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Pandas Groupby: Groupby conditional statement

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I am trying to identify the location of stops from gps data but need to account for some gps drift.

I have identified stops and isolated them into a new dataframe:

df['Stopped'] = (df.groupby('DAY')['LAT'].diff().abs() <= 0.0005) & (df.groupby('DAY')['LNG'].diff().abs() <= 0.0005)

df2 = df.loc[(df['Stopped'] == True)]

Now I can label groups that have the exact match in coordinates using:

df2['StoppedEvent'] = df2.groupby(['LAT','LNG']).ngroup() 

But I want to group by the same conditions of Stopped. Something like this but that works:

df2['StoppedEvent'] = df2.groupby((['LAT','LNG']).diff().fillna(0).abs() <= 0.0005).ngroup() 

pandas group-by pandas-groupby latitude-longitude

share|improve this question

asked Mar 22 at 18:39

acrowacrow

62

add a comment | 

0

I am trying to identify the location of stops from gps data but need to account for some gps drift.

I have identified stops and isolated them into a new dataframe:

df['Stopped'] = (df.groupby('DAY')['LAT'].diff().abs() <= 0.0005) & (df.groupby('DAY')['LNG'].diff().abs() <= 0.0005)

df2 = df.loc[(df['Stopped'] == True)]

Now I can label groups that have the exact match in coordinates using:

df2['StoppedEvent'] = df2.groupby(['LAT','LNG']).ngroup() 

But I want to group by the same conditions of Stopped. Something like this but that works:

df2['StoppedEvent'] = df2.groupby((['LAT','LNG']).diff().fillna(0).abs() <= 0.0005).ngroup() 

pandas group-by pandas-groupby latitude-longitude

share|improve this question

asked Mar 22 at 18:39

acrowacrow

62

add a comment | 

I am trying to identify the location of stops from gps data but need to account for some gps drift.

I have identified stops and isolated them into a new dataframe:

df['Stopped'] = (df.groupby('DAY')['LAT'].diff().abs() <= 0.0005) & (df.groupby('DAY')['LNG'].diff().abs() <= 0.0005)

df2 = df.loc[(df['Stopped'] == True)]

Now I can label groups that have the exact match in coordinates using:

df2['StoppedEvent'] = df2.groupby(['LAT','LNG']).ngroup() 

But I want to group by the same conditions of Stopped. Something like this but that works:

df2['StoppedEvent'] = df2.groupby((['LAT','LNG']).diff().fillna(0).abs() <= 0.0005).ngroup() 

pandas group-by pandas-groupby latitude-longitude

share|improve this question

asked Mar 22 at 18:39

acrowacrow

62

df['Stopped'] = (df.groupby('DAY')['LAT'].diff().abs() <= 0.0005) & (df.groupby('DAY')['LNG'].diff().abs() <= 0.0005)

df2 = df.loc[(df['Stopped'] == True)]

df2['StoppedEvent'] = df2.groupby(['LAT','LNG']).ngroup() 

df2['StoppedEvent'] = df2.groupby((['LAT','LNG']).diff().fillna(0).abs() <= 0.0005).ngroup() 

share|improve this question

asked Mar 22 at 18:39

acrowacrow

62

share|improve this question

asked Mar 22 at 18:39

acrowacrow

62

asked Mar 22 at 18:39

acrowacrow

62

asked Mar 22 at 18:39

acrowacrow

62

1 Answer
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I would do something like the following:

df['Stopped'] = (df.groupby('DAY')['LAT'].diff().abs() <= 0.0005)
 & (df.groupby('DAY')['LNG'].diff().abs() <= 0.0005)
df["Stopped_Group"] = (~df["Stopped"]).cumsum()
df2 = df.loc[df['Stopped']]

Now you'll have a column, "Stopped_Group", which is constant within a set of rows that are close to each other as determined by your logic. In the original dataframe, df, this column won't have any meaning for rows that correspond to motion.

To get your desired output (if I understand you correctly), do something like the following:

df2["Stopped_Duration"] = df2.groupby("Stopped_Group").transform("size")

share|improve this answer

answered Mar 22 at 18:48

PMendePMende

1,9101613

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0

I would do something like the following:

df['Stopped'] = (df.groupby('DAY')['LAT'].diff().abs() <= 0.0005)
 & (df.groupby('DAY')['LNG'].diff().abs() <= 0.0005)
df["Stopped_Group"] = (~df["Stopped"]).cumsum()
df2 = df.loc[df['Stopped']]

Now you'll have a column, "Stopped_Group", which is constant within a set of rows that are close to each other as determined by your logic. In the original dataframe, df, this column won't have any meaning for rows that correspond to motion.

To get your desired output (if I understand you correctly), do something like the following:

df2["Stopped_Duration"] = df2.groupby("Stopped_Group").transform("size")

share|improve this answer

answered Mar 22 at 18:48

PMendePMende

1,9101613

add a comment | 

0

I would do something like the following:

df['Stopped'] = (df.groupby('DAY')['LAT'].diff().abs() <= 0.0005)
 & (df.groupby('DAY')['LNG'].diff().abs() <= 0.0005)
df["Stopped_Group"] = (~df["Stopped"]).cumsum()
df2 = df.loc[df['Stopped']]

Now you'll have a column, "Stopped_Group", which is constant within a set of rows that are close to each other as determined by your logic. In the original dataframe, df, this column won't have any meaning for rows that correspond to motion.

To get your desired output (if I understand you correctly), do something like the following:

df2["Stopped_Duration"] = df2.groupby("Stopped_Group").transform("size")

share|improve this answer

answered Mar 22 at 18:48

PMendePMende

1,9101613

add a comment | 

I would do something like the following:

df['Stopped'] = (df.groupby('DAY')['LAT'].diff().abs() <= 0.0005)
 & (df.groupby('DAY')['LNG'].diff().abs() <= 0.0005)
df["Stopped_Group"] = (~df["Stopped"]).cumsum()
df2 = df.loc[df['Stopped']]

Now you'll have a column, "Stopped_Group", which is constant within a set of rows that are close to each other as determined by your logic. In the original dataframe, df, this column won't have any meaning for rows that correspond to motion.

To get your desired output (if I understand you correctly), do something like the following:

df2["Stopped_Duration"] = df2.groupby("Stopped_Group").transform("size")

share|improve this answer

answered Mar 22 at 18:48

PMendePMende

1,9101613

df['Stopped'] = (df.groupby('DAY')['LAT'].diff().abs() <= 0.0005)
 & (df.groupby('DAY')['LNG'].diff().abs() <= 0.0005)
df["Stopped_Group"] = (~df["Stopped"]).cumsum()
df2 = df.loc[df['Stopped']]

df2["Stopped_Duration"] = df2.groupby("Stopped_Group").transform("size")

share|improve this answer

answered Mar 22 at 18:48

PMendePMende

1,9101613

answered Mar 22 at 18:48

PMendePMende

1,9101613

answered Mar 22 at 18:48

PMendePMende

1,9101613