explode list and get index values for each list

Question

original DataFrame df

  label  value
0     a      1
1     a      2
2     b      3
3     a      4
4     b      5

ds = df.groupby('label')['value'].apply(list)
```
label
a    [1, 2, 4]
b       [3, 5]
```

ds.explode() generate

What I need is a new column to represents the position of the item in the value column in the original list.

 label value position
 a    1      0
 a    2      1
 a    4      2
 b    3      0
 b    5      1

update:

Example above is not appropriate, suppose the original dataframe is

  label   value
0     a  [1, 2]
1     a     [2]
2     b     [3]
3     a     [4]
4     b     [5]

df.explode('value') gives

  label value
0     a     1
0     a     2
1     a     2
2     b     3
3     a     4
4     b     5

how I get

  label value position
0     a     1  0
0     a     2  1
1     a     2  0
2     b     3  0
3     a     4  0
4     b     5  0

BTW, do you really need explode? How about just df.sort_values("label")? — Chris
– Chris, Commented Jan 5, 2021 at 3:53

Ch3steR · Accepted Answer · 2021-01-05 04:21:06Z

2

First, get length of list in value then apply np.arange or range on it using pd.Series.map or pd.Series.apply. Now, explode value

df['value'] = df['value'].str.len().map(np.arange)
#                                 _.map(range) # Alternative
df.explode('value')

  label value
0     a     0
0     a     1
1     a     0
2     b     0
3     a     0
4     b     0

answered Jan 5, 2021 at 4:21

Ch3steR

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Comments

wwnde · Accepted Answer · 2021-01-05 03:52:22Z

0

Lets Try:

 df['position']=df.groupby(['label'])['value'].cumcount()

answered Jan 5, 2021 at 3:52

wwnde

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