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Pandas subset by index closest to a particular value

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1

I was wondering how to subset a DataFrame for the index closest to a particular value. For example:

import pandas as pd
import numpy as np

np.random.seed(24)
df = pd.DataFrame('A': np.linspace(1, 10, 10))
df = pd.concat([df, pd.DataFrame(np.random.randn(10, 4), columns=list('BCDE'))],
 axis=1)

df.index = np.random.randn(len(df.index))
closest_to = 0
df.loc[df.index.difference([closest_to]).min()]

This subsets for the index furthest below 0, but I am looking for the absolute difference closest to zero.

EDIT:
Adding df.loc[abs(df.index.difference([closest_to])).min()] works when the closest value is positive, but obviously gives a KeyError for when a negative value is the closest.

python pandas

share|improve this question

edited Mar 22 at 13:00

Cr1064

asked Mar 22 at 12:41

Cr1064Cr1064

897

add a comment | 

1

I was wondering how to subset a DataFrame for the index closest to a particular value. For example:

import pandas as pd
import numpy as np

np.random.seed(24)
df = pd.DataFrame('A': np.linspace(1, 10, 10))
df = pd.concat([df, pd.DataFrame(np.random.randn(10, 4), columns=list('BCDE'))],
 axis=1)

df.index = np.random.randn(len(df.index))
closest_to = 0
df.loc[df.index.difference([closest_to]).min()]

This subsets for the index furthest below 0, but I am looking for the absolute difference closest to zero.

EDIT:
Adding df.loc[abs(df.index.difference([closest_to])).min()] works when the closest value is positive, but obviously gives a KeyError for when a negative value is the closest.

python pandas

share|improve this question

edited Mar 22 at 13:00

Cr1064

asked Mar 22 at 12:41

Cr1064Cr1064

897

add a comment | 

I was wondering how to subset a DataFrame for the index closest to a particular value. For example:

import pandas as pd
import numpy as np

np.random.seed(24)
df = pd.DataFrame('A': np.linspace(1, 10, 10))
df = pd.concat([df, pd.DataFrame(np.random.randn(10, 4), columns=list('BCDE'))],
 axis=1)

df.index = np.random.randn(len(df.index))
closest_to = 0
df.loc[df.index.difference([closest_to]).min()]

This subsets for the index furthest below 0, but I am looking for the absolute difference closest to zero.

EDIT:
Adding df.loc[abs(df.index.difference([closest_to])).min()] works when the closest value is positive, but obviously gives a KeyError for when a negative value is the closest.

python pandas

share|improve this question

edited Mar 22 at 13:00

Cr1064

asked Mar 22 at 12:41

Cr1064Cr1064

897

import pandas as pd
import numpy as np

np.random.seed(24)
df = pd.DataFrame('A': np.linspace(1, 10, 10))
df = pd.concat([df, pd.DataFrame(np.random.randn(10, 4), columns=list('BCDE'))],
 axis=1)

df.index = np.random.randn(len(df.index))
closest_to = 0
df.loc[df.index.difference([closest_to]).min()]

share|improve this question

edited Mar 22 at 13:00

Cr1064

asked Mar 22 at 12:41

Cr1064Cr1064

897

share|improve this question

edited Mar 22 at 13:00

Cr1064

asked Mar 22 at 12:41

Cr1064Cr1064

897

asked Mar 22 at 12:41

Cr1064Cr1064

897

asked Mar 22 at 12:41

Cr1064Cr1064

897

1 Answer
1

active

oldest

votes

0

I found a solution which isn't very neat but does the job.

closest_to = 0
idx = abs(df.index.difference([closest_to])).min()
if idx in df.index:
 pass
else:
 idx = -idx
df.loc[idx]

share|improve this answer

answered Mar 22 at 13:14

Cr1064Cr1064

897

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You got to the same solution than I, but a different implementation. ;-)
  
  – Jurgen Strydom
  Mar 22 at 13:15
  

  
  
  
  

add a comment | 

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0

I found a solution which isn't very neat but does the job.

closest_to = 0
idx = abs(df.index.difference([closest_to])).min()
if idx in df.index:
 pass
else:
 idx = -idx
df.loc[idx]

share|improve this answer

answered Mar 22 at 13:14

Cr1064Cr1064

897

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You got to the same solution than I, but a different implementation. ;-)
  
  – Jurgen Strydom
  Mar 22 at 13:15
  

  
  
  
  

add a comment | 

0

I found a solution which isn't very neat but does the job.

closest_to = 0
idx = abs(df.index.difference([closest_to])).min()
if idx in df.index:
 pass
else:
 idx = -idx
df.loc[idx]

share|improve this answer

answered Mar 22 at 13:14

Cr1064Cr1064

897

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You got to the same solution than I, but a different implementation. ;-)
  
  – Jurgen Strydom
  Mar 22 at 13:15
  

  
  
  
  

add a comment | 

I found a solution which isn't very neat but does the job.

closest_to = 0
idx = abs(df.index.difference([closest_to])).min()
if idx in df.index:
 pass
else:
 idx = -idx
df.loc[idx]

share|improve this answer

answered Mar 22 at 13:14

Cr1064Cr1064

897

closest_to = 0
idx = abs(df.index.difference([closest_to])).min()
if idx in df.index:
 pass
else:
 idx = -idx
df.loc[idx]

share|improve this answer

answered Mar 22 at 13:14

Cr1064Cr1064

897

answered Mar 22 at 13:14

Cr1064Cr1064

897

answered Mar 22 at 13:14

Cr1064Cr1064

897