Adding a column thats result of difference in consecutive rows in pandas

Question

Lets say I have a dataframe like this

    A   B
0   a   b
1   c   d
2   e   f 
3   g   h

0,1,2,3 are times, a, c, e, g is one time series and b, d, f, h is another time series. I need to be able to add two columns to the orignal dataframe which is got by computing the differences of consecutive rows for certain columns.

So i need something like this

    A   B   dA
0   a   b  (a-c)
1   c   d  (c-e)
2   e   f  (e-g)
3   g   h   Nan

I saw something called diff on the dataframe/series but that does it slightly differently as in first element will become Nan.

Answer 1

Use shift.

df['dA'] = df['A'] - df['A'].shift(-1)

Answer 2

You could use diff and pass -1 as the periods argument:

>>> df = pd.DataFrame({"A": [9, 4, 2, 1], "B": [12, 7, 5, 4]})
>>> df["dA"] = df["A"].diff(-1)
>>> df
   A   B  dA
0  9  12   5
1  4   7   2
2  2   5   1
3  1   4 NaN

[4 rows x 3 columns]

Answer 3

When using data in CSV, this would work perfectly:

my_data = pd.read_csv('sale_data.csv')
df = pd.DataFrame(my_data)
df['New_column'] = df['target_column'].diff(1)
print(df) #for the console but not necessary 

Answer 4

Rolling differences can also be calculated this way:

df=pd.DataFrame(my_data)
my_data = pd.read_csv('sales_data.csv')
i=0
j=1
while j < len(df['Target_column']):
    j=df['Target_column'][i+1] - df['Target_column'][i] #the difference btwn two values in a column.
    i+=1 #move to the next value in the column.
    j+=1 #next value in the new column.
    print(j)