Dropping Empty Lists from Nested List

Question

So I have a function which returns a List which contains either empty lists or Series. I loop through a list of tickers and for each it will return a empty list or Series and store them inside one list.

However, after looping through all I want to be able to drop the empty lists and only have the Series within the list.

def get_revenue_growth(ticker) -> pd.DataFrame:

income_statement_annually = fa.financial_statement_growth(ticker, FA_API_KEY, period="annual")

if  'revenueGrowth' in income_statement_annually.index:
    
    revenue_growth = income_statement_annually.loc['revenueGrowth']
    exchange_df = pd.DataFrame({ticker : revenue_growth})        
    exchange_df.index = pd.to_datetime(pd.Series(exchange_df.index))
    exchange_df = exchange_df[exchange_df.index.year >= 1998]
    exchange_df = exchange_df.sort_index()
    
    print('Getting Revenue Growth for ' + ticker + ': Passed')
    
else:
    print('Getting Revenue Growth for ' + ticker + ': Failed')
    
    exchange_df = []
    
return exchange_df

This is the function I am calling via this:

revenue_growth = [get_revenue_growth(t) for t in tickers]

Here is what the output looks like...

So what I am trying to achieve is to remove all the empty lists. I tried this list2 = [x for x in list1 if x != []] but it did not work.

@JonSG I tried this earlier, sadly get the error ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all(). — RHH
– RHH, Commented Jun 3, 2021 at 13:55

Cristofor · Accepted Answer · 2021-06-03 13:57:20Z

1

You can simply solve it via:

list2 = [x for x in list1 if len(x)>0]

answered Jun 3, 2021 at 13:57

Cristofor

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PCM · Accepted Answer · 2021-06-03 14:02:08Z

1

Look at this Example -

mylist = [] 

if len(mylist) == 0:
    del mylist # Deletes the Empty List
else:
    # Do Something else

Modify this piece for your program

answered Jun 3, 2021 at 14:02

PCM

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