'How to create an iteration for a df

How to create a loop for these statements by incrementing 0 one by one up to 25.(including the incrementation of all the df parameters eg.ES_0_BME680_Temp to ES_1_BME680_TEMP etc up to 25) and produce output for all the calculations.

df['0_680ph20']=611.2*np.exp((17.625*df[['ES_0_BME680_TEMP']])/(243.12+df[['ES_0_BME680_TEMP']]))

df['0_680aH']=(df['ES_0_BME680_RH'] /100)*(df['0_680ph20']/(461.52*(df['ES_0_BME680_TEMP']+273.15)))*1000

df['0_680LN']=np.log(((df['0_680aH']/1000)*461.52*(df['ES_0_BME680_TEMP']+273.15))/(0.5*611.2))

df['0_680T_tar']=(df['0_680LN']*243.12)/(17.625-df['0_680LN'])

df['0_688ph20']=611.2*np.exp((17.625*df[['ES_0_BME688_TEMP']])/(243.12+df[['ES_0_BME688_TEMP']]))

df['0_688aH']=(df['ES_0_BME688_RH'] /100)*(df['0_688ph20']/(461.52*(df['ES_0_BME688_TEMP']+273.15)))*1000

df['0_688LN']=np.log(((df['0_688aH']/1000)*461.52*(df['ES_0_BME688_TEMP']+273.15))/(0.5*611.2))

df['0_688T_tar']=(df['0_688LN']*243.12)/(17.625-df['0_688LN'])

thank you.

pandasdataframeloopsmathiteration


Solution 1:[1]

you could do a for loop, using f-strings to create your string

something like:

for n in range(26):
    df[f'{n}_680ph20']=611.2*np.exp((17.625*df[[f'ES_{n}_BME680_TEMP']])/(243.12+df[[f'ES_{n}_BME680_TEMP']]))
    df[f'{n}_680aH']=(df[f'ES_{n}_BME680_RH'] /100)*(df[f'{n}_680ph20']/(461.52*(df[f'ES_{n}_BME680_TEMP']+273.15)))*1000
    df[f'{n}_680LN']=np.log(((df[f'{n}_680aH']/1000)*461.52*(df[f'ES_{n}_BME680_TEMP']+273.15))/(0.5*611.2))
    df[f'{n}_680T_tar']=(df[f'{n}_680LN']*243.12)/(17.625-df[f'{n}_680LN'])
    df[f'{n}_688ph20']=611.2*np.exp((17.625*df[[f'ES_{n}_BME688_TEMP']])/(243.12+df[[f'ES_{n}_BME688_TEMP']]))
    df[f'{n}_688aH']=(df[f'ES_{n}_BME688_RH'] /100)*(df[f'{n}_688ph20']/(461.52*(df[f'ES_{n}_BME688_TEMP']+273.15)))*1000
    df[f'{n}_688LN']=np.log(((df[f'{n}_688aH']/1000)*461.52*(df[f'ES_{n}_BME688_TEMP']+273.15))/(0.5*611.2))
    df[f'{n}_688T_tar']=(df[f'{n}_688LN']*243.12)/(17.625-df['0_688LN'])

Also, I see that you are doing the same 4 operations for two different digits. You could also create a function to do that, something like

def expandata(df, digits, nitems):
    for n in range(nitems+1):
        df[f'{n}_{digits}ph20']=611.2*np.exp((17.625*df[[f'ES_{n}_BME{digits}_TEMP']])/(243.12+df[[f'ES_{n}_BME{digits}_TEMP']]))
        df[f'{n}_{digits}aH']=(df[f'ES_{n}_BME{digits}_RH'] /100)*(df[f'{n}_{digits}ph20']/(461.52*(df[f'ES_{n}_BME{digits}_TEMP']+273.15)))*1000
        df[f'{n}_{digits}LN']=np.log(((df[f'{n}_{digits}aH']/1000)*461.52*(df[f'ES_{n}_BME{digits}_TEMP']+273.15))/(0.5*611.2))
        df[f'{n}_{digits}T_tar']=(df[f'{n}_{digits}LN']*243.12)/(17.625-df[f'{n}_{digits}LN'])

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Solution Source
Solution 1

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