Don't drop unique value with dropna() pandas

Question

what’s up?

I am having a little problem, where I need to use the pandas dropna function to remove rows from my dataframe. However, I need it to not delete the unique values from my dataframe.

Let me explain better. I have the following dataframe:

id	birthday
0102-2	09/03/2020
0103-2	14/03/2020
0104-2	NaN
0105-2	NaN
0105-2	25/03/2020
0108-2	07/04/2020

In the case above, I need to delete the row from my dataframe based on the NaN values in the birthday column. However, as you can see the id "0104-2" is unique unlike the id "0105-2" where it has a NaN value and another with a date. So I would like to keep track of all the lines that have NaN that are unique.

Is it feasible to do this with dropna, or would I have to pre-process the information beforehand?

Asked By: Douglas F

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Source

Answer 1

I hope this is what you wanted:

import numpy as np
import pandas as pd

df = pd.DataFrame({'id': [1, 2, 3, 3, 4], 'birthday': [1994, np.nan, np.nan, 2019, 2020]})
df["id_count"] = df.groupby('id')['id'].transform('count') # helper column to identify unique ids
df_filtered = df[(df["id_count"] > 1) & df["birthday"].isna() == False] # filter out nans within non-unique ids
df_filtered.drop(["id_count"], axis=1) # drop helper column

output

    id  birthday
0   1   1994.0
1   2   NaN
3   3   2019.0
4   4   2020.0

Answered By: hannez

Answer 2

Sorry I read your requirement wrongly.. Should sort first then using shift()

df.sort_values(by='id', na_position='first', inplace=True)
df.loc[(df['birthday'].notna()) | (df['id'].shift(-1) != df['id'])]

Original Answer:

You can filter using notna() or duplicated()

df.loc[(df['birthday'].notna()) | (df.duplicated(subset='id', keep=False))]

Answered By: crx91

Answer 3

I could do this by grouping by id and aggregating with .agg('first'):

# added a couple more rows to show the behaviour
df = pd.DataFrame({'id': ['0102-2', '0102-2', '0103-2', '0103-2', '0104-2', '0105-2', '0105-2', '0108-2'],
                   'birthday': ['09/03/2020', np.nan, '14/03/2020', '12/03/2020', np.nan, np.nan, '25/03/2020', '07/04/2020']})
df

# Output:
    id      birthday
0   0102-2  09/03/2020
1   0102-2  NaN         # duplicate id with NaN goes second
2   0103-2  14/03/2020
3   0103-2  12/03/2020  # duplicate id with an earlier date
4   0104-2  NaN         # unique with NaN
5   0105-2  NaN         # duplicate with NaN goes first
6   0105-2  25/03/2020
7   0108-2  07/04/2020

df.groupby('id').agg('first').reset_index()

# Output:
    id      birthday
0   0102-2  09/03/2020
1   0103-2  14/03/2020
2   0104-2  None
3   0105-2  25/03/2020
4   0108-2  07/04/2020

Answered By: n.shabankin

Answer 4

You could sort by the birthday column and then drop duplicates keeping the first out of the two, by doing the following:

The complete code would look like this:

import pandas as pd
import numpy as np

data = {
    "id": ['102-2','103-2','104-2', '105-2', '105-2', '108-2'],
    "birthday":['09/03/2020', '14/03/2020', np.nan, np.nan, '25/03/2020', '07/04/2020']
}

df = pd.DataFrame(data)
df.sort_values(['birthday'], inplace=True)
df.drop_duplicates(subset="id", keep='first', inplace=True)
df.sort_values(['id'], inplace=True)

CODE EXPLANATION:
Here is the original dataframe:

import pandas as pd
import numpy as np

data = {
    "id": ['102-2','103-2','104-2', '105-2', '105-2', '108-2'],
    "birthday":['09/03/2020', '14/03/2020', np.nan, np.nan, '25/03/2020', '07/04/2020']
}

df = pd.DataFrame(data)

Now sort the dataframe:

df.sort_values(['birthday'], inplace=True)

Then drop the duplicates based on the id column. Keeping only the first value.

df.drop_duplicates(subset="id", keep='first', inplace=True)

Answered By: ScottC

Don't drop unique value with dropna() pandas

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