How to get only top 5 peer companies list for a given isin list

uma

Hi, I am using below code in python to pull the peers list for the given isin list. The api return more number of peers for each isin and there is no mapping for isin and peers. In the response, i need only top 5 peers for each isin and i want to map which peers belong to which isin. How to update the code ?

for isin in isin_list:
peer_list.append('Peers(' + isin + ')')

df1, err = ek.get_data(instruments=peer_list, fields=["TR.PrimaryInstrument", "TR.ExchangeTicker","TR.CompanyName", "TR.ISIN"],raw_output=False);

Zhenya Kovalyov

@uma, this is more of a python/pandas usage question, I strongly encourage you to research both. What you want can be achieved with this:

import eikon as ek 

import pandas as pd

First, let's design a function that is going to take top five peers, I have chosen the market cap as the selection parameter. After getting the data I sort by Market Cap, create a new column with my original company's ISIN code, and setting it as the index of my data frame before returning it:

def get_top_five_peers(code): 
	df, e = ek.get_data(f'PEERS({code})', ['TR.PrimaryInstrument', 'TR.ExchangeTicker','TR.CompanyName', 'TR.ISIN', 'TR.CompanyMarketCap(ShType=OUT)']) 
	res = df.sort_values(by=['Company Market Cap'], ascending=False).head(5).copy() 
	res['Peer of'] = code 
	res.set_index('Peer of', inplace=True) 
	return res 

Here is how you can use the `get_top_five_peers' function:

isins = ['US0378331005', 'GB00BH4HKS39', 'RU0009024277', 'US46625H1005'] 

df = pd.concat([get_top_five_peers(isin) for isin in isins]) 

Here is the result: