I have used great_circle function to find the distance between two coordinates.For example, if i have the following data:
lat1= ['-77.85']
lon1= ['80.23']
lat2= [-22.4532']
lon2= ['62.45']
And to find the distance i use the following code:
import pandas as pd
from geopy.distance import great_circle
from datetime import datetime,timedelta as dt
distance = df.apply(lambda x: great_circle((x['lat1'], x['lon1']),(x['lat2'], x['lon2'])), axis = 1) # gives distance in km
And to find the time from the distance, as we know Time = Distance/Speed. I assume speed to be 80km/hr and I use:
time= distance/80
The obtained 'time' has the unit 'km' . I want to convert this function to time.
df['time'] = pd.to_timedelta(time,unit='h')
I obtain the following error:
Error : ValueError: Invalid type for timedelta scalar: class 'geopy.distance.great_circle'
Could you tell me how to obtain 'time' as time function and not as distance 'km'?
Consider using .km attribute.
Demo:
In [33]: df
Out[33]:
lat1 lon1 lat2 lon2
0 -77.85 80.23 -22.4532 62.45
In [34]: distance = x.apply(lambda x: great_circle((x['lat1'], x['lon1']),
(x['lat2'], x['lon2'])).km,
axis = 1)
# ----> ^^^
In [35]: distance
Out[35]:
0 6233.235674
dtype: float64
In [36]: df['time'] = pd.to_timedelta(distance/80,unit='h')
In [37]: df
Out[37]:
lat1 lon1 lat2 lon2 time
0 -77.85 80.23 -22.4532 62.45 3 days 05:54:55.605600