Mixing
Resetting outliers in a timeseries dataframe to 3 SD
Domain: Python & Pandas
I have a time series data frame which has the total number of customers for each day for the last 10 years.
The columns are:
- date
- total customers
There are outliers in my total customers column.
I wanted to reset the outliers outside of 3 standard deviations above the mean to a value as defined by the formula below.
Outlier which is above 3SD = Mean + 3 S.D.
python dataframe statistics
asked Nov 21 '18 at 21:39
zoshzosh
267
add a comment |
Domain: Python & Pandas
I have a time series data frame which has the total number of customers for each day for the last 10 years.
The columns are:
- date
- total customers
There are outliers in my total customers column.
I wanted to reset the outliers outside of 3 standard deviations above the mean to a value as defined by the formula below.
Outlier which is above 3SD = Mean + 3 S.D.
python dataframe statistics
asked Nov 21 '18 at 21:39
zoshzosh
267
add a comment |
Domain: Python & Pandas
I have a time series data frame which has the total number of customers for each day for the last 10 years.
The columns are:
- date
- total customers
There are outliers in my total customers column.
I wanted to reset the outliers outside of 3 standard deviations above the mean to a value as defined by the formula below.
Outlier which is above 3SD = Mean + 3 S.D.
python dataframe statistics
asked Nov 21 '18 at 21:39
zoshzosh
267
Domain: Python & Pandas
I have a time series data frame which has the total number of customers for each day for the last 10 years.
The columns are:
- date
- total customers
There are outliers in my total customers column.
I wanted to reset the outliers outside of 3 standard deviations above the mean to a value as defined by the formula below.
Outlier which is above 3SD = Mean + 3 S.D.
python dataframe statistics
python dataframe statistics
asked Nov 21 '18 at 21:39
zoshzosh
267
asked Nov 21 '18 at 21:39
zoshzosh
267
edited Nov 21 '18 at 21:51
zosh
asked Nov 21 '18 at 21:39
zoshzosh
267
asked Nov 21 '18 at 21:39
zoshzosh
267
asked Nov 21 '18 at 21:39
zoshzosh
267
267
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
You could use the .clip_upper() method to limit values in the customers column to mean+3*sd.
m = df['total customers'].mean()
sd = df['total customers'].std()
df['total customers'] = df['total_customers'].clip_upper(m + 3*sd)
Here's the documentation for clip_upper.
answered Nov 21 '18 at 21:44
CraigCraig
2,1961819
Thank you so much for your reply
– zosh
Nov 21 '18 at 21:52
1
This function does exactly what you are asking for. It replaces any values that exceed the 'clip' value with the 'clip' value. It does not remove anything.
– Craig
Nov 21 '18 at 21:54
got it thank you so much
– zosh
Nov 21 '18 at 21:55
Hey Craig, sorry to bother you again: What if I wanted to completely remove all the rows with outliers?
– zosh
Nov 21 '18 at 22:13
@zosh - That's a new question, but the answer is to use boolean indexing as described in stackoverflow.com/a/23200666/7517724
– Craig
Nov 22 '18 at 0:13
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
You could use the .clip_upper() method to limit values in the customers column to mean+3*sd.
m = df['total customers'].mean()
sd = df['total customers'].std()
df['total customers'] = df['total_customers'].clip_upper(m + 3*sd)
Here's the documentation for clip_upper.
answered Nov 21 '18 at 21:44
CraigCraig
2,1961819
Thank you so much for your reply
– zosh
Nov 21 '18 at 21:52
1
This function does exactly what you are asking for. It replaces any values that exceed the 'clip' value with the 'clip' value. It does not remove anything.
– Craig
Nov 21 '18 at 21:54
got it thank you so much
– zosh
Nov 21 '18 at 21:55
Hey Craig, sorry to bother you again: What if I wanted to completely remove all the rows with outliers?
– zosh
Nov 21 '18 at 22:13
@zosh - That's a new question, but the answer is to use boolean indexing as described in stackoverflow.com/a/23200666/7517724
– Craig
Nov 22 '18 at 0:13
add a comment |
You could use the .clip_upper() method to limit values in the customers column to mean+3*sd.
m = df['total customers'].mean()
sd = df['total customers'].std()
df['total customers'] = df['total_customers'].clip_upper(m + 3*sd)
Here's the documentation for clip_upper.
answered Nov 21 '18 at 21:44
CraigCraig
2,1961819
Thank you so much for your reply
– zosh
Nov 21 '18 at 21:52
1
This function does exactly what you are asking for. It replaces any values that exceed the 'clip' value with the 'clip' value. It does not remove anything.
– Craig
Nov 21 '18 at 21:54
got it thank you so much
– zosh
Nov 21 '18 at 21:55
Hey Craig, sorry to bother you again: What if I wanted to completely remove all the rows with outliers?
– zosh
Nov 21 '18 at 22:13
@zosh - That's a new question, but the answer is to use boolean indexing as described in stackoverflow.com/a/23200666/7517724
– Craig
Nov 22 '18 at 0:13
add a comment |
You could use the .clip_upper() method to limit values in the customers column to mean+3*sd.
m = df['total customers'].mean()
sd = df['total customers'].std()
df['total customers'] = df['total_customers'].clip_upper(m + 3*sd)
Here's the documentation for clip_upper.
answered Nov 21 '18 at 21:44
CraigCraig
2,1961819
You could use the .clip_upper() method to limit values in the customers column to mean+3*sd.
m = df['total customers'].mean()
sd = df['total customers'].std()
df['total customers'] = df['total_customers'].clip_upper(m + 3*sd)
Here's the documentation for clip_upper.
answered Nov 21 '18 at 21:44
CraigCraig
2,1961819
answered Nov 21 '18 at 21:44
CraigCraig
2,1961819
answered Nov 21 '18 at 21:44
CraigCraig
2,1961819
answered Nov 21 '18 at 21:44
CraigCraig
2,1961819
2,1961819
Thank you so much for your reply
– zosh
Nov 21 '18 at 21:52
1
This function does exactly what you are asking for. It replaces any values that exceed the 'clip' value with the 'clip' value. It does not remove anything.
– Craig
Nov 21 '18 at 21:54
got it thank you so much
– zosh
Nov 21 '18 at 21:55
Hey Craig, sorry to bother you again: What if I wanted to completely remove all the rows with outliers?
– zosh
Nov 21 '18 at 22:13
@zosh - That's a new question, but the answer is to use boolean indexing as described in stackoverflow.com/a/23200666/7517724
– Craig
Nov 22 '18 at 0:13
add a comment |
Thank you so much for your reply
– zosh
Nov 21 '18 at 21:52
1
This function does exactly what you are asking for. It replaces any values that exceed the 'clip' value with the 'clip' value. It does not remove anything.
– Craig
Nov 21 '18 at 21:54
got it thank you so much
– zosh
Nov 21 '18 at 21:55
Hey Craig, sorry to bother you again: What if I wanted to completely remove all the rows with outliers?
– zosh
Nov 21 '18 at 22:13
@zosh - That's a new question, but the answer is to use boolean indexing as described in stackoverflow.com/a/23200666/7517724
– Craig
Nov 22 '18 at 0:13
Thank you so much for your reply
– zosh
Nov 21 '18 at 21:52
Thank you so much for your reply
– zosh
Nov 21 '18 at 21:52
1
1
This function does exactly what you are asking for. It replaces any values that exceed the 'clip' value with the 'clip' value. It does not remove anything.
– Craig
Nov 21 '18 at 21:54
This function does exactly what you are asking for. It replaces any values that exceed the 'clip' value with the 'clip' value. It does not remove anything.
– Craig
Nov 21 '18 at 21:54
got it thank you so much
– zosh
Nov 21 '18 at 21:55
got it thank you so much
– zosh
Nov 21 '18 at 21:55
Hey Craig, sorry to bother you again: What if I wanted to completely remove all the rows with outliers?
– zosh
Nov 21 '18 at 22:13
Hey Craig, sorry to bother you again: What if I wanted to completely remove all the rows with outliers?
– zosh
Nov 21 '18 at 22:13
@zosh - That's a new question, but the answer is to use boolean indexing as described in stackoverflow.com/a/23200666/7517724
– Craig
Nov 22 '18 at 0:13
@zosh - That's a new question, but the answer is to use boolean indexing as described in stackoverflow.com/a/23200666/7517724
– Craig
Nov 22 '18 at 0:13
add a comment |
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