我正在尝试从列df['date_of_admission']
中减去列df['DOB']
,以找出两者之间的差值,并将年龄值存储在df['age']
列中,但是,出现此错误:
OverflowError:int64加法中的溢出
DOB date_of_admission age
2000-05-07 2019-01-19 12:26:00
1965-01-30 2019-03-21 02:23:12
NaT 2018-11-02 18:30:10
1981-05-01 2019-05-08 12:26:00
1957-01-10 2018-12-31 04:01:15
1968-07-14 2019-01-28 15:05:09
NaT 2018-04-13 06:20:01
NaT 2019-02-15 01:01:57
2001-02-10 2019-03-21 08:22:00
1990-03-29 2018-11-29 03:05:03
..... ......
..... .....
..... .....
我已经尝试了以下方法:
import numpy as np
import pandas as pd
from datetime import dt
df['age'] = (df['date_of_admission'] - df['DOB']).dt.days // 365
在找到两者之间的差异后,预计将获得以下年龄列:
age
26
69
NaN
58
.
.
.
答案 0 :(得分:3)
OP很可能使用医疗MIMIC数据集,其中日期被加乱以保护患者的身份。具体来说,对于89岁以上的患者,they shifted the date of birth by 300 years。
这样的较长时间跨度在使用熊猫timedelta时会溢出:
pd.to_timedelta(300, unit="Y", box=False)
> numpy.timedelta64(-8979658473709551616,'ns')
在数据框操作中发生这种情况时,您会遇到错误。根据@tawab_shakeel的答案改编而成:
df = pd.DataFrame(data={"DOB":['2000-05-07','1965-01-30','1700-01-01'],
"date_of_admission":["2019-01-19 12:26:00","2019-03-21 02:23:12", "2000-01-01 02:23:23"]})
df['DOB'] = pd.to_datetime(df['DOB']).dt.date
df['date_of_admission'] = pd.to_datetime(df['date_of_admission']).dt.date
# Gives AttributeError: Can only use .dt accessor with datetimelike values
df['age'] = ((df['date_of_admission']-df['DOB']).dt.days) //365
# Gives OverflowError: long too big to convert
pd.to_timedelta(df['date_of_admission']-df['DOB'])
任何转换为timedelta64[ns]
数据类型的计算都会出现此问题。
作为一种解决方法,您可以改用apply
操作,直接计算每个元素的年龄元素:
df['age'] = df.apply(lambda e: (e['date_of_admission'] - e['DOB']).days/365, axis=1)
答案 1 :(得分:1)
1)。您做得正确,但是textLabel
包含唯一的日期,而text
包含日期和时间。操纵override func tableView(_ tableView: UITableView, cellForRowAt indexPath: IndexPath) -> UITableViewCell {
let cell = tableView.dequeueReusableCell(withIdentifier: "cellTypeIdentifier", for: indexPath)
// Configure the cell’s contents.
// I'm assuming you have a Realm object called `Idea`
// And `ideas` is an array of those `Idea` objects
let idea = ideas[indexPath.row]
cell.textLabel?.text = idea.name
return cell
}
,使其仅包含日期,然后您将得到结果。
2)。在这里,我向您的代码中添加了DOB
,以便您获得结果。
date_of_admission
希望它能对您有所帮助。
答案 2 :(得分:1)
将两列都转换为日期,然后减去
import pandas as pd
df['date_of_admission'] = pd.to_datetime(df['date_of_admission']).dt.date
df['DOB'] = pd.to_datetime(df['DOB']).dt.date
df['age'] = ((df['date_of_admission']-df['DOB']).dt.days) //365
第二次测试
#Now I have use DOB AND date_of_admission data from the question and it is working fine
df = pd.DataFrame(data={"DOB":['2000-05-07','1965-01-30','NaT'],
"date_of_admission":["2019-01-19 12:26:00","2019-03-21 02:23:12", "2018-11-02 18:30:10"]})
df['DOB'] = pd.to_datetime(df['DOB']).dt.date
df['date_of_admission'] = pd.to_datetime(df['date_of_admission']).dt.date
df['age'] = ((df['date_of_admission']-df['DOB']).dt.days) //365
结果:
DOB date_of_admission age
2000-05-07 2019-01-19 18.0
1965-01-30 2019-03-21 54.0
NaT 2018-11-02 NaN