How to do operations with different columns based on "first" condition using pandas Dataframe?

I have this dataframe:

I need to perform some addictions and substractions based on conditions.

If we first have a non-NaN value in the tp, then do this: tp-entry

if, instead, the first non-NaN value is contained inside "sl" column,
then do this sl-entry.

We will store these values inside a new column called "pl", so the final datafram
will look like this:

I tried (with no success) this (reproducible code):

tbl = {"date" :["2022-02-27", "2022-02-27", "2022-02-27", "2022-02-27", "2022-02-28", 
                  "2022-02-28","2022-02-28", "2022-02-28"],
      "entry" : ["NaN", "NaN", 1.2, "NaN", "NaN", 1.3, "NaN", "NaN"],
      "tp" : ["NaN", "NaN", "NaN", 1.4, "NaN", "NaN", "NaN", "NaN"],
      "sl" : ["NaN", "NaN", "NaN", "NaN", "NaN", "NaN", "NaN", 1.15]}

df = pd.DataFrame(tbl)


df.sort_values(by = "date", inplace=True)



df['pl'] = np.where(df["entry"])  #i don't know how to continue...

Any ideas? Do you know better way?

Edit

In the photo entry-sl is wrong, I need sl-entry

>Solution :

IIUC, you can combine "tp" and "pl", then bfill per group. Also bfill "entry" per group. Then assign the difference only on the first (i.e. non duplicate) row per date:

group = df['date']
s1 = df['tp'].fillna(df['sl']).groupby(group).bfill()
s2 = df['entry'].groupby(group).bfill()

df.loc[~group.duplicated(), 'pl'] = s1-s2

NB. if you really have dates with times, use instead as group:

group = pd.to_datetime(df['date']).dt.date

output:

         date  entry   tp    sl    pl
0  2022-02-27    NaN  NaN   NaN  0.20
1  2022-02-27    NaN  NaN   NaN   NaN
2  2022-02-27    1.2  NaN   NaN   NaN
3  2022-02-27    NaN  1.4   NaN   NaN
4  2022-02-28    NaN  NaN   NaN -0.15
5  2022-02-28    1.3  NaN   NaN   NaN
6  2022-02-28    NaN  NaN   NaN   NaN
7  2022-02-28    NaN  NaN  1.15   NaN