Python 基于不同类型的值派生日期列
我有一个如下所示的数据帧Python 基于不同类型的值派生日期列,python,pandas,list,dataframe,python-datetime,Python,Pandas,List,Dataframe,Python Datetime,我有一个如下所示的数据帧 df = pd.DataFrame({'subject_id' :[1,2,3,4,5], 'date_of_interview':['2007-05-27','2008-03-13','2010-11-19','2011-10-05','2004-11-02'], 'Age':[31,35,78,72,43], 'value'
df = pd.DataFrame({'subject_id' :[1,2,3,4,5],
'date_of_interview':['2007-05-27','2008-03-13','2010-11-19','2011-10-05','2004-11-02'],
'Age':[31,35,78,72,43],
'value':[6,0.33,1990,np.nan,2001],
'age_detected':[25,35,98,65,40]})
df['date_of_interview'] = pd.to_datetime(df['date_of_interview'])
我想根据value
和age\u-detected
列创建一个名为dis\u-date
的新列
例:受试者id=1的访谈日期为2007-05-27。如果我们看一下他的价值栏,我们可以看到他的价值是6,这意味着我们必须从采访日期减去6年,才能将2001-05-27
作为dis\u日期
然而,如果你看主体id=3,他在值列中有一个年份值,所以他的dis_日期将是1990-11-19
当值列中有NA
时,我们必须查看他的age\u detected
列,并从age
中减去它以获得年数
例:受试者id=4的年龄为72岁,检测到年龄为65岁。现在差是7,他的离店日期将是2004-10-05
如果少于6位代表年份数,请注意“值”列中的值。如果是1,就意味着减去1年。如果它是0.33,意味着减去4个月。1年=12个月。0.33=3.96个月(4个月)
我试过这样的方法,但没用
for i in range(len(df['value'])):
if (len(str(df['value'][i]))) < 6:
df['dis_date'] = df['date_of_interview'] - pd.DateOffset(years=df['value'][i])
范围内i的(len(df['value']):
如果(len(str(df['value'][i]))<6:
df['dis_date']=df['date_of_session']-pd.DateOffset(年份=df['value'][i])
我希望我的输出如下所示
df = pd.DataFrame({'subject_id' :[1,2,3,4,5],
'date_of_interview':['2007-05-27','2008-03-13','2010-11-19','2011-10-05','2004-11-02'],
'Age':[31,35,78,72,43],
'value':[6,0.33,1990,np.nan,2001],
'age_detected':[25,35,98,65,40]})
df['date_of_interview'] = pd.to_datetime(df['date_of_interview'])
在该解决方案中,为验证替换年份或减去月份创建了辅助列:
#if value less like 1 multiple by 12, another values set to NaNs
df['m1'] = np.where(df['value'].lt(1), df['value'].mul(12).round(), np.nan)
#if values more like 1000 it is year
df['y1'] = df['value'].where(df['value'].gt(1000))
#if values between 1, 1000 is necessary subtract years from value column
y1 = df['Age'].sub(df['age_detected'])
df['y2'] = np.where(y1.between(1, 1000), df['date_of_interview'].dt.year.sub(y1), np.nan)
#joined years to one column
df['y'] = df['y1'].fillna(df['y2'])
#replaced years by another column
f1 = lambda x: x['date_of_interview'] - pd.DateOffset(year=(int(x['y'])))
df['dis_date1'] = df.dropna(subset=['date_of_interview','y']).apply(f1, axis=1)
#subtracted months if non missing values
f1 = lambda x: x['date_of_interview'] - pd.DateOffset(months=(int(x['m1'])))
df['dis_date2'] = df.dropna(subset=['m1']).apply(f1, axis=1)
#join together
df['dis_date'] = df['dis_date1'].fillna(df['dis_date2'])
print (df)
subject_id date_of_interview Age value age_detected m1 y1 \
0 1 2007-05-27 31 6.00 25 NaN NaN
1 2 2008-03-13 35 0.33 35 4.0 NaN
2 3 2010-11-19 78 1990.00 98 NaN 1990.0
3 4 2011-10-05 72 NaN 65 NaN NaN
4 5 2004-11-02 43 2001.00 40 NaN 2001.0
y2 y dis_date1 dis_date2 dis_date
0 2001.0 2001.0 2001-05-27 NaT 2001-05-27
1 NaN NaN NaT 2007-11-13 2007-11-13
2 NaN 1990.0 1990-11-19 NaT 1990-11-19
3 2004.0 2004.0 2004-10-05 NaT 2004-10-05
4 2001.0 2001.0 2001-11-02 NaT 2001-11-02
您如何定义0.33
年<代码>12/4
?1年=12个月。0.33=3.96个月(4个月)我今晚到家时会看看这个。当然。那真的很有帮助。谢谢。2007-12-13
正确吗?因为如果减去4个月得到2007-11-13