CODEWITHSUNDEEP
pythonpandas
doomdaam
doomdaam

Reputation: 783

For loop through multiple rows by the index using pandas

Using the following data in train_data_sample and the below code, how can I iterate through each index latitude and longitude? (see below for wished results)

   latitude longitude price
0   55.6632 12.6288 2595000
1   55.6637 12.6291 2850000
2   55.6637 12.6291 2850000
3   55.6632 12.6290 3198000
4   55.6632 12.6290 2995000
5   55.6638 12.6294 2395000
6   55.6637 12.6291 2995000
7   55.6642 12.6285 4495000
8   55.6632 12.6285 3998000
9   55.6638 12.6294 3975000

from numpy import cos, sin, arcsin, sqrt
from math import radians

def haversine(row):
   
    for index in train_data_sample.index:
        lon1 = train_data_sample["longitude"].loc[train_data_sample.index==index]
        lat1 = train_data_sample["latitude"].loc[train_data_sample.index==index]
        lon2 = row['longitude']
        lat2 = row['latitude']
        lon1, lat1, lon2, lat2 = map(radians, [lon1, lat1, lon2, lat2])
        dlon = lon2 - lon1 
        dlat = lat2 - lat1 
        a = sin(dlat/2)**2 + cos(lat1) * cos(lat2) * sin(dlon/2)**2
        c = 2 * arcsin(sqrt(a)) 
        km = 6367 * c
    return km

def insert_dist(df):
    df["distance"+str(index)] = df.apply(lambda row: haversine(row), axis=1)
    return df

print(insert_dist(train_data_sample))

This is the result for index 0. It looks at the coordinates for index 0 versus every other row and returns the distance in meters. So the distance between coordinates for index 0 and 1 are ~50 meters.

latitude    longitude   price   distance0
0   55.6632 12.6288 2595000    0.000000
1   55.6637 12.6291 2850000    0.058658
2   55.6637 12.6291 2850000    0.058658
3   55.6632 12.6290 3198000    0.012536
4   55.6632 12.6290 2995000    0.012536
5   55.6638 12.6294 2395000    0.076550
6   55.6637 12.6291 2995000    0.058658
7   55.6642 12.6285 4495000    0.112705
8   55.6632 12.6285 3998000    0.018804
9   55.6638 12.6294 3975000    0.076550

The end result should return not only distance0, but also distance1, distance2, etc.

Upvotes: 0

Views: 855

Answers (2)

Andre
Andre

Reputation: 788

It seems like your making things a bit more complicating than necessary. By nesting a for loop in another for loop you can achieve what you want in a more straightforward way.

from numpy import cos, sin, arcsin, sqrt
from math import radians
import pandas as pd
import numpy as np


# recreate your dataframe
data = [[55.6632, 12.6288, 2595000],
        [55.6637, 12.6291, 2850000],
        [55.6637, 12.6291, 2850000], 
        [55.6632, 12.6290, 3198000]]

data = np.array(data)

train_data_sample = pd.DataFrame(data, columns = ["latitude", "longitude", "price"])


# copied  "distance calculating" code here
def GetDistance(lon1, lat1, lon2, lat2):
    lon1, lat1, lon2, lat2 = map(radians, [lon1, lat1, lon2, lat2])
    dlon = lon2 - lon1 
    dlat = lat2 - lat1 
    a = sin(dlat/2)**2 + cos(lat1) * cos(lat2) * sin(dlon/2)**2
    c = 2 * arcsin(sqrt(a)) 
    km = 6367 * c
    return km

# loop over every row with iterrows
for index, row in train_data_sample.iterrows():
    
    distances = []
    
    lat1, lon1 = row[["longitude", "longitude"]]
    
    # loop again over every row with iterrows
    for index_2, row_2 in train_data_sample.iterrows():
        lat2, lon2 = row_2[["longitude", "longitude"]]
        # get the distance
        distances.append( GetDistance(lon1, lat1, lon2, lat2) )
        
    # add the column to the dataframe    
    train_data_sample["distance"+str(index)] = distances

Upvotes: 1

EddyG
EddyG

Reputation: 685

I would not use apply here as it works row wise, but would go for a matrix approach using numpy instead.

First convert all degrees into radians:

df['latitude'] *= np.pi/180
df['longitude'] *= np.pi/180

Then turn the latitude and longitude vectors into matrices by duplicating the vector as many times as the length of the vector. For lat2/lon2 take the transpose.

lat1 = np.tile(df['latitude'].values.reshape([-1,1]),(1,df.shape[0]))
lon1 = np.tile(df['longitude'].values.reshape([-1,1]),(1,df.shape[0]))

lat2 = np.transpose(lat1)
lon2 = np.transpose(lon1)

Now you have 4 matrices, which contain all combinations between the lat/lon pairs on which you can simply apply your function all at once to get all the distances in one go:

dlon = lon2 - lon1 
dlat = lat2 - lat1 
a = sin(dlat/2)**2 + cos(lat1) * cos(lat2) * sin(dlon/2)**2
c = 2 * arcsin(sqrt(a)) 
km = 6367 * c

This result can be stitched to your original dataframe:

result = pd.concat([df,pd.DataFrame(km,columns=df.index)],axis=1)

Upvotes: 0

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