Getting error while using the random choices function

Question

I'm trying to open random links using

webbrowser.open(random.choices("link1","link2","link3")

but it's showing error Check it out and help me resolve this issue.

My code:

webbrowser.open(random.choices("https://www.youtube.com/watch?v=YGf8JJZM_Yg", "https://www.youtube.com/watch?v=upsF9NULamA"))

Error I am facing:

    total = cum_weights[-1] + 0.0   # convert to float
TypeError: can only concatenate str (not "float") to str

` Traceback (most recent call last): File "c:\PYTHON PROGRAMMES\DUSTIN- THE VOICE ASSISTANT\dustin.py", line 233, in <module> webbrowser.open(random.choices("youtube.com/watch?v=YGf8JJZM_Yg", "youtube.com/watch?v=upsF9NULamA")) File "C:\Users\socia\AppData\Local\Programs\Python\Python39\lib\random.py", line 500, in choices total = cum_weights[-1] + 0.0 # convert to float TypeError: can only concatenate str (not "float") to str ` — Ashutosh Singh
– Ashutosh Singh, Commented Aug 24, 2021 at 5:09

U13-Forward · Accepted Answer · 2021-08-24 05:15:50Z

2

Instead of:

total = cum_weights[-1] + 0.0

Try:

total = float(cum_weights[-1])

Edit:

For your other error with random.choices, try this instead:

webbrowser.open(random.choice(["https://www.youtube.com/watch?v=YGf8JJZM_Yg", "https://www.youtube.com/watch?v=upsF9NULamA"]))

random.choices take in a single argument with a parameter of a list, it isn't a *args expression in the implementation.

edited Aug 24, 2021 at 5:15

answered Aug 24, 2021 at 4:54

U13-Forward

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3 Comments

Divyesh Peshavaria Over a year ago

This line where the exception is raised is in random.py. It is not editable. The exception is happening because arguments to random.choices() are passed incorrectly.

Ashutosh Singh Over a year ago

ERROR ` Traceback (most recent call last): webbrowser.open(random.choices(["youtube.com/watch?v=YGf8JJZM_Yg", "youtube.com/watch?v=upsF9NULamA"])) File "C:\Users\socia\AppData\Local\Programs\Python\Python39\lib\webbrowser.py", line 86, in open if browser.open(url, new, autoraise): File "C:\Users\socia\AppData\Local\Programs\Python\Python39\lib\webbrowser.py", line 603, in open os.startfile(url) TypeError: startfile: filepath should be string, bytes or os.PathLike, not list `

U13-Forward Over a year ago

@AshutoshSingh Edited my answer, should work now, it was as typo

Divyesh Peshavaria · Accepted Answer · 2021-08-24 05:23:48Z

1

The arguments to random.choices() must be an iterable like list, tuple, etc like this:

random.choices(
("https://www.youtube.com/watch?v=YGf8JJZM_Yg", "https://www.youtube.com/watch?v=upsF9NULamA")
)

Notice the extra pair of parentheses. You can pass as many arguments you want but they should be given like this:

random.choices((arg1, arg2, arg3, ...))

instead of:

random.choices(arg1, arg2, arg3, ...)

Both of the below options will work:

Use random.choice() over random.choices() to select only 1 option out of multiple options.

1.

webbrowser.open(random.choice(("https://www.youtube.com/watch?v=YGf8JJZM_Yg", "https://www.youtube.com/watch?v=upsF9NULamA")))

webbrowser.open(random.choice(["https://www.youtube.com/watch?v=YGf8JJZM_Yg", "https://www.youtube.com/watch?v=upsF9NULamA"]))

edited Aug 24, 2021 at 5:23

answered Aug 24, 2021 at 4:56

Divyesh Peshavaria

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3 Comments

Divyesh Peshavaria Over a year ago

@AshutoshSingh edited answer to add code samples, check if it helps you

Ashutosh Singh Over a year ago

no. it's not working :( ` Traceback (most recent call last) webbrowser.open(random.choices(["youtube.com/watch?v=YGf8JJZM_Yg", "youtube.com/watch?v=upsF9NULamA"])) File "C:\Users\socia\AppData\Local\Programs\Python\Python39\lib\webbrowser.py", line 86, in open if browser.open(url, new, autoraise): File "C:\Users\socia\AppData\Local\Programs\Python\Python39\lib\webbrowser.py", line 603, in open os.startfile(url) TypeError: startfile: filepath should be string, bytes or os.PathLike, not list `

Divyesh Peshavaria Over a year ago

@AshutoshSingh can you try using random.choice() function instead of random.choices()? It is more suitable when you need only 1 option out of a list of options. Updating my answer accordingly,

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