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python - 如何创建一个循环来计算日期内列的重复次数？

问题描述

我正在处理一个大型医疗数据集，现在我面临一个问题。

我想添加一个新列“再入院”，它表示截至入院日期 6 个月前进行的手术次数。我有这个：

Patient_ID  Surgery_Date
1838        2017-01-05
1838        2018-04-26
87          2017-01-11
1838        2017-07-06
87          2017-03-17
1838        2018-08-02
87          2017-11-15
1838        2018-11-22
87          2017-02-01
87          2017-06-21
1838        2018-06-14

我想要这个：

Patient_ID Surgery_Date  Readmission
1838       2017-01-05        0
1838       2018-04-26        0
087        2017-01-11        0
1838       2017-07-06        0
087        2017-03-17        2
1838       2018-08-02        2
087        2017-11-15        1
1838       2018-11-22        2
087        2017-02-01        1
087        2017-06-21        3
1838       2018-06-14        1

我已经在这里问过一个类似的问题：如何找到重复的患者并添加一个新的列来帮助我处理这段代码：

import pandas as pd
import datetime as dt
import numpy as np

# Your data plus a new patient that comes often                                                                                                                                                                    
data = {'Patient_ID':[12,1352,55,1352,12,6,1352,100,100,100,100] ,
        'Surgery_Date': ['25/01/2009', '28/01/2009','29/01/2009','12/12/2008','23/02/2008','2/02/2009','12/01/2009','01/01/2009','01/02/2009','01/01/2010','01/02/2010']}

df = pd.DataFrame(data,columns = ['Patient_ID','Surgery_Date'])
readmissions = pd.Series(np.zeros(len(df),dtype=int),index=df.index))

# Loop through all unique ids                                                                                                                                                                                      
all_id = df['Patient_ID'].unique()
id_admissions = {}
for pid in all_id:
    # These are all the times a patient with a given ID has had surgery                                                                                                                                            
    patient = df.loc[df['Patient_ID']==pid]
    admissions_sorted = pd.to_datetime(patient['Surgery_Date'], format='%d/%m/%Y').sort_values()

    # This checks if the previous surgery was longer than 180 days ago                                                                                                                                              
    frequency = admissions_sorted.diff()<dt.timedelta(days=180)

    # Compute the readmission                                                                                                                                                                                      
    n_admissions = [0]
    for v in frequency.values[1:]:
       n_admissions.append((n_admissions[-1]+1)*v)

    # Add these value to the time series                                                                                                                                                                           
    readmissions.loc[admissions_sorted.index] = n_admissions


df['Readmission'] = readmissions

然而，结果并不适合每个患者和每个日期。是这样的：

Patient_ID Surgery_Date  Readmission
1838       2017-01-05        0
1838       2018-04-26        0
087        2017-01-11        0
1838       2017-07-06        0
087        2017-03-17        2
1838       2018-08-02        2
087        2017-11-15        4 (It's wrong because in the last 6 months there was 1 surgery for this ID)
1838       2018-11-22        3 (It's wrong because in the last 6 months there were 2 surgeries for this ID)
087        2017-02-01        1
087        2017-06-21        3
1838       2018-06-14        1

任何人都可以帮助我吗？

标签： pythonloopsdatebinarydiff

解决方案

这

admissions_sorted.diff()

计算连续的 diff() 并且不超过一个索引。此代码将累积计算 diff() 并与 180 天进行比较：

import pandas as pd
import datetime as dt
import numpy as np

# Your data plus a new patient that comes often                                                                                                                                                                    
data = {'Patient_ID':[1838,1838,87,1838,87,1838,87,1838,87,87,1838], 
        'Surgery_Date': ['2017-01-05','2018-04-26','2017-01-11','2017-07-06','2017-03-17','2018-08-02','2017-11-15','2018-11-22','2017-02-01','2017-06-21','2018-06-14']}

df = pd.DataFrame(data,columns = ['Patient_ID','Surgery_Date'])
readmissions = pd.Series(np.zeros(len(df),dtype=int),index=df.index)

# Loop through all unique ids                                                                                                                                                                                      
all_id = df['Patient_ID'].unique()
id_admissions = {}
for pid in all_id:
    # These are all the times a patient with a given ID has had surgery                                                                                                                                            
    patient = df.loc[df['Patient_ID']==pid]
    admissions_sorted = pd.to_datetime(patient['Surgery_Date'], format='%Y-%m-%d').sort_values()
    # This checks if the previous surgery was longer than 180 days ago                                                                                                                                              
    frequency = admissions_sorted.diff()/np.timedelta64(1, 'D')
    n_admissions = []
    for i in range(frequency.shape[0]-1):
      cumulative_diff = frequency.iloc[::-1][i:-1].cumsum().astype(int)
      #Compute the readmission                                                                                                                                                                                      
      n_admissions.append(cumulative_diff[cumulative_diff<180].count())
    n_admissions.append(0)
    n_admissions.reverse()

    # Add these value to the time series                                                                                                                                                                           
    readmissions.loc[admissions_sorted.index] = n_admissions


df['Readmission'] = readmissions

输出是：

    Patient_ID Surgery_Date  Readmission
0         1838   2017-01-05            0
1         1838   2018-04-26            0
2           87   2017-01-11            0
3         1838   2017-07-06            0
4           87   2017-03-17            2
5         1838   2018-08-02            2
6           87   2017-11-15            1
7         1838   2018-11-22            2
8           87   2017-02-01            1
9           87   2017-06-21            3
10        1838   2018-06-14            1

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