().date(), x['DOB']).years if x.notnull().all() else pd.NaT, axis=1)
print(df33)

DOB   Age
0 1986-11-29  35.0
1        NaT   NaT
2 2006-02-23  16.0

Answer 3

备注:下面的答案只是为了解决你的问题。我建议使用像relativedelta这样的库来正确计算年龄。

"NaN"并不是真的麻木。nan，您应该将dob字典修改为:

dob = {'DOB': ['11/29/1986', pd.NaT, '02/23/2006']}

对于日期时间类型，最好使用pandas的NaT值来表示不是时间值。

那么你可以用它来转换成pd。日期和时间，然后做其他事情。

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但是不修改字典dob的快速修复是:

在开头包含这个检查:

if born == 'NaN':
return 'NaN'