加快在大 pandas dataframe 中搜索最近的上限值和下限值

Question

My dataframe looks similar to this example below (just with way more entries). I want to obtain the nearest upper and lower number for a given value, for each group.

a    b  
600  10
600  12
600  15
600  17
700   8
700  11
700  19

For example for a value of 13. I would like to obtain a new dataframe similar to:

a    b  
600  12
600  15
700  11
700  19

I already tried the solution from Ivo Merchiers in How do I find the closest values in a Pandas series to an input number? using groupby and apply to run it for the different groups.

def find_neighbours(value):
  exactmatch=df[df.num==value]
  if !exactmatch.empty:
      return exactmatch.index
  else:
      lowerneighbour_ind = df[df.num<value].num.idxmax()
      upperneighbour_ind = df[df.num>value].num.idxmin()
      return [lowerneighbour_ind, upperneighbour_ind]

df=df.groupby('a').apply(find_neighbours, 13)

But since my dataset has around 16 million lines this procedure takes extremely long. Is there possibly a faster way to obtain a solution?

Edit Thanks for your answers. I forgot to add some info. If a close number appears multiple times I would like to have all lines transfered to the new dataframe. And when there is only one upper (lower) and no lower (upper) neighbour, this lines should be ignored.

a    b  
600  10
600  12
600  15
600  17
700   8
700  11
700  19
800  14
800  15
900  12
900  14
900  14

Leads for 13 to this:

a    b  
600  12
600  15
700  11
700  19
900  12
900  14
900  14

Thanks for your help!

Answer 1

Yes we can speed it up

v=13

s=(df.b-v)
t=s.abs().groupby([df.a,np.sign(s)]).transform('min')
df1=df.loc[s.abs()==t]
df1=df1[df1.b.sub(v).groupby(df.a).transform('nunique')>1]
df1
Out[102]: 
      a   b
1   600  12
2   600  15
5   700  11
6   700  19
9   900  12
10  900  14
11  900  14

Answer 2

try this

def neighbours(x):
    d = (df.b-x)
    return df.loc[[d[d==d[d>0].min()].index[0], d[d==d[d<0].max()].index[0]]]
neighbours(13)