How to remove number along from list?

Example my list will be [“12.345”,Test_1,Test_2,“45.456”,Test_3,Test_4,Test_5,“67.789”]

Now I want remove all numbers.

expected list: [Test_1,Test_2,Test_3,Test_4,Test_5]

The list which we are getting with number ( fractio number) dynamic index.

I tried the below way nothing is working.

put [“12.345”,Test_1,Test_2,“45.456”,Test_3,Test_4,Test_5,“67.789”] into notes
delete each item of notes which is Number
log notes

put [“12.345”,Test_1,Test_2,“45.456”,Test_3,Test_4,Test_5,“67.789”] into notes
delete each item of notes which is [“1”,“2”,“3”, “4”,“.”]
log notes

put [“12.345”,Test_1,Test_2,“45.456”,Test_3,Test_4,Test_5,“67.789”] into notes
delete [“1”,“2”,“3”,“4”,“.”] from notes
log notes

Is there any alternate method?

Your first attempt was very very close to being correct, but you were missing the word “a”! As written:

delete each item of notes which is Number

the is operator compares each item of the list to see if it is (is equal to) the variable Number. Instead, you need to use the is a operator to check whether each item of the list is a number:

delete each item of notes which is a Number
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Hello Sense TalkDoug,

Do you have any idea how to remove the words based on specified letter?
Example:
put [“2023-05-31T23:59:59Z”,“2021-12-31T23:59:59Z”,“2021-12-31T23:59:59Z”,“2017-12-31T23:59:59Z”,“2017-12-31T23:59:59Z”,“2017-12-31T23:59:59Z”,“2017-12-31T23:59:59Z”,“2017-12-31T23:59:59Z”,“2017-12-31T23:59:59Z”,“2017-12-31T23:59:59Z”] into TestDta
Here I need to remove Txx:xx:xxZ from all date in the list and log only the date which is greater than or equal to today

Something like this will probably do what you’re looking for:

put ["2023-05-31T23:59:59Z","2021-12-31T23:59:59Z","2021-12-31T23:59:59Z","2017-12-31T23:59:59Z","2017-12-31T23:59:59Z","2017-12-31T23:59:59Z","2017-12-31T23:59:59Z","2017-12-31T23:59:59Z","2027-12-31T23:59:59Z","2017-12-31T23:59:59Z"] into TestData

delete <"T", chars, "Z"> from each item of TestData
put TestData

put each item of TestData where date(each) is later than today
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