从宽到长返回空输出-Python数据框 [英] Wide to long returns empty output - Python dataframe

问题描述

我有一个数据框，可以从下面给出的代码中生成

I have a dataframe which can be generated from the code as given below

df = pd.DataFrame({'person_id' :[1,2,3],'date1': 
['12/31/2007','11/25/2009','10/06/2005'],'val1': 
[2,4,6],'date2': ['12/31/2017','11/25/2019','10/06/2015'],'val2':[1,3,5],'date3': 
['12/31/2027','11/25/2029','10/06/2025'],'val3':[7,9,11]})

我按照以下解决方案将其从宽转换为长

I followed the below solution to convert it from wide to long

pd.wide_to_long(df, stubnames=['date', 'val'], i='person_id', 
j='grp').sort_index(level=0)

尽管这适用于如下所示的示例数据，但不适用于我的实际数据，该数据有200列以上.我的真实数据不是subject_id，而是我的真实数据，它的subject_ID是DC0001，DC0002等值."I"是否总是必须是数字?而是将存根值添加为数据集中的新列，并且行数为零

Though this works with sample data as shown below, it doesn't work with my real data which has more than 200 columns. Instead of person_id, my real data has subject_ID which is values like DC0001,DC0002 etc. Does "I" always have to be numeric? Instead it adds the stub values as new columns in my dataset and has zero rows

这就是我的真实专栏的样子

This is how my real columns looks like

我的真实数据可能还包含NA.那么，是否必须使用默认值填充它们才能使wide_to_long工作?

My real data might contains NA's as well. So do I have to fill them with default values for wide_to_long to work?

您能帮忙解决一下什么问题吗?或任何其他获得相同结果的方法也很有帮助.

Can you please help as to what can be the issue? Or any other approach to achieve the same result is also helpful.

推荐答案

问题出在您的列名上，用于从宽到长的转换的数字必须在列名的末尾，或者您需要指定一个后缀为groupby.我认为最简单的解决方案是创建一个接受正则表达式和数据框的函数.

The issue is with your column names, the numbers used to convert from wide to long need to be at the end of your column names or you need to specify a suffix to groupby. I think the easiest solution is to create a function that accepts regex and the dataframe.

import pandas as pd
import re

def change_names(df, regex):
    # Select one of three column groups
    old_cols = df.filter(regex = regex).columns
    # Create list of new column names
    new_cols = []
    for col in old_cols:
        # Get the stubname of the original column
        stub = ''.join(re.split(r'\d', col))
        # Get the time point
        num = re.findall(r'\d+', col) # returns a list like ['1']
        # Make new column name
        new_col = stub + num[0]
        new_cols.append(new_col)

    # Create dictionary mapping old column names to new column names
    dd = {oc: nc for oc, nc in zip(old_cols, new_cols)}
    # Rename columns
    df.rename(columns = dd, inplace = True)

    return df


tdf = pd.DataFrame({'person_id' :[1,2,3],'h1date': ['12/31/2007','11/25/2009','10/06/2005'],'t1val': [2,4,6],'h2date': ['12/31/2017','11/25/2019','10/06/2015'],'t2val':[1,3,5],'h3date': ['12/31/2027','11/25/2029','10/06/2025'],'t3val':[7,9,11]})

# Change date columns
tdf = change_names(tdf, 'date$')
tdf = change_names(tdf, 'val$')

print(tdf)
    person_id      hdate1  tval1      hdate2  tval2      hdate3  tval3
0          1  12/31/2007      2  12/31/2017      1  12/31/2027      7
1          2  11/25/2009      4  11/25/2019      3  11/25/2029      9
2          3  10/06/2005      6  10/06/2015      5  10/06/2025     11

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