Unused parameter

in ungraded lab 1 in module 1 in this function:
def generate_chart_code(instruction: str, model: str, out_path_v1: str) → str:
why we are passing out_path_v1 ?
we are instructing llm to save the figure to this path whereas it does not have tools for this.

prompt = f"“”
You are a data visualization expert.

Return your answer *strictly* in this format:

<execute_python>
# valid python code here
</execute_python>

Do not add explanations, only the tags and the code.

The code should create a visualization from a DataFrame 'df' with these columns:
- date (M/D/YY)
- time (HH:MM)
- cash_type (card or cash)
- card (string)
- price (number)
- coffee_name (string)
- quarter (1-4)
- month (1-12)
- year (YYYY)

User instruction: {instruction}

Requirements for the code:
1. Assume the DataFrame is already loaded as 'df'.
2. Use matplotlib for plotting.
3. Add clear title, axis labels, and legend if needed.
4. Save the figure as '{out_path_v1}' with dpi=300.
5. Do not call plt.show().
6. Close all plots with plt.close().
7. Add all necessary import python statements

Return ONLY the code wrapped in <execute_python> tags.
"""

i removed this parameter and everything worked just like before so i assume inclusion of this parameter is of no use

Hi arashbahariye,

In cell 4, ‘chart_v1.png’ is passed to out_path_v1, which is next included in the prompt through {out_path_v1}.

In my case, the code produced by the LLM includes plt.savefig(‘chart_v1.png’, dpi=300). So the LLM uses plt.savefig as the tool to save the figure to the working directory.

After running cell 5, if I check File->Open, I can see that the file chart_v1.png has been created. If you run the code again without using out_path_v1, if the file chart_v1.png already exists in the working directory, that file will be used as it will not be overwritten (which would happen if out_path_v1 were used).

I hope this clarifies.