Which Splinterlands cards have the highest spread %? || Ep. #48 || Splinterlands

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(Edited)

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Previously I did a post about a major mistake I made when I started buying cards on the market. Since doing that post, I have always wondered which cards have the highest spread regarding USD value between its lowest and lowest per base card XP. In this post, we will explore which splinterlands cards have the highest spread.

Top 10 Monster Cards with the highest spread %

idnamegoldeditionlow_pricelow_price_bcxspread %
238Mighty DrickenFALSEPromo17617,032933,349
130Archmage AriusFALSEPromo69066,774933,3363
205Prince JulianFALSEPromo35033,871933,3323
117Red DragonTRUEPromo1299324,75300
130Archmage AriusTRUEPromo10500030000250
205Prince JulianTRUEPromo70002000250
50Haunted SpiritFALSEAlpha1,70,509233,9882
34Divine HealerFALSEAlpha0,480,175174,2857
47Skeleton AssassinTRUEBeta3,351,289159,8914
34Divine HealerFALSEBeta0,1760,069155,0725

I am not surprised with the top 5 cards with the highest spread, but I am surprised to see the likes of Divine Healer and Skeleton Assassin's in the top 10.

Python Code

I am a hobbyist programmer and used python to query and pull the data. You can run the following to produce the table above.

import requests
from pandas import json_normalize
import numpy as np
import pandas as pd

# splinterlands grouped market summary
url = 'https://api.splinterlands.io/market/for_sale_grouped'
response = requests.get(url)
data_grouped = response.json()
df_grouped = json_normalize(data_grouped)

# splinterlands card details market summary
url = 'https://api.splinterlands.io/cards/get_details'
response = requests.get(url)
data_card_details = response.json()
df_card_details = json_normalize(data_card_details)

# merge
df_grouped.rename(columns={'card_detail_id':'id'}, inplace=True)
df_merged = df_grouped.merge(df_card_details[['id', 'name']], on='id', how='inner')

# update columns
df_merged.loc[df_merged['edition'] == 0, ['edition']] = 'Alpha'
df_merged.loc[df_merged['edition'] == 1, ['edition']] = 'Beta'
df_merged.loc[df_merged['edition'] == 2, ['edition']] = 'Promo'
df_merged.loc[df_merged['edition'] == 3, ['edition']] = 'Reward'
df_merged.loc[df_merged['edition'] == 4, ['edition']] = 'Untamed'
df_merged.loc[df_merged['edition'] == 5, ['edition']] = 'Dice'

# drop columns
df_merged.drop(['qty','high_price'], axis=1,inplace=True)

# order columns
df_merged = df_merged[['id','name','gold','edition','low_price','low_price_bcx']]

# add new column: spread
df_merged['spread %'] = ((df_merged['low_price'] / df_merged['low_price_bcx']) - 1) * 100

# sort by spread
df_merged.sort_values('spread %', inplace=True, ascending=False)

# export to excel
df_merged.to_excel(r'C:\Development\splinterlands\splinterlands.xlsx')

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I recently did a compilation post titled, The New Players Guide to Splinterlands: A Collection of Articles and Guides. This post serves as a summary of articles and guides I have written for new players starting fresh on Splinterlands. If you enjoy reading my Splinterlands content, please follow and support me by signing up to playing Splinterlands through my affiliate link: https://splinterlands.com?ref=mercurial9.


Thank you for reading and hope you have a good rest of the day!

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This post has been upvoted by @opgaming. We like what you're posting, and we want to reward our members who produce quality content. Keep up the good work!
If this sounds like the sort of group you want to be a part of, click the image below! We're too OP to be put down!

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Thanks for sharing! - @ashikstd

Well, as I'm not a programmer and I don't understand things that programmer does actually so it's also beyond my understanding.
But it seems a nice one.

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