Extracting the numeric part from a cell

2021 11 17
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更新时间：2021 11 17
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Hi everyone,
I have a column string in this format
"injection-0:94577)"
"injection-0:94578)"
"injection-0:94579)"
"injection-0:94580)"
"injection-0:94581)"
"injection-0:94582)"
"injection-0:94583)"
I want to extract the numeric part that starts with 9 like 94577 from roow one for instance. Can anyone help please? Thank you.

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Paul
Paul 2021-11-17
Ran in:

0 个投票

Ran in:
str = ["injection-0:94577)"
"injection-0:94578)"
"injection-0:94579)"
"injection-0:94580)"
"injection-0:94581)"
"injection-0:94582)"
"injection-0:94583)"]
str = 7×1 string array
"injection-0:94577)" "injection-0:94578)" "injection-0:94579)" "injection-0:94580)" "injection-0:94581)" "injection-0:94582)" "injection-0:94583)"
newstr = extractBefore(extractAfter(str,":"),")")
newstr = 7×1 string array
"94577" "94578" "94579" "94580" "94581" "94582" "94583"

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Mahdi Khademishamami
编辑：Mahdi Khademishamami 2021-11-17
Thank you! What about converting the newstr to an array? Should I use str2num?
Paul
Paul 2021-11-17
Ran in:
Ran in:
I would just use double()
str = ["injection-0:94577)"
"injection-0:94578)"
"injection-0:94579)"
"injection-0:94580)"
"injection-0:94581)"
"injection-0:94582)"
"injection-0:94583)"];
newstr = extractBefore(extractAfter(str,":"),")");
d = double(newstr)
d = 7×1
94577 94578 94579 94580 94581 94582 94583

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