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get value from pandas segment and subtract in place

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1

i have a table with values similar to

val1 val2 val3 segVal
 0 12.3 88.2
 20 0 0
 50 14.5 88.7
 70 0 0
 85 0 0
 90 18.2 88.9

for my segVal, i need to use the differences from my val1 columns where val2 is known. so my first segment would be zero to 50, i'm subtracting from 0 and applying that to all segVal rows. my next segment is at 90 so i would subtract that from 50 and apply that.

So my output table would be

val1 val2 val3 segVal
 0 12.3 88.2 50
 20 0 0 50
 50 14.5 88.7 50
 70 0 0 40
 85 0 0 40
 90 18.2 88.9 40

my current somewhat working method is

df1 = df[df.val2 != 0]
df1 = df1.copy()
df1.segVal=(df1['val1'].diff(-1))*1

so i'm creating a additional df and calculating the values this way, then merging back the values with the original df.

It seems there has got to be a better way to do this, I mean, my method works, but doesn't' seem too efficient creating additional df's

python pandas

share|improve this question

edited Mar 8 at 16:34

Messak

asked Mar 8 at 16:25

MessakMessak

1098

add a comment | 

1

i have a table with values similar to

val1 val2 val3 segVal
 0 12.3 88.2
 20 0 0
 50 14.5 88.7
 70 0 0
 85 0 0
 90 18.2 88.9

for my segVal, i need to use the differences from my val1 columns where val2 is known. so my first segment would be zero to 50, i'm subtracting from 0 and applying that to all segVal rows. my next segment is at 90 so i would subtract that from 50 and apply that.

So my output table would be

val1 val2 val3 segVal
 0 12.3 88.2 50
 20 0 0 50
 50 14.5 88.7 50
 70 0 0 40
 85 0 0 40
 90 18.2 88.9 40

my current somewhat working method is

df1 = df[df.val2 != 0]
df1 = df1.copy()
df1.segVal=(df1['val1'].diff(-1))*1

so i'm creating a additional df and calculating the values this way, then merging back the values with the original df.

It seems there has got to be a better way to do this, I mean, my method works, but doesn't' seem too efficient creating additional df's

python pandas

share|improve this question

edited Mar 8 at 16:34

Messak

asked Mar 8 at 16:25

MessakMessak

1098

add a comment | 

i have a table with values similar to

val1 val2 val3 segVal
 0 12.3 88.2
 20 0 0
 50 14.5 88.7
 70 0 0
 85 0 0
 90 18.2 88.9

for my segVal, i need to use the differences from my val1 columns where val2 is known. so my first segment would be zero to 50, i'm subtracting from 0 and applying that to all segVal rows. my next segment is at 90 so i would subtract that from 50 and apply that.

So my output table would be

val1 val2 val3 segVal
 0 12.3 88.2 50
 20 0 0 50
 50 14.5 88.7 50
 70 0 0 40
 85 0 0 40
 90 18.2 88.9 40

my current somewhat working method is

df1 = df[df.val2 != 0]
df1 = df1.copy()
df1.segVal=(df1['val1'].diff(-1))*1

so i'm creating a additional df and calculating the values this way, then merging back the values with the original df.

It seems there has got to be a better way to do this, I mean, my method works, but doesn't' seem too efficient creating additional df's

python pandas

share|improve this question

edited Mar 8 at 16:34

Messak

asked Mar 8 at 16:25

MessakMessak

1098

val1 val2 val3 segVal
 0 12.3 88.2
 20 0 0
 50 14.5 88.7
 70 0 0
 85 0 0
 90 18.2 88.9

val1 val2 val3 segVal
 0 12.3 88.2 50
 20 0 0 50
 50 14.5 88.7 50
 70 0 0 40
 85 0 0 40
 90 18.2 88.9 40

df1 = df[df.val2 != 0]
df1 = df1.copy()
df1.segVal=(df1['val1'].diff(-1))*1

share|improve this question

edited Mar 8 at 16:34

Messak

asked Mar 8 at 16:25

MessakMessak

1098

share|improve this question

edited Mar 8 at 16:34

Messak

asked Mar 8 at 16:25

MessakMessak

1098

asked Mar 8 at 16:25

MessakMessak

1098

asked Mar 8 at 16:25

MessakMessak

1098

1 Answer
1

active

oldest

votes

1

Here's one way:

df['segVal'] = df.where(df.val2.ne(0)).val1.dropna().diff().reindex(df.index).bfill()

 val1 val2 val3 segVal
0 0 12.3 88.2 50.0 
1 20 0.0 0.0 50.0 
2 50 14.5 88.7 50.0 
3 70 0.0 0.0 40.0 
4 85 0.0 0.0 40.0 
5 90 18.2 88.9 40.0 

share|improve this answer

edited Mar 8 at 16:36

answered Mar 8 at 16:31

ALollzALollz

16.8k41838

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you very much, i knew there was a cleaner and better way.
  
  – Messak
  Mar 8 at 16:38
  
  

  
  
  
  

add a comment | 

Your Answer

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1

Here's one way:

df['segVal'] = df.where(df.val2.ne(0)).val1.dropna().diff().reindex(df.index).bfill()

 val1 val2 val3 segVal
0 0 12.3 88.2 50.0 
1 20 0.0 0.0 50.0 
2 50 14.5 88.7 50.0 
3 70 0.0 0.0 40.0 
4 85 0.0 0.0 40.0 
5 90 18.2 88.9 40.0 

share|improve this answer

edited Mar 8 at 16:36

answered Mar 8 at 16:31

ALollzALollz

16.8k41838

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you very much, i knew there was a cleaner and better way.
  
  – Messak
  Mar 8 at 16:38
  
  

  
  
  
  

add a comment | 

1

Here's one way:

df['segVal'] = df.where(df.val2.ne(0)).val1.dropna().diff().reindex(df.index).bfill()

 val1 val2 val3 segVal
0 0 12.3 88.2 50.0 
1 20 0.0 0.0 50.0 
2 50 14.5 88.7 50.0 
3 70 0.0 0.0 40.0 
4 85 0.0 0.0 40.0 
5 90 18.2 88.9 40.0 

share|improve this answer

edited Mar 8 at 16:36

answered Mar 8 at 16:31

ALollzALollz

16.8k41838

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you very much, i knew there was a cleaner and better way.
  
  – Messak
  Mar 8 at 16:38
  
  

  
  
  
  

add a comment | 

Here's one way:

df['segVal'] = df.where(df.val2.ne(0)).val1.dropna().diff().reindex(df.index).bfill()

 val1 val2 val3 segVal
0 0 12.3 88.2 50.0 
1 20 0.0 0.0 50.0 
2 50 14.5 88.7 50.0 
3 70 0.0 0.0 40.0 
4 85 0.0 0.0 40.0 
5 90 18.2 88.9 40.0 

share|improve this answer

edited Mar 8 at 16:36

answered Mar 8 at 16:31

ALollzALollz

16.8k41838

df['segVal'] = df.where(df.val2.ne(0)).val1.dropna().diff().reindex(df.index).bfill()

 val1 val2 val3 segVal
0 0 12.3 88.2 50.0 
1 20 0.0 0.0 50.0 
2 50 14.5 88.7 50.0 
3 70 0.0 0.0 40.0 
4 85 0.0 0.0 40.0 
5 90 18.2 88.9 40.0 

share|improve this answer

edited Mar 8 at 16:36

answered Mar 8 at 16:31

ALollzALollz

16.8k41838

answered Mar 8 at 16:31

ALollzALollz

16.8k41838

answered Mar 8 at 16:31

ALollzALollz

16.8k41838