You can use 2 times cumsum(): # reset val desired_col #0 0 1 1 #1 0 5 6 #2 0 4 10 #3 1 2 2 #4 1 -1 -1 #5 0 6 5 #6 0 4 9 #7 1 2 2 df[‘cumsum’] = df[‘reset’].cumsum() #cumulative sums of groups to column des df[‘des’]= df.groupby([‘cumsum’])[‘val’].cumsum() print … Read more
Here’s a very fast way: result = ((((12 * n + 45) * n + 50) * n + 15) * n – 2) * n // 120 How I got there: Rewrite the inner sum as the well-known x*(x+1)//2. So the whole thing becomes sum(x**2 * x*(x+1)//2 for x in range(n+1)). Rewrite to sum(x**4 … Read more
This does it: np.cumsum(x[::-1])[::-1]
Just apply cumsum on the pandas.Series df[‘SUM_C’] and assign it to a new column: df[‘CUMSUM_C’] = df[‘SUM_C’].cumsum() Result: df Out[34]: A B SUM_C CUMSUM_C 0 1 1 10 10 1 1 2 20 30
You can use groupby by custom Series: df = pd.DataFrame({‘a’: [1, 1, -1, 1, -1, -1]}) print (df) a 0 1 1 1 2 -1 3 1 4 -1 5 -1 print ((df.a != df.a.shift()).cumsum()) 0 1 1 1 2 2 3 3 4 4 5 4 Name: a, dtype: int32 for i, g in … Read more
This should do it, need groupby() twice: df.groupby([‘name’, ‘day’]).sum() \ .groupby(level=0).cumsum().reset_index() Explanation: print(df) name day no 0 Jack Monday 10 1 Jack Tuesday 20 2 Jack Tuesday 10 3 Jack Wednesday 50 4 Jill Monday 40 5 Jill Wednesday 110 # sum per name/day print( df.groupby([‘name’, ‘day’]).sum() ) no name day Jack Monday 10 Tuesday … Read more