pandas 基于函数返回单独的DataFrame值 [英] Pandas Return Separate DataFrame Values Based on Function

问题描述

我有两个DataFrame，df1是场所的位置，df2是站的位置.我正在尝试找到一种更有效的方法来应用距离函数，以查找在特定范围内的站点并返回站点的名称.如果距离函数是+/- 1的纬度差，则这是我的预期结果:

I have two DataFrames, df1 are locations of places and df2 are locations of stations. I am trying to find a more efficient way to apply a distance function to find which stations are within a certain range and return the station's name. If the distance function is a Latitude Difference of +/- 1 this is my expected outcome:

# df1
   Lat  Long 
0   30    31    
1   37    48    
2   54    62    
3   67    63     

# df2
   Station_Lat  Station_Long Station
0           30            32     ABC    
1           43            48     DEF    
2           84            87     GHI    
3           67            62     JKL    

# ....Some Code that compares df1 and df2....

# result
   Lat  Long  Station_Lat  Station_Long Station
    30    31           30            32     ABC
    67    63           67            62     JKL

我有一个使用笛卡尔积/Cross Join在其上应用功能的解决方案一个DataFrame.此解决方案有效，但是我在一个真实的数据集中有数百万行，这使笛卡尔积非常慢.

I have a solution that uses a cartesian product/Cross Join to apply a function on a single DataFrame. This solution works, but I have millions of rows in a true dataset which makes a cartesian product very slow.

import pandas as pd
df1 = pd.DataFrame({'Lat' : [30, 37, 54, 67],
                    'Long' : [31, 48, 62, 63]})

df2 = pd.DataFrame({'Station_Lat' : [30, 43, 84, 67],
                    'Station_Long' : [32, 48, 87, 62],
                    'Station':['ABC', 'DEF','GHI','JKL']})

# creating a 'key' for a cartesian product
df1['key'] = 1
df2['key'] = 1

# Creating the cartesian Join
df3 = pd.merge(df1, df2, on='key')

# some distance function that returns True or False
# assuming the distance function I want is +/- 1 of two values
def some_distance_func(x,y):
    return x-y >= -1 and x-y <= 1

# applying the function to a column using vectorized approach
# https://stackoverflow.com/questions/52673285/performance-of-pandas-apply-vs-np-vectorize-to-create-new-column-from-existing-c
df3['t_or_f'] =  list(map(some_distance_func,df3['Lat'],df3['Station_Lat']))

# result
print(df3.loc[df3['t_or_f']][['Lat','Long','Station_Lat','Station_Long','Station']].reset_index(drop=True))

我还尝试了使用 iterrows() ，但这比交叉联接方法要慢.是否有更pythonic/更有效的方法来实现我想要的?

I have also tried a looping approach with iterrows(), but that is slower than the cross join method. Is there a more pythonic/efficient way to achieve what I am looking for?

推荐答案

您可以使用

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)

此解决方案比您的解决方案要快一些. 10.2 ms ± 201 µs per loop与12.2 ms ± 1.34 ms per loop.

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

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