How to convert string to a dataframe name pandas/python

Question

I have 2 dataframes (dfA, dfB)

I have another dataframe dfC which has a column called 'SOURCE'. Values of SOURCE will either be dfA or dfB.

I am trying to go row by row on dfC and depending on the SOURCE field, either update dfA or dfB.

To do this I tried to create a variable 'a' which will take the value of SOURCE. The issue is, 'a will equal 'dfA' or 'dfB' (reads it as a string). I cannot use this string to call the correct dataframe (error thrown)

for i in [0, len(dfC.index)-1]:
    a = dfC.loc[i,'Source']
    temp = a[['colA', 'colB', colC']]][(a['movement_id']) == dfC.loc[i,'TEMP']]

can you post your dataframes so we can see the columns and sample data? use df.head() — Kyle
– Kyle, Commented Oct 16, 2018 at 17:55
Rather than passing the DF as a string, why not just pass the variable itself? 'dfA', 'dfB' to dfA, dfB — G. Anderson
– G. Anderson, Commented Oct 16, 2018 at 17:59
The goal is not to hardcode as dfC will continuously grow. Is there a way I can convert a string to the variable name? That would solve the problem — babz
– babz, Commented Oct 16, 2018 at 18:00
You can use locals(), but keep in mind that it is often a sign of bad design. — 0x5453
– 0x5453, Commented Oct 16, 2018 at 18:08

quest · Accepted Answer · 2018-10-16 18:56:39Z

1

Just use if-else in your loop;

for ind in range(dfC.shape[0]):
   if dfC.loc[ind, 'SOURCE'] == 'dfA':
       "do what you want to dfA"
   if dfC.loc[ind, 'SOURCE'] == 'dfB':
           "do what you want to dfB"

Does this sound reasonable?

answered Oct 16, 2018 at 18:56

quest

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PyJan · Accepted Answer · 2018-10-17 10:55:51Z

1

I suggest to use pandas.DataFrame.update which updates dataframe based on index.

dfa = DataFrame([[2],[4]], index=[1,2], columns=['val'])
dfb = DataFrame([[4],[8]], index=[1,2], columns=['val'])
dfc = DataFrame([['dfa',-1],['dfb',-2],['dfa',-3]], index=[1,1,2], columns='dest','val'])

dfa.update(dfc[dfc['dest']=='dfa'])
dfb.update(dfc[dfc['dest']=='dfb'])

Using for loop in pandas is usually discouraged.

answered Oct 17, 2018 at 10:55

PyJan

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