How to match between columns in a dataframe and keep another column

Question

I would like to get the matches from one column with the other columns in a dataframe. Below is an example:

  date        tableNameFrom   tableNameJoin   attributeName
1 29-03-2019  film            language        [film.languageId, language.languageID]
2 30-03-2019  inventory       rental          [invetory.inventoryId, rental.filmId]

In the example above, I would like to match between tablenameFrom and tablenameJoin with the attributeName. But, I want to keep the date column. Here desired output:

  date        tableName    attributeName
1 29-03-2019  film         languageId
2 29-03-2019  language     languageID
3 30-03-2019  inventory    inventoryId
4 30-03-2019  rental       filmId

Any idea what should I do? Thank you.

Answer 1

This may not be what you want, but yields the expected output for the given dataframe:

(df.set_index('date').attributeName.apply(pd.Series).stack()
 .reset_index().drop('level_1', axis=1).set_index('date')[0]
 .str.split('.').apply(pd.Series)
 )

It ignores tableNameFrome and tableNameJoin and assumes that these are given in attributeName.

+----+------------+----------+-------------+
|    |       date |        0 |           1 |
|----+------------+----------+-------------|
|  0 | 29-03-2019 |     film |  languageId |
|  1 | 29-03-2019 | language |  languageID |
|  2 | 30-03-2019 | invetory | inventoryId |
|  3 | 30-03-2019 |   rental |      filmId |
+----+------------+----------+-------------+

Answer 2

df.drop(["tableNameFrom","tableNameJoin"],inplace=True)
df2=pd.DataFrame()
for row in df.itertuples():
    for i in range(2):
        df2=df2.append(pd.Series(
            [row[1],"".join(row[2][i].split('.')[0]),"".join(row[2][i].split('.')[1])]).T,ignore_index=True)
df2.columns=["date","tableName","tableAttribute"]
df2

don't see why you need the middle two columns so i dropped them.

Answer 3

Here is one possible approach that avoids using apply

Raw DataFrame

         date tableNameFrom tableNameJoin                           attributeName
0  29-03-2019          film      language  [film.languageId, language.languageID]
1  30-03-2019     inventory        rental   [invetory.inventoryId, rental.filmId]

Step 1 - (1) Split attributeName into 2 separate columns using , as the separator, (2) remove unwanted square brackets ([ or ]), (3) drop unwanted columns

# 1
df[['tableName','attributeName2']] = df['attributeName'].str.split(',', expand=True)

# 2
df['tableName'] = df['tableName'].str.strip('[')
df['attributeName2'] = df['attributeName2'].str.strip(']')

# 3
df.drop(['attributeName','tableNameFrom','tableNameJoin'], axis=1, inplace=True)

print(df)
         date             tableName        attributeName2
0  29-03-2019       film.languageId   language.languageID
1  30-03-2019  invetory.inventoryId         rental.filmId

Step 2 - Finally, use this SO post to split the rows from the columns tableName and attributeName2 into separate columns

df_match = (df.set_index(['date'])
               .stack()
               .str.split('.', expand=True)
               .stack()
               .unstack(-1)
               .reset_index(-1, drop=True)
               .reset_index()
            )
df_match.columns = ['date','tableName','attributeName']

print(df_match)
         date  tableName attributeName
0  29-03-2019       film    languageId
1  29-03-2019   language    languageID
2  30-03-2019   invetory   inventoryId
3  30-03-2019     rental        filmId

Details

• .set_index(['date'] - set the columns that are to be kept as the index of the DataFrame
• .stack() - stack rows
• .str.split('.', expand=True) - call str.split on period (.) to split those cells into separate columns
• .stack() - get rid of NULL values, so call stack again
• .unstack(-1)
  • since we want the last level of the index to become our columns, so unstack using unstack(-1) (unstack on the last level)
  • I have shown this in detail below¹
• .reset_index(-1, drop=True) - eliminate superfluous last level using reset_index(-1)

---

¹ Here is the output before the .unstack(-1) step and an explanation of why we need to use -1 inside unstack()

df_intermediate = (df.set_index(['date'])
                       .stack()
                       .str.split('.', expand=True)
                       .stack()
                )

print(df_intermediate)
date                         
29-03-2019  tableName       0           film
                            1     languageId
            attributeName2  0       language
                            1     languageID
30-03-2019  tableName       0       invetory
                            1    inventoryId
            attributeName2  0         rental
                            1         filmId
dtype: object

• this shows that we want the last index level (with values 0, 1) to become the columns so we choose -1 in .unstack(-1)

Answer 4

This is basically a melt problem but with an intermediate step to clean up your attributeName column:

---

# Clean up attributeName column by converting it to string and expanding it by splitting
df = pd.concat([df, df.attributeName.apply(lambda x: ' '.join(x).split('.')[1]).str.split(expand=True)], axis=1)\
     .drop('attributeName', axis=1)\
     .rename({0:'attribute1', 1:'attribute2'}, axis=1)

Which gives us the following dataframe:

print(df)
         date tableNameFrom tableNameJoin   attribute1 attribute2
0  29-03-2019          film      language   languageId   language
1  30-03-2019     inventory        rental  inventoryId     rental

---

Then we can use melt twice to get the desired output:

df1 = df.melt(id_vars='date', value_vars=['tableNameFrom', 'tableNameJoin'], value_name='tableName').drop('variable', axis=1)
df2 = df.melt(id_vars='date', value_vars=['attribute1', 'attribute2'], value_name='attributeName').drop(['variable', 'date'], axis=1)

df = pd.concat([df1, df2], axis=1)

Which gives us the final output:

print(df)
        date  tableName attributeName
0  29-03-2019       film    languageId
1  30-03-2019  inventory   inventoryId
2  29-03-2019   language      language
3  30-03-2019     rental        rental