How to find the index of a row for a particular value in a particular column and then create a new column with that starting point? - python-3.x

Question

Example of dataframe
What I'm trying to do with my dataframe...
Locate the first 0 value in a certain column (G in the example photo).
Create a new column (Time) with the value 0 lining up on the same row with the same 0 value in column (G).
And then each row after the 0 in column (Time) +(1/60) until the end of the data.
And -(1/60) before the 0 in (Time) column until the beginning of data.
What is the best method to achieve this?
Any advice would be appreciated. Thank you.

Answer 1

Pretty straight forward
identify index of row that contains the value you are looking for
then construct an array where start is negative, zero will be index row and end at value for end of series
import numpy as np
df = pd.DataFrame({"Time":np.full(25, 0), "G":[i if i>0 else 0 for i in range(10,-15,-1)]})
# find the where value is zero
idx = df[df["G"]==0].index[0]
intv = round(1/60, 3)
# construct a numpy array from a range of values (start, stop, increment)
df["Time"] = np.arange(-idx*intv, (len(df)-idx)*intv, intv)
df.loc[idx, "Time"] = 0 # just remove any rounding at zero point
print(df.to_string(index=False))
output
Time G
-0.170 10
-0.153 9
-0.136 8
-0.119 7
-0.102 6
-0.085 5
-0.068 4
-0.051 3
-0.034 2
-0.017 1
0.000 0
0.017 0
0.034 0
0.051 0
0.068 0
0.085 0
0.102 0
0.119 0
0.136 0
0.153 0
0.170 0
0.187 0
0.204 0
0.221 0
0.238 0