Pandas Return Separate DataFrame Values Based on Function

You can use pd.cut function to specify proper intervals in which latitudes are contained and simply merge two dataframes to obtain the result:

bins = [(i-1,i+1) for i in df1['Lat']]
bins = [item for subbins in bins for item in subbins]

df1['Interval'] = pd.cut(df1['Lat'], bins=bins)
df2['Interval'] = pd.cut(df2['Station_Lat'], bins=bins)

pd.merge(df1,df2)

This solution is slightly faster than yours. 10.2 ms ± 201 µs per loop vs 12.2 ms ± 1.34 ms per loop.

Maybe it is faster:

df2= df2.sort_values("Station_Lat")

After sorting, you can use 'searchsorted":

df1["idx"]=df2.Station_Lat.searchsorted(df1.Lat)

"idx" is the 'nearest' station lat. index, or idx+1 is this. Maybe you need duplicate the last row in df2 (see the "searchsorted doc) to avoid over indexing it. The use "apply" with this custom function:

def dist(row): 
    if  abs(row.Lat-df2.loc[row.idx].Station_Lat)<=1: 
            return df2.loc[row.idx].Station 
    elif abs(row.Lat-df2.loc[row.idx+1].Station_Lat)<=1: 
            return df2.loc[row.idx+1].Station 

    return False 

df1.apply(dist,axis=1)                                                                                               

0      ABC
1    False
2    False
3      JKL
dtype: object

Edit: Because in 'dist()' it is assumed that df2.index is ordered and monotonic increasing (see: roww.idx+1), the 1st code line must be corrected:

df2= df2.sort_values("Station_Lat").reset_index(drop=True)

And 'dist()' is somewhat faster that way (but doesn't beat the Cartesian product method):

def dist(row):  
          idx=row.idx 
          lat1,lat2= df2.loc[idx:idx+1,"Station_Lat"] 
          if  abs(row.Lat-lat1)<=1:  
                 return df2.loc[idx,"Station"] 
          elif abs(row.Lat-lat2)<=1:  
                 return df2.loc[idx+1,"Station"] 
          return False 

How about a lambda?

df3[df3.apply(lambda x, col1='Lat', col2='Station_Lat': x[col1]-x[col2] >= -1 and x[col1]-x[col2] <= 1, axis=1)]['Station']

Output:

0     ABC
15    JKL

Edit: Here's a second solution. (Note: This also uses abs() since >=-1 and <= 1 seems redundant.)

for i in df1.index:
    for j in df2.index:
        if abs(df1.loc[i, 'Lat'] - df2.loc[j, 'Station_Lat']) <=1:
            print(df2.loc[j, 'Station'])

Or, in list comprehension form:

df2.loc[[i for i in df1.index for j in df2.index if abs(df1.loc[i, 'Lat'] - df2.loc[j, 'Station_Lat']) <=1], 'Station']

Output:

ABC
JKL