Retain the highest absolute value and return an average value from rows with duplicate index
2
I have a set of values with duplicate index in a DataFrame:
value
CDE 2.318620
CDE -3.097715
LXU -3.791043
LXU 4.818995
SWN 3.059964
SWN -4.349304
OAS -3.336539
LPI -3.037097
LPI -5.701044
LPI -3.519923
CZR -3.270018
CZR -3.056712
The required result is to retain only the highest absolute values and return the average values in a new column:
value average
CDE -3.097715 -0.389547
LXU 4.818995 0.513976
SWN -4.349304 -0.644670
OAS -3.336539 -3.336539
LPI -5.701044 -4.086021
CZR -3.270018 -3.163365
I have tried to .apply(lambda) to the duplicate rows but got an "axis" error:
max_absolute = lambda x: max(x.min(), x.max(), key=abs)
df_duplicate_absmax = df.groupby(df.index).apply(max_absolute, axis=1)
ps: Adapting Abhi's solution to work with NaN:
df1 = df.groupby(df.index)['value'].agg([lambda x: max(x[~np.isnan(x)], key=abs), 'mean'])
python python-3.x pandas dataframe pandas-groupby
asked Dec 9 '18 at 1:19
artDecoartDeco
185214
add a comment |
2
I have a set of values with duplicate index in a DataFrame:
value
CDE 2.318620
CDE -3.097715
LXU -3.791043
LXU 4.818995
SWN 3.059964
SWN -4.349304
OAS -3.336539
LPI -3.037097
LPI -5.701044
LPI -3.519923
CZR -3.270018
CZR -3.056712
The required result is to retain only the highest absolute values and return the average values in a new column:
value average
CDE -3.097715 -0.389547
LXU 4.818995 0.513976
SWN -4.349304 -0.644670
OAS -3.336539 -3.336539
LPI -5.701044 -4.086021
CZR -3.270018 -3.163365
I have tried to .apply(lambda) to the duplicate rows but got an "axis" error:
max_absolute = lambda x: max(x.min(), x.max(), key=abs)
df_duplicate_absmax = df.groupby(df.index).apply(max_absolute, axis=1)
ps: Adapting Abhi's solution to work with NaN:
df1 = df.groupby(df.index)['value'].agg([lambda x: max(x[~np.isnan(x)], key=abs), 'mean'])
python python-3.x pandas dataframe pandas-groupby
asked Dec 9 '18 at 1:19
artDecoartDeco
185214
add a comment |
2
2
2
0
I have a set of values with duplicate index in a DataFrame:
value
CDE 2.318620
CDE -3.097715
LXU -3.791043
LXU 4.818995
SWN 3.059964
SWN -4.349304
OAS -3.336539
LPI -3.037097
LPI -5.701044
LPI -3.519923
CZR -3.270018
CZR -3.056712
The required result is to retain only the highest absolute values and return the average values in a new column:
value average
CDE -3.097715 -0.389547
LXU 4.818995 0.513976
SWN -4.349304 -0.644670
OAS -3.336539 -3.336539
LPI -5.701044 -4.086021
CZR -3.270018 -3.163365
I have tried to .apply(lambda) to the duplicate rows but got an "axis" error:
max_absolute = lambda x: max(x.min(), x.max(), key=abs)
df_duplicate_absmax = df.groupby(df.index).apply(max_absolute, axis=1)
ps: Adapting Abhi's solution to work with NaN:
df1 = df.groupby(df.index)['value'].agg([lambda x: max(x[~np.isnan(x)], key=abs), 'mean'])
python python-3.x pandas dataframe pandas-groupby
asked Dec 9 '18 at 1:19
artDecoartDeco
185214
I have a set of values with duplicate index in a DataFrame:
value
CDE 2.318620
CDE -3.097715
LXU -3.791043
LXU 4.818995
SWN 3.059964
SWN -4.349304
OAS -3.336539
LPI -3.037097
LPI -5.701044
LPI -3.519923
CZR -3.270018
CZR -3.056712
The required result is to retain only the highest absolute values and return the average values in a new column:
value average
CDE -3.097715 -0.389547
LXU 4.818995 0.513976
SWN -4.349304 -0.644670
OAS -3.336539 -3.336539
LPI -5.701044 -4.086021
CZR -3.270018 -3.163365
I have tried to .apply(lambda) to the duplicate rows but got an "axis" error:
max_absolute = lambda x: max(x.min(), x.max(), key=abs)
df_duplicate_absmax = df.groupby(df.index).apply(max_absolute, axis=1)
ps: Adapting Abhi's solution to work with NaN:
df1 = df.groupby(df.index)['value'].agg([lambda x: max(x[~np.isnan(x)], key=abs), 'mean'])
python python-3.x pandas dataframe pandas-groupby
python python-3.x pandas dataframe pandas-groupby
asked Dec 9 '18 at 1:19
artDecoartDeco
185214
asked Dec 9 '18 at 1:19
artDecoartDeco
185214
edited Dec 30 '18 at 5:18
artDeco
asked Dec 9 '18 at 1:19
artDecoartDeco
185214
asked Dec 9 '18 at 1:19
artDecoartDeco
185214
asked Dec 9 '18 at 1:19
artDecoartDeco
185214
185214
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
2
Use:
df1 = df.groupby(df.index)['value'].agg([lambda x: max(x,key=abs), 'mean'])
df1.columns = ['value', 'average']
print (df1)
value average
CDE -3.097715 -0.389547
CZR -3.270018 -3.163365
LPI -5.701044 -4.086021
LXU 4.818995 0.513976
OAS -3.336539 -3.336539
SWN -4.349304 -0.644670
answered Dec 9 '18 at 1:27
AbhiAbhi
2,480320
1
There are two reasons why this solution is inefficient: (1) using pure Python built-ins (max) with Pandas series / NumPy arrays, (2) usinggroupbytwice when a singlegroupbyis sufficient.
– jpp
Dec 9 '18 at 1:56
3
Good solution, Abhi with thanks :) I appreciate the use of max_absolute lambda function for simplicity!
– artDeco
Dec 9 '18 at 2:10
2
Nice!. Note that columns can be named directly in .agg like.agg([('value', lambda x: max(x,key=abs)), ('average', 'mean')])
– Andrew Lavers
Dec 9 '18 at 12:05
1
Excellent, Andrew Lavers for adding value to a good solution! Thanks :)
– artDeco
Dec 9 '18 at 12:58
add a comment |
1
Here's a solution using groupby + agg with two functions, one to calculate the maximum by absolute value, the other to calculate the mean:
def max_abs(x):
return x.iloc[x.abs().values.argmax()]
res = df.groupby(level=0).agg([max_abs, 'mean'])
.xs('value', axis=1, drop_level=True)
print(res)
max_abs mean
CDE -3.097715 -0.389547
CZR -3.270018 -3.163365
LPI -5.701044 -4.086021
LXU 4.818995 0.513976
OAS -3.336539 -3.336539
SWN -4.349304 -0.644670
answered Dec 9 '18 at 1:53
jppjpp
97.9k2159109
1
That's great, jpp - very Pythonic!
– artDeco
Dec 9 '18 at 2:11
1
I also like your use of .xs() and the MultiIndex address, jpp :)
– artDeco
Dec 10 '18 at 1:00
add a comment |
1
from io import StringIO
import pandas as pd
df = pd.read_fwf(StringIO("""
cod value
CDE 2.318620
CDE -3.097715
LXU -3.791043
LXU 4.818995
SWN 3.059964
SWN -4.349304
OAS -3.336539
LPI -3.037097
LPI -5.701044
LPI -3.519923
CZR -3.270018
CZR -3.056712
"""), header=1, Index=None)
# Create a new column with absoulte value
df['abs_value'] = df['value'].abs()
# Calulate the mean in new data farame, grouped by code using
# pandas groupped aggregation naming the column average
df_avg = df.groupby("cod").value.agg([('average', 'mean')])
# Choose the row within group with largest abs value
df_abs = df.sort_values("abs_value").groupby("cod").tail(1)[["cod", "value"]]
# Join the average and the max
df_abs.join(df_avg, on="cod")
Result:
cod value average
1 CDE -3.097715 -0.389547
10 CZR -3.270018 -3.163365
6 OAS -3.336539 -3.336539
5 SWN -4.349304 -0.644670
3 LXU 4.818995 0.513976
8 LPI -5.701044 -4.086021
answered Dec 9 '18 at 1:59
Andrew LaversAndrew Lavers
3,0511713
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
Use:
df1 = df.groupby(df.index)['value'].agg([lambda x: max(x,key=abs), 'mean'])
df1.columns = ['value', 'average']
print (df1)
value average
CDE -3.097715 -0.389547
CZR -3.270018 -3.163365
LPI -5.701044 -4.086021
LXU 4.818995 0.513976
OAS -3.336539 -3.336539
SWN -4.349304 -0.644670
answered Dec 9 '18 at 1:27
AbhiAbhi
2,480320
1
There are two reasons why this solution is inefficient: (1) using pure Python built-ins (max) with Pandas series / NumPy arrays, (2) usinggroupbytwice when a singlegroupbyis sufficient.
– jpp
Dec 9 '18 at 1:56
3
Good solution, Abhi with thanks :) I appreciate the use of max_absolute lambda function for simplicity!
– artDeco
Dec 9 '18 at 2:10
2
Nice!. Note that columns can be named directly in .agg like.agg([('value', lambda x: max(x,key=abs)), ('average', 'mean')])
– Andrew Lavers
Dec 9 '18 at 12:05
1
Excellent, Andrew Lavers for adding value to a good solution! Thanks :)
– artDeco
Dec 9 '18 at 12:58
add a comment |
2
Use:
df1 = df.groupby(df.index)['value'].agg([lambda x: max(x,key=abs), 'mean'])
df1.columns = ['value', 'average']
print (df1)
value average
CDE -3.097715 -0.389547
CZR -3.270018 -3.163365
LPI -5.701044 -4.086021
LXU 4.818995 0.513976
OAS -3.336539 -3.336539
SWN -4.349304 -0.644670
answered Dec 9 '18 at 1:27
AbhiAbhi
2,480320
1
There are two reasons why this solution is inefficient: (1) using pure Python built-ins (max) with Pandas series / NumPy arrays, (2) usinggroupbytwice when a singlegroupbyis sufficient.
– jpp
Dec 9 '18 at 1:56
3
Good solution, Abhi with thanks :) I appreciate the use of max_absolute lambda function for simplicity!
– artDeco
Dec 9 '18 at 2:10
2
Nice!. Note that columns can be named directly in .agg like.agg([('value', lambda x: max(x,key=abs)), ('average', 'mean')])
– Andrew Lavers
Dec 9 '18 at 12:05
1
Excellent, Andrew Lavers for adding value to a good solution! Thanks :)
– artDeco
Dec 9 '18 at 12:58
add a comment |
2
2
2
Use:
df1 = df.groupby(df.index)['value'].agg([lambda x: max(x,key=abs), 'mean'])
df1.columns = ['value', 'average']
print (df1)
value average
CDE -3.097715 -0.389547
CZR -3.270018 -3.163365
LPI -5.701044 -4.086021
LXU 4.818995 0.513976
OAS -3.336539 -3.336539
SWN -4.349304 -0.644670
answered Dec 9 '18 at 1:27
AbhiAbhi
2,480320
Use:
df1 = df.groupby(df.index)['value'].agg([lambda x: max(x,key=abs), 'mean'])
df1.columns = ['value', 'average']
print (df1)
value average
CDE -3.097715 -0.389547
CZR -3.270018 -3.163365
LPI -5.701044 -4.086021
LXU 4.818995 0.513976
OAS -3.336539 -3.336539
SWN -4.349304 -0.644670
answered Dec 9 '18 at 1:27
AbhiAbhi
2,480320
edited Dec 9 '18 at 2:14
answered Dec 9 '18 at 1:27
AbhiAbhi
2,480320
answered Dec 9 '18 at 1:27
AbhiAbhi
2,480320
answered Dec 9 '18 at 1:27
AbhiAbhi
2,480320
2,480320
1
There are two reasons why this solution is inefficient: (1) using pure Python built-ins (max) with Pandas series / NumPy arrays, (2) usinggroupbytwice when a singlegroupbyis sufficient.
– jpp
Dec 9 '18 at 1:56
3
Good solution, Abhi with thanks :) I appreciate the use of max_absolute lambda function for simplicity!
– artDeco
Dec 9 '18 at 2:10
2
Nice!. Note that columns can be named directly in .agg like.agg([('value', lambda x: max(x,key=abs)), ('average', 'mean')])
– Andrew Lavers
Dec 9 '18 at 12:05
1
Excellent, Andrew Lavers for adding value to a good solution! Thanks :)
– artDeco
Dec 9 '18 at 12:58
add a comment |
1
There are two reasons why this solution is inefficient: (1) using pure Python built-ins (max) with Pandas series / NumPy arrays, (2) usinggroupbytwice when a singlegroupbyis sufficient.
– jpp
Dec 9 '18 at 1:56
3
Good solution, Abhi with thanks :) I appreciate the use of max_absolute lambda function for simplicity!
– artDeco
Dec 9 '18 at 2:10
2
Nice!. Note that columns can be named directly in .agg like.agg([('value', lambda x: max(x,key=abs)), ('average', 'mean')])
– Andrew Lavers
Dec 9 '18 at 12:05
1
Excellent, Andrew Lavers for adding value to a good solution! Thanks :)
– artDeco
Dec 9 '18 at 12:58
1
1
There are two reasons why this solution is inefficient: (1) using pure Python built-ins (
max) with Pandas series / NumPy arrays, (2) using groupby twice when a single groupby is sufficient.– jpp
Dec 9 '18 at 1:56
There are two reasons why this solution is inefficient: (1) using pure Python built-ins (
max) with Pandas series / NumPy arrays, (2) using groupby twice when a single groupby is sufficient.– jpp
Dec 9 '18 at 1:56
3
3
Good solution, Abhi with thanks :) I appreciate the use of max_absolute lambda function for simplicity!
– artDeco
Dec 9 '18 at 2:10
Good solution, Abhi with thanks :) I appreciate the use of max_absolute lambda function for simplicity!
– artDeco
Dec 9 '18 at 2:10
2
2
Nice!. Note that columns can be named directly in .agg like
.agg([('value', lambda x: max(x,key=abs)), ('average', 'mean')])– Andrew Lavers
Dec 9 '18 at 12:05
Nice!. Note that columns can be named directly in .agg like
.agg([('value', lambda x: max(x,key=abs)), ('average', 'mean')])– Andrew Lavers
Dec 9 '18 at 12:05
1
1
Excellent, Andrew Lavers for adding value to a good solution! Thanks :)
– artDeco
Dec 9 '18 at 12:58
Excellent, Andrew Lavers for adding value to a good solution! Thanks :)
– artDeco
Dec 9 '18 at 12:58
add a comment |
1
Here's a solution using groupby + agg with two functions, one to calculate the maximum by absolute value, the other to calculate the mean:
def max_abs(x):
return x.iloc[x.abs().values.argmax()]
res = df.groupby(level=0).agg([max_abs, 'mean'])
.xs('value', axis=1, drop_level=True)
print(res)
max_abs mean
CDE -3.097715 -0.389547
CZR -3.270018 -3.163365
LPI -5.701044 -4.086021
LXU 4.818995 0.513976
OAS -3.336539 -3.336539
SWN -4.349304 -0.644670
answered Dec 9 '18 at 1:53
jppjpp
97.9k2159109
1
That's great, jpp - very Pythonic!
– artDeco
Dec 9 '18 at 2:11
1
I also like your use of .xs() and the MultiIndex address, jpp :)
– artDeco
Dec 10 '18 at 1:00
add a comment |
1
Here's a solution using groupby + agg with two functions, one to calculate the maximum by absolute value, the other to calculate the mean:
def max_abs(x):
return x.iloc[x.abs().values.argmax()]
res = df.groupby(level=0).agg([max_abs, 'mean'])
.xs('value', axis=1, drop_level=True)
print(res)
max_abs mean
CDE -3.097715 -0.389547
CZR -3.270018 -3.163365
LPI -5.701044 -4.086021
LXU 4.818995 0.513976
OAS -3.336539 -3.336539
SWN -4.349304 -0.644670
answered Dec 9 '18 at 1:53
jppjpp
97.9k2159109
1
That's great, jpp - very Pythonic!
– artDeco
Dec 9 '18 at 2:11
1
I also like your use of .xs() and the MultiIndex address, jpp :)
– artDeco
Dec 10 '18 at 1:00
add a comment |
1
1
1
Here's a solution using groupby + agg with two functions, one to calculate the maximum by absolute value, the other to calculate the mean:
def max_abs(x):
return x.iloc[x.abs().values.argmax()]
res = df.groupby(level=0).agg([max_abs, 'mean'])
.xs('value', axis=1, drop_level=True)
print(res)
max_abs mean
CDE -3.097715 -0.389547
CZR -3.270018 -3.163365
LPI -5.701044 -4.086021
LXU 4.818995 0.513976
OAS -3.336539 -3.336539
SWN -4.349304 -0.644670
answered Dec 9 '18 at 1:53
jppjpp
97.9k2159109
Here's a solution using groupby + agg with two functions, one to calculate the maximum by absolute value, the other to calculate the mean:
def max_abs(x):
return x.iloc[x.abs().values.argmax()]
res = df.groupby(level=0).agg([max_abs, 'mean'])
.xs('value', axis=1, drop_level=True)
print(res)
max_abs mean
CDE -3.097715 -0.389547
CZR -3.270018 -3.163365
LPI -5.701044 -4.086021
LXU 4.818995 0.513976
OAS -3.336539 -3.336539
SWN -4.349304 -0.644670
answered Dec 9 '18 at 1:53
jppjpp
97.9k2159109
answered Dec 9 '18 at 1:53
jppjpp
97.9k2159109
answered Dec 9 '18 at 1:53
jppjpp
97.9k2159109
answered Dec 9 '18 at 1:53
jppjpp
97.9k2159109
97.9k2159109
1
That's great, jpp - very Pythonic!
– artDeco
Dec 9 '18 at 2:11
1
I also like your use of .xs() and the MultiIndex address, jpp :)
– artDeco
Dec 10 '18 at 1:00
add a comment |
1
That's great, jpp - very Pythonic!
– artDeco
Dec 9 '18 at 2:11
1
I also like your use of .xs() and the MultiIndex address, jpp :)
– artDeco
Dec 10 '18 at 1:00
1
1
That's great, jpp - very Pythonic!
– artDeco
Dec 9 '18 at 2:11
That's great, jpp - very Pythonic!
– artDeco
Dec 9 '18 at 2:11
1
1
I also like your use of .xs() and the MultiIndex address, jpp :)
– artDeco
Dec 10 '18 at 1:00
I also like your use of .xs() and the MultiIndex address, jpp :)
– artDeco
Dec 10 '18 at 1:00
add a comment |
1
from io import StringIO
import pandas as pd
df = pd.read_fwf(StringIO("""
cod value
CDE 2.318620
CDE -3.097715
LXU -3.791043
LXU 4.818995
SWN 3.059964
SWN -4.349304
OAS -3.336539
LPI -3.037097
LPI -5.701044
LPI -3.519923
CZR -3.270018
CZR -3.056712
"""), header=1, Index=None)
# Create a new column with absoulte value
df['abs_value'] = df['value'].abs()
# Calulate the mean in new data farame, grouped by code using
# pandas groupped aggregation naming the column average
df_avg = df.groupby("cod").value.agg([('average', 'mean')])
# Choose the row within group with largest abs value
df_abs = df.sort_values("abs_value").groupby("cod").tail(1)[["cod", "value"]]
# Join the average and the max
df_abs.join(df_avg, on="cod")
Result:
cod value average
1 CDE -3.097715 -0.389547
10 CZR -3.270018 -3.163365
6 OAS -3.336539 -3.336539
5 SWN -4.349304 -0.644670
3 LXU 4.818995 0.513976
8 LPI -5.701044 -4.086021
answered Dec 9 '18 at 1:59
Andrew LaversAndrew Lavers
3,0511713
add a comment |
1
from io import StringIO
import pandas as pd
df = pd.read_fwf(StringIO("""
cod value
CDE 2.318620
CDE -3.097715
LXU -3.791043
LXU 4.818995
SWN 3.059964
SWN -4.349304
OAS -3.336539
LPI -3.037097
LPI -5.701044
LPI -3.519923
CZR -3.270018
CZR -3.056712
"""), header=1, Index=None)
# Create a new column with absoulte value
df['abs_value'] = df['value'].abs()
# Calulate the mean in new data farame, grouped by code using
# pandas groupped aggregation naming the column average
df_avg = df.groupby("cod").value.agg([('average', 'mean')])
# Choose the row within group with largest abs value
df_abs = df.sort_values("abs_value").groupby("cod").tail(1)[["cod", "value"]]
# Join the average and the max
df_abs.join(df_avg, on="cod")
Result:
cod value average
1 CDE -3.097715 -0.389547
10 CZR -3.270018 -3.163365
6 OAS -3.336539 -3.336539
5 SWN -4.349304 -0.644670
3 LXU 4.818995 0.513976
8 LPI -5.701044 -4.086021
answered Dec 9 '18 at 1:59
Andrew LaversAndrew Lavers
3,0511713
add a comment |
1
1
1
from io import StringIO
import pandas as pd
df = pd.read_fwf(StringIO("""
cod value
CDE 2.318620
CDE -3.097715
LXU -3.791043
LXU 4.818995
SWN 3.059964
SWN -4.349304
OAS -3.336539
LPI -3.037097
LPI -5.701044
LPI -3.519923
CZR -3.270018
CZR -3.056712
"""), header=1, Index=None)
# Create a new column with absoulte value
df['abs_value'] = df['value'].abs()
# Calulate the mean in new data farame, grouped by code using
# pandas groupped aggregation naming the column average
df_avg = df.groupby("cod").value.agg([('average', 'mean')])
# Choose the row within group with largest abs value
df_abs = df.sort_values("abs_value").groupby("cod").tail(1)[["cod", "value"]]
# Join the average and the max
df_abs.join(df_avg, on="cod")
Result:
cod value average
1 CDE -3.097715 -0.389547
10 CZR -3.270018 -3.163365
6 OAS -3.336539 -3.336539
5 SWN -4.349304 -0.644670
3 LXU 4.818995 0.513976
8 LPI -5.701044 -4.086021
answered Dec 9 '18 at 1:59
Andrew LaversAndrew Lavers
3,0511713
from io import StringIO
import pandas as pd
df = pd.read_fwf(StringIO("""
cod value
CDE 2.318620
CDE -3.097715
LXU -3.791043
LXU 4.818995
SWN 3.059964
SWN -4.349304
OAS -3.336539
LPI -3.037097
LPI -5.701044
LPI -3.519923
CZR -3.270018
CZR -3.056712
"""), header=1, Index=None)
# Create a new column with absoulte value
df['abs_value'] = df['value'].abs()
# Calulate the mean in new data farame, grouped by code using
# pandas groupped aggregation naming the column average
df_avg = df.groupby("cod").value.agg([('average', 'mean')])
# Choose the row within group with largest abs value
df_abs = df.sort_values("abs_value").groupby("cod").tail(1)[["cod", "value"]]
# Join the average and the max
df_abs.join(df_avg, on="cod")
Result:
cod value average
1 CDE -3.097715 -0.389547
10 CZR -3.270018 -3.163365
6 OAS -3.336539 -3.336539
5 SWN -4.349304 -0.644670
3 LXU 4.818995 0.513976
8 LPI -5.701044 -4.086021
answered Dec 9 '18 at 1:59
Andrew LaversAndrew Lavers
3,0511713
answered Dec 9 '18 at 1:59
Andrew LaversAndrew Lavers
3,0511713
answered Dec 9 '18 at 1:59
Andrew LaversAndrew Lavers
3,0511713
answered Dec 9 '18 at 1:59
Andrew LaversAndrew Lavers
3,0511713
3,0511713
add a comment |
add a comment |
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