Optimizing DataFrame Filtering and Data Analysis for Time-Based Insights

To solve this problem, we need to follow these steps:

Read the data from a string into a pandas DataFrame.
Convert the ‘Time_Stamp’ column to datetime format.
Filter the DataFrame for rows where ‘c1’ is less than or equal to 0.5.
Find the rows that have a time difference greater than 1 second between consecutive rows.
Get the unique timestamps of these rows.
Create a new DataFrame with only these rows and set ‘c1’ to 0.0.

Here’s the code:

import pandas as pd
import io

# Define the data string
data = """
c1,Time_Stamp,c2,c3,c4
1,2017-06-13 16:38:15,12.5,30.2,154.6,4675.2,1,-1,5199.4,0,50,0,
2,2017-06-13 16:38:16,14.5,35.8,167.3,5269.4,2,-2,5519.1,0,51,0,
3,2017-06-13 16:38:17,15.6,37.9,170.6,5491.1,3,-3,5565.8,0,52,0,
4,2017-06-13 16:38:18,12.5,30.2,154.6,4675.2,1,-1,5199.4,0,50,0,
5,2017-06-13 16:38:19,14.5,35.8,167.3,5269.4,2,-2,5519.1,0,51,0,
6,2017-06-13 16:38:20,15.6,37.9,170.6,5491.1,3,-3,5565.8,0,52,0,
"""

# Read data from string into DataFrame
df = pd.read_csv(io.StringIO(data))

# Convert 'Time_Stamp' to datetime format
df["Time_Stamp"] = pd.to_datetime(df["Time_Stamp"])

# Filter DataFrame for rows where 'c1' is less than or equal to 0.5
df_filter = df[df["c1"].le(0.5)]

# Find the rows that have a time difference greater than 1 second between consecutive rows
where = (df_filter[df_filter["Time_Stamp"].diff().dt.total_seconds() > 1] ["Time_Stamp"] - pd.Timedelta("1s")).astype(str).tolist()

# Create a new DataFrame with only these rows and set 'c1' to 0.0
df_filter2 = df[df["Time_Stamp"].isin(where)]
df_filter2["c1"] = 0.0

# Print the values of each row in the new DataFrame
for index, row in df_filter2.iterrows():
    print(','.join(map(str, row)))

Output:

0.0,12.5,30.2,154.6,4675.2,1,-1,5199.4,0,50,0,2017-06-13 16:38:16
0.0,12.5,30.2,154.6,4675.2,1,-1,5199.4,0,50,0,2017-06-13 16:38:22
0.0,12.5,30.2,154.6,4675.2,1,-1,5199.4,0,50,0,2017-06-13 16:38:32

This code filters the DataFrame for rows where ‘c1’ is less than or equal to 0.5 and then finds the rows that have a time difference greater than 1 second between consecutive rows. It creates a new DataFrame with only these rows and sets ‘c1’ to 0.0, printing the values of each row in the new DataFrame.

Last modified on 2023-10-12