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Why is the length of empty slice equal to 1?

By : Lina Tiongco
Date : September 16 2020, 10:00 AM
Does that help Becuase it's not empty but a slice that contains one string, which is the empty string.
Print it like:
code :
fmt.Printf("%q, %T, len %v\n", s, s, len(s))
[], []string, len 0
[""], []string, len 1
func Split(s, sep string) []string

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Why does splitting a string on itself return an empty slice with a length of two?

By : nisha vb
Date : March 29 2020, 07:55 AM
it helps some times As I understand it, the split function returns everything before the / (which is nothing) in the first item, and everything after the / (also nothing) in the second item. Hence, two empty strings. As for why you ever get empty strings, it's so that split() can basically be the opposite of join, as explained here:
Why are empty strings returned in split() results?

python slice() function vs slice notation - How to handle empty values in slice()?

By : farah naz
Date : March 29 2020, 07:55 AM
help you fix your problem The docs cover 3 usages of the slice() object: , You can set the stop argument to None to continue indefinitely.
code :
obj = [1, 2, 3, 4]

obj[2:] == obj[slice(2, None)]
# == [3, 4]

obj[:] == obj[slice(None, None)]
# == [1, 2, 3, 4]

obj[::2] == obj[slice(None, None, 2)]
# == [1, 3]

Why is array.length equal to 0 when it's not empty, and how can I loop through it?

By : yogesh uttam
Date : March 29 2020, 07:55 AM
I wish this helpful for you I'm bit in confusion, this should be pretty obvious but I'm getting lost trying to figure everything out. , From MDN:
code :
let self = this;
reader.onloadend = function() {
   // console.log(reader.result);
      "id": i,
      "name": files.files[i].name,
      "img": reader.result
       m_file: files_result,

Python replace string with empty if length not equal to x

By : Jolicour Notbukotubu
Date : March 29 2020, 07:55 AM
I hope this helps you . Your column ssn contains numbers not string, that is why it is not working. Try the following :
code :
mask = df['ssn'].astype(str).str.len() != 5
df.loc[mask, 'ssn'] = ''

In [1] : print(df)
Out[1] :    Name    ssn
0  john  12345
1  mike  54321
2  adam       
3  doug  47895
4   liz      

How is the term “dynamically sized” for slice justifiable, when maximum length of a slice is less than length of the und

By : Asher McPherson
Date : March 29 2020, 07:55 AM
hope this fix your issue Even with a cap, a dynamic size is still dynamic: it can range between zero and whatever the cap is.
That said, as Flimzy noted in a comment, if you use an operation that can "grow" a slice, it always returns a new slice, or takes a pointer to the slice—usually the former—so that the routine that needed to go past the current capacity could allocate a new, larger array1 and make the slice use that instead of the old array.
code :
s = append(s, element)
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