Python function to calculate balance for every row corresponding to individual transactions - numpy

Question

I am working on a bank statement, corresponding to the output dataframe and an ending balance corresponding to the output['balance'][0] I would like to calculate all balance values for the individual transactions as described below. It's a very straightforward calculation and yet it doesn't seem to be working - is there something quite obvious I am missing? Thanks in advance!
output['balance'] = ''
output['balance'][0] = 21.15
if len(output[amount]) > 0:
return output[balance][i+1].append((output[balance][i]-output[amount][i+1]))
else:
output[balance].append((output[balance][0]))
output[['balance']] = output['Amount'].apply(lambda amount: bal_calc(output, amount))```
---------------------------------------------------------------------------
KeyError Traceback (most recent call last)
~\Anaconda3\lib\site-packages\pandas\core\indexes\base.py in get_loc(self, key, method, tolerance)
2896 try:
-> 2897 return self._engine.get_loc(key)
2898 except KeyError:
pandas/_libs/index.pyx in pandas._libs.index.IndexEngine.get_loc()
pandas/_libs/index.pyx in pandas._libs.index.IndexEngine.get_loc()
pandas/_libs/hashtable_class_helper.pxi in pandas._libs.hashtable.PyObjectHashTable.get_item()
pandas/_libs/hashtable_class_helper.pxi in pandas._libs.hashtable.PyObjectHashTable.get_item()
KeyError: 4.95
During handling of the above exception, another exception occurred:
KeyError Traceback (most recent call last)
<ipython-input-271-b85947935fca> in <module>
----> 1 output[['balance']] = output['Amount'].apply(lambda amount: bal_calc(output, amount))
~\Anaconda3\lib\site-packages\pandas\core\series.py in apply(self, func, convert_dtype, args, **kwds)
4040 else:
4041 values = self.astype(object).values
-> 4042 mapped = lib.map_infer(values, f, convert=convert_dtype)
4043
4044 if len(mapped) and isinstance(mapped[0], Series):
pandas/_libs/lib.pyx in pandas._libs.lib.map_infer()
<ipython-input-271-b85947935fca> in <lambda>(amount)
----> 1 output[['balance']] = output['Amount'].apply(lambda amount: bal_calc(output, amount))
<ipython-input-270-cbf5ac20716d> in bal_calc(output, amount)
2 output['balance'] = ''
3 output['balance'][0] = 21.15
----> 4 if len(output[amount]) > 0:
5 return output[balance][i+1].append((output[balance][i]-output[amount][i+1]))
6 else:
~\Anaconda3\lib\site-packages\pandas\core\frame.py in __getitem__(self, key)
2978 if self.columns.nlevels > 1:
2979 return self._getitem_multilevel(key)
-> 2980 indexer = self.columns.get_loc(key)
2981 if is_integer(indexer):
2982 indexer = [indexer]
~\Anaconda3\lib\site-packages\pandas\core\indexes\base.py in get_loc(self, key, method, tolerance)
2897 return self._engine.get_loc(key)
2898 except KeyError:
-> 2899 return self._engine.get_loc(self._maybe_cast_indexer(key))
2900 indexer = self.get_indexer([key], method=method, tolerance=tolerance)
2901 if indexer.ndim > 1 or indexer.size > 1:
pandas/_libs/index.pyx in pandas._libs.index.IndexEngine.get_loc()
pandas/_libs/index.pyx in pandas._libs.index.IndexEngine.get_loc()
pandas/_libs/hashtable_class_helper.pxi in pandas._libs.hashtable.PyObjectHashTable.get_item()
pandas/_libs/hashtable_class_helper.pxi in pandas._libs.hashtable.PyObjectHashTable.get_item()
KeyError: 4.95

Answer 1

It will be easier to understand your problem you can post your existing dataframe and intended dataframe. From your description I think you can approach calculating balance like this
import pandas as pd
## creating dummy dataframe for testing
arr = np.random.choice(range(500,1000),(10,2))
debit_credit = np.random.choice([0,1], (10))
arr[:,0] = arr[:,0] * debit_credit
arr[:,1] = arr[:,1] * (1-debit_credit)
df=pd.DataFrame(arr, columns=["Debit", "Credit"])
display(df)
## calculating Balance
df["Balance"] = (df.Debit-df.Credit).cumsum()
display(df)
Output
Debit Credit Balance
0 957 0 957
1 0 698 259
2 608 0 867
3 0 969 -102
4 0 766 -868
5 0 551 -1419
6 985 0 -434
7 861 0 427
8 927 0 1354
9 0 923 431

Answer 2

bs['balance'][0] = 21.15
for i in range(1, len(bs)):
bs.loc[i, 'balance'] = bs.loc[i-1, 'balance'] + bs.loc[i, 'Credit'] -bs.loc[i, 'Debit']