Replace a value in a column if that value appears only once
$begingroup$
I have a dataframe, I want to replace the values in one column by "other" if the value count of that value in that column is exactly 1
i Food_group
0 Flake
1 Flake
2 Flake
3 Almond
4 Drink
5 Drink
6 Flake
I have tried,
data["food_group"] = data.food_group.apply(lambda x: "other" if x.value_counts()==1 else x)
I got error
AttributeError: 'str' object has no attribute 'value_counts'
python
$endgroup$
add a comment |
$begingroup$
I have a dataframe, I want to replace the values in one column by "other" if the value count of that value in that column is exactly 1
i Food_group
0 Flake
1 Flake
2 Flake
3 Almond
4 Drink
5 Drink
6 Flake
I have tried,
data["food_group"] = data.food_group.apply(lambda x: "other" if x.value_counts()==1 else x)
I got error
AttributeError: 'str' object has no attribute 'value_counts'
python
$endgroup$
add a comment |
$begingroup$
I have a dataframe, I want to replace the values in one column by "other" if the value count of that value in that column is exactly 1
i Food_group
0 Flake
1 Flake
2 Flake
3 Almond
4 Drink
5 Drink
6 Flake
I have tried,
data["food_group"] = data.food_group.apply(lambda x: "other" if x.value_counts()==1 else x)
I got error
AttributeError: 'str' object has no attribute 'value_counts'
python
$endgroup$
I have a dataframe, I want to replace the values in one column by "other" if the value count of that value in that column is exactly 1
i Food_group
0 Flake
1 Flake
2 Flake
3 Almond
4 Drink
5 Drink
6 Flake
I have tried,
data["food_group"] = data.food_group.apply(lambda x: "other" if x.value_counts()==1 else x)
I got error
AttributeError: 'str' object has no attribute 'value_counts'
python
python
asked Apr 21 at 9:24
KHAN irfanKHAN irfan
12110
12110
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is easier if you store the value count separately to avoid redoing it inside the apply loop. You can do it like this:
food_count = data["food_group"].value_counts()
data["food_group"] = data["food_group"].apply(lambda x: "other" if food_count[x]==1 else x)
$endgroup$
$begingroup$
Have you tried the code? it does not work
$endgroup$
– KHAN irfan
Apr 21 at 9:43
$begingroup$
Test it now. :) Tried on another dataset and forgot to change back to "food_group"
$endgroup$
– Simon Larsson
Apr 21 at 9:45
$begingroup$
Tested it on your data now, so it should work.
$endgroup$
– Simon Larsson
Apr 21 at 9:51
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is easier if you store the value count separately to avoid redoing it inside the apply loop. You can do it like this:
food_count = data["food_group"].value_counts()
data["food_group"] = data["food_group"].apply(lambda x: "other" if food_count[x]==1 else x)
$endgroup$
$begingroup$
Have you tried the code? it does not work
$endgroup$
– KHAN irfan
Apr 21 at 9:43
$begingroup$
Test it now. :) Tried on another dataset and forgot to change back to "food_group"
$endgroup$
– Simon Larsson
Apr 21 at 9:45
$begingroup$
Tested it on your data now, so it should work.
$endgroup$
– Simon Larsson
Apr 21 at 9:51
add a comment |
$begingroup$
It is easier if you store the value count separately to avoid redoing it inside the apply loop. You can do it like this:
food_count = data["food_group"].value_counts()
data["food_group"] = data["food_group"].apply(lambda x: "other" if food_count[x]==1 else x)
$endgroup$
$begingroup$
Have you tried the code? it does not work
$endgroup$
– KHAN irfan
Apr 21 at 9:43
$begingroup$
Test it now. :) Tried on another dataset and forgot to change back to "food_group"
$endgroup$
– Simon Larsson
Apr 21 at 9:45
$begingroup$
Tested it on your data now, so it should work.
$endgroup$
– Simon Larsson
Apr 21 at 9:51
add a comment |
$begingroup$
It is easier if you store the value count separately to avoid redoing it inside the apply loop. You can do it like this:
food_count = data["food_group"].value_counts()
data["food_group"] = data["food_group"].apply(lambda x: "other" if food_count[x]==1 else x)
$endgroup$
It is easier if you store the value count separately to avoid redoing it inside the apply loop. You can do it like this:
food_count = data["food_group"].value_counts()
data["food_group"] = data["food_group"].apply(lambda x: "other" if food_count[x]==1 else x)
edited Apr 21 at 9:45
answered Apr 21 at 9:40
Simon LarssonSimon Larsson
1,140216
1,140216
$begingroup$
Have you tried the code? it does not work
$endgroup$
– KHAN irfan
Apr 21 at 9:43
$begingroup$
Test it now. :) Tried on another dataset and forgot to change back to "food_group"
$endgroup$
– Simon Larsson
Apr 21 at 9:45
$begingroup$
Tested it on your data now, so it should work.
$endgroup$
– Simon Larsson
Apr 21 at 9:51
add a comment |
$begingroup$
Have you tried the code? it does not work
$endgroup$
– KHAN irfan
Apr 21 at 9:43
$begingroup$
Test it now. :) Tried on another dataset and forgot to change back to "food_group"
$endgroup$
– Simon Larsson
Apr 21 at 9:45
$begingroup$
Tested it on your data now, so it should work.
$endgroup$
– Simon Larsson
Apr 21 at 9:51
$begingroup$
Have you tried the code? it does not work
$endgroup$
– KHAN irfan
Apr 21 at 9:43
$begingroup$
Have you tried the code? it does not work
$endgroup$
– KHAN irfan
Apr 21 at 9:43
$begingroup$
Test it now. :) Tried on another dataset and forgot to change back to "food_group"
$endgroup$
– Simon Larsson
Apr 21 at 9:45
$begingroup$
Test it now. :) Tried on another dataset and forgot to change back to "food_group"
$endgroup$
– Simon Larsson
Apr 21 at 9:45
$begingroup$
Tested it on your data now, so it should work.
$endgroup$
– Simon Larsson
Apr 21 at 9:51
$begingroup$
Tested it on your data now, so it should work.
$endgroup$
– Simon Larsson
Apr 21 at 9:51
add a comment |
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