Count the values after an underscore in a Pandas Series

Use filter and groupby:

>>> df.filter(like='event') \
      .groupby(lambda x: x.rsplit('_', 1)[1], axis=1) \
      .sum().sum()

negative    11
neutral      5
positive     9
dtype: int64

You could create new columns where you add all the ones with similar effects, e.g. df["negative"] = df["event_2_negative"] + df["event_4_negative"] and then delete the other columns and then do a df["negative"].value_counts() for example or df["negative"].sum()

This is a rare case where accounting in a loop seems practical as there's a manageable count of columns

# start a dictionary to pack with results
categories = {k: 0 for k in ("negative", "positive", "neutral")}

for colname in df:  # dataframes are iterable by-column name!
    if "_" not in colname:
        continue  # or consider alerting on this condition
    for category in categories.keys():
        if category in colname:  # substring compare
            categories[category] += df[colname].sum()
            break  # successfully found category -> next colname
    else:  # didn't find and break
        raise ValueError(f"no category for {colname}")

# create a collection of results
df_results = pd.Series(categories).to_frame().reset_index()

    index      0
0   negative   11
1   positive   9
2   neutral    5

One option is to reshape the data, where you have all the negatives in one column, same for positives and neutral. Your columns have a pattern (some end in positives, some columns end in neutral, ...).

We can use that pattern in pivot_longer from pyjanitor to reshape the data, then aggregate(reduce):

# pip install git+https://github.com/pyjanitor-devs/pyjanitor.git
import pandas as pd
import janitor
(df.pivot_longer(index='ID', 
                 names_to=('positive', 'negative', 'neutral'), 
                 names_pattern=('positive', 'negative', 'neutral'))
   .iloc[:, 1:]
   .sum()
) 
positive     9.0
negative    11.0
neutral      5.0
dtype: float64

df. \
drop("ID", axis=1). \
rename(columns=lambda x: x.split("_")[2]). \
unstack(). \
droplevel(1). \
reset_index(). \
groupby("index"). \
sum()

#           0
# index     
# negative  11
# neutral   5
# positive  9