So I'm a beginner at Python and I have a dataframe with Country, avgTemp and year. What I want to do is calculate new rows on each country where the year adds 20 and avgTemp is multiplied by a variable called tempChange. I don't want to remove the previous values though, I just want to append the new values.
This is how the dataframe looks:
Preferably I would also want to create a loop that runs the code a certain number of times Super grateful for any help!
If you need to copy the values from the dataframe as an example you can have it here:
Country avgTemp year
0 Afghanistan 14.481583 2012
1 Africa 24.725917 2012
2 Albania 13.768250 2012
3 Algeria 23.954833 2012
4 American Samoa 27.201417 2012
243 rows × 3 columns
If you want to repeat the rows, I'd create a new dataframe, perform any operation in the new dataframe (sum 20 years, multiply the temperature by a constant or an array, etc...) and use then use concat()
to append it to the original dataframe:
import pandas as pd
tempChange=1.15
data = {'Country':['Afghanistan','Africa','Albania','Algeria','American Samoa'],'avgTemp':[14,24,13,23,27],'Year':[2012,2012,2012,2012,2012]}
df = pd.DataFrame(data)
df_2 = df.copy()
df_2['avgTemp'] = df['avgTemp']*tempChange
df_2['Year'] = df['Year']+20
df = pd.concat([df,df_2]) #ignore_index=True if you wish to not repeat the index value
print(df)
Output:
Country avgTemp Year
0 Afghanistan 14.00 2012
1 Africa 24.00 2012
2 Albania 13.00 2012
3 Algeria 23.00 2012
4 American Samoa 27.00 2012
0 Afghanistan 16.10 2032
1 Africa 27.60 2032
2 Albania 14.95 2032
3 Algeria 26.45 2032
4 American Samoa 31.05 2032
where df is your data frame name:
df['tempChange'] = df['year']+ 20 * df['avgTemp']
This will add a new column to your df with the logic above. I'm not sure if I understood your logic correct so the math may need some work
I believe that what you're looking for is
dfName['newYear'] = dfName.apply(lambda x: x['year'] + 20,axis=1)
dfName['tempDiff'] = dfName.apply(lambda x: x['avgTemp']*tempChange,axis=1)
This is how you apply to each row.
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