如何在熊猫数据框中将单元格的值拆分为多行？

Question

I have a following data frame, which was obtained using the code:

     df1=df.groupby('id')['x,y'].apply(lambda x: rdp(x.tolist(), 5.0)).reset_index()

Refer here

The resultant data frame obtained :

      id          x,y
  0   1    [(0, 0), (1, 2)]
  1   2    [(1, 3), (1, 2)]
  2   3    [(2, 5), (4, 6)]  

Is it possible to get something like this:

         id      x,y
     0   1      (0, 0)
     1   1      (1, 2)
     2   2      (1, 3)
     3   2      (1, 2)
     4   3      (2, 5)
     5   3      (4, 6)

Here, the list of coordinates obtained as a result in previous df is split into new rows against their respective ids.

Answer 1

You can use DataFrame constructor with stack :

df2 = pd.DataFrame(df1['x,y'].values.tolist(), index=df1['id'])
        .stack()
        .reset_index(level=1, drop=True)
        .reset_index(name='x,y')
print (df2)

   id     x,y
0   1  (0, 0)
1   1  (1, 2)
2   2  (1, 3)
3   2  (1, 2)
4   3  (2, 5)
5   3  (4, 6)

numpy solution use numpy.repeat by lengths of values by str.len , x,y column is flattenig by numpy.ndarray.sum :

df2 = pd.DataFrame({'id': np.repeat(df1['id'].values, df1['x,y'].str.len()), 
                   'x,y': df1['x,y'].values.sum()})

print (df2)
   id     x,y
0   1  (0, 0)
0   1  (1, 2)
1   2  (1, 3)
1   2  (1, 2)
2   3  (2, 5)
2   3  (1, 9)
2   3  (4, 6)

Timings :

In [54]: %timeit pd.DataFrame(df1['x,y'].values.tolist(), index=df1['id']).stack().reset_index(level=1, drop=True).reset_index(name='x,y')
1000 loops, best of 3: 1.49 ms per loop

In [55]: %timeit pd.DataFrame({'id': np.repeat(df1['id'].values, df1['x,y'].str.len()), 'x,y': df1['x,y'].values.sum()})
1000 loops, best of 3: 562 µs per loop

#piRSquared solution
In [56]: %timeit pd.DataFrame({'id': df1['id'].repeat(df1['x,y'].str.len()), 'x,y': df1['x,y'].sum() })
1000 loops, best of 3: 712 µs per loop

Answer 2

• Calculating the new 'id' column
  • We can use pandas str.len method to quickly count the number of elements in each element's sub-list. This is convenient as we can directly pass this result to the repeat method of df1['id'] which will repeat each element by a corresponding amount from the lengths we passed.
• Calculating the new 'x,y' column
  • typically, I like to use np.concatenate to push all the sub-lists together. However, in this case, the sub-lists are lists of tuples. np.concatenate will not treat these as lists of objects. So instead, I use the sum method and that will use the underlying sum method on lists, which will in turn concatenate.

pandas

if we stick with pandas we can keep the code cleaner
Use repeat with str.len and sum

pd.DataFrame({
        'id': df1['id'].repeat(df1['x,y'].str.len()),
        'x,y': df1['x,y'].sum()
    })

   id     x,y
0   1  (0, 0)
0   1  (1, 2)
1   2  (1, 3)
1   2  (1, 2)
2   3  (2, 5)
2   3  (4, 6)

numpy

we can quicken this approach up by using the underlying numpy arrays and equivalent numpy methods
NOTE: this is equivalent logic!

pd.DataFrame({
        'id': df1['id'].values.repeat(df1['x,y'].str.len()),
        'x,y': df1['x,y'].values.sum()
    })

We can speed it up even more by skipping the the str.len method and calculating the lengths with a list comprehension.

pd.DataFrame({
        'id': df1['id'].values.repeat([len(w) for w in df1['x,y'].values.tolist()]),
        'x,y': df1['x,y'].values.sum()
    })

Time Tests

small data

%%timeit
pd.DataFrame({
        'id': df1['id'].values.repeat([len(w) for w in df1['x,y'].values.tolist()]),
        'x,y': df1['x,y'].values.sum()
    })
1000 loops, best of 3: 351 µs per loop

%%timeit
pd.DataFrame({
        'id': df1['id'].repeat(df1['x,y'].str.len()),
        'x,y': df1['x,y'].sum()
    })
1000 loops, best of 3: 590 µs per loop

%%timeit 
pd.DataFrame({'id': np.repeat(df1['id'].values, df1['x,y'].str.len()), 
                   'x,y': df1['x,y'].values.sum()})
​
1000 loops, best of 3: 498 µs per loop

larger data

df1 = pd.concat([df1.head(3)] * 100, ignore_index=True)

%%timeit
pd.DataFrame({
        'id': df1['id'].values.repeat([len(w) for w in df1['x,y'].values.tolist()]),
        'x,y': df1['x,y'].values.sum()
    })
1000 loops, best of 3: 579 µs per loop

%%timeit
pd.DataFrame({
        'id': df1['id'].repeat(df1['x,y'].str.len()),
        'x,y': df1['x,y'].sum()
    })
1000 loops, best of 3: 841 µs per loop

%%timeit 
pd.DataFrame({'id': np.repeat(df1['id'].values, df1['x,y'].str.len()), 
                   'x,y': df1['x,y'].values.sum()})
​
1000 loops, best of 3: 704 µs per loop