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Counting Consecutive Identical Elements in a List of 1s and 0s: A Python Guide

Learn how to effectively count consecutive identical elements in a list of 1s and 0s using Python. This detailed guide will break down the solution with easy-to-understand examples. --- This video is based on the question https://stackoverflow.com/q/71854448/ asked by the user 'GradStudent' ( https://stackoverflow.com/u/3461324/ ) and on the answer https://stackoverflow.com/a/71854687/ provided by the user 'robinood' ( https://stackoverflow.com/u/8814229/ ) at 'Stack Overflow' website. Thanks to these great users and Stackexchange community for their contributions. Visit these links for original content and any more details, such as alternate solutions, latest updates/developments on topic, comments, revision history etc. For example, the original title of the Question was: Consecutive repeated identical elements Also, Content (except music) licensed under CC BY-SA https://meta.stackexchange.com/help/l... The original Question post is licensed under the 'CC BY-SA 4.0' ( https://creativecommons.org/licenses/... ) license, and the original Answer post is licensed under the 'CC BY-SA 4.0' ( https://creativecommons.org/licenses/... ) license. If anything seems off to you, please feel free to write me at vlogize [AT] gmail [DOT] com. --- Counting Consecutive Identical Elements in a List of 1s and 0s: A Python Guide When working with lists in Python, you might encounter scenarios where you need to analyze patterns, such as counting consecutive identical elements. In this guide, we’ll address a common problem faced by many Python developers: how to count the number of times there are consecutive 1s in a binary list (a list containing only 0s and 1s). The Problem at Hand Consider the following binary list: [[See Video to Reveal this Text or Code Snippet]] The goal is to find out how many times there are consecutive sequences of 1s. In this case, the expected output is 2, as there are two distinct blocks of consecutive 1s before the block of 0s interrupts the sequence. Current Approach You might start with a function, such as the one provided below, that looks something like this: [[See Video to Reveal this Text or Code Snippet]] Despite its intentions, this code will return 1 instead of the desired 2. Let's explore why this happens and how we can fix it. Issue Explanation The underlying issue is that while the current code is attempting to track pairs of consecutive 1s, it fails to distinguish between different sequences. Instead, it collects all 1s, which reduces to 1 when converted to a set (the set only retains unique values). The Solution: A Revised Approach To correctly count the number of consecutive sequences of 1s, we can adjust the approach. Here’s how: Track the Start of Each Sequence: Instead of appending all consecutive 1s to the list, we need to recognize only the start of each new group of 1s. Use Boolean Flags: Introduce a boolean flag to track whether we are currently in a sequence of 1s. Here is a refined version of the function: [[See Video to Reveal this Text or Code Snippet]] How This Works Initialization: We start by initializing an empty list res and a boolean flag started. Iteration Through the List: We iterate through the list while using indices to check for identical consecutive elements. Recording Sequences: If we find a 1 that is part of a consecutive sequence and we haven't already counted the start of that sequence (started is False), we append it to res and set started to True. If we hit a different number, we reset started back to False. Final Output When you run this revised function with the specified list, you will correctly receive the output 2, indicating the two sequences of consecutive 1s. Conclusion Counting consecutive identical elements in a list can be tricky, but with the right approach, it becomes manageable. By tracking the start of each sequence and using boolean flags, we can accurately determine the number of sequences of 1s in a binary list. Feel free to test this method with different lists and explore how it can be adapted to various similar problems!