I am trying to mask SSn and want show it on label caption.
lblSPTINTo.Caption = rsMM("SPTIN")
lblCPTINTo.Caption = rsMM("CPTIN")
i am trying to use substring function to get last 4 characters but i not am to able to use it as it throws compile error .
lblSPTINTo.Caption = rsMM("SPTIN").sutbstring(4,4)
Replace sutbstring with Substring.
But it won't work that way because the first parameter is the index and the second parameter in Substring is the length, if you want the last 4 characters:
Dim last4 As String = rsMM("SPTIN")
If last4.Length > 4 Then last4 = last4.Substring(last4.Length - 4)
Related
I have this dataframe:
p=pd.DataFrame({'text':[2,'string']})
and trying to replace digit 2 by an 'a' using this code:
p['text']=p['text'].str.replace('\d+', 'a')
But instead of letter a and get NaN?
What am I doing wrong here?
In your dataframe, the first value of the text column is actually a number, not a string, thus the NaN error when you try to call .str. Just convert it to a string first:
p['text'] = p['text'].astype(str).str.replace('\d+', 'a')
Output:
>>> p
text
0 a
1 string
(Note that .str.replace is soon going to change the default value of regex from True to False, so you won't be able to use regular expressions without passing regex=True, e.g. .str.replace('\d+', 'a', regex=True))
1 - 2 of 2
Above is my text. This is from paging of a web application. How do i extract the last number of the above text. SO i will get the count of list in that page and i can run a loop with respect to the number.
You can use substring
Let's consider your example. You have a String 1 - 2 of 2 (pagination probably)
Each of individual character is a specified index of a String
1 = 0
space = 1
- = 2
space = 3
etc.
String has a set of methods to perform various tasks. One of them is length() which gives you number of characters in your String
What you can do is to pass your length of String to substring.
Example:
myString.substring(0,1) will give you results of 1
myString.substring(0,myString.length()) wil give you results of 1 - 2 of 5
Additional info: myString.length() is an int type so you can perform math operations like + or -
myString.substring(0,myString.length()-1) will give you results of 1 - 2 of
I gave you the tools, now it's time for you to find the solutions.
You could just split the string using the spaces and then grab the last element of the split array. That should cover you even if the last number has more than one digit. Throw in a trim, just in case, to remove any leading/trailing white space.
String[] splitter = pageCount.trim().split(" ");
System.out.println(splitter[splitter.length - 1]);
if we write 12wkd3, how to choose/filter 123 as integer in octave?
example in octave:
A = input("A?\n")
A?
12wkd3
A = 123
while 12wkd3 is user keyboard input and A = 123 is the expected answer.
assuming that the general form you're looking for is taking an arbitrary string from the user input, removing anything non-numeric, and storing the result it as an integer:
A = input("A? /n",'s');
A = int32(str2num(A(isdigit(A))));
example output:
A?
324bhtk.p89u34
A = 3248934
to clarify what's written above:
in the input statement, the 's' argument causes the answer to get stored as a string, otherwise it's evaluated by Octave first. most inputs would produce errors, others may be interpreted as functions or variables.
isdigit(A) produces a logical array of values for A with a 1 for any character that is a 0-9 number, and 0 otherwise.
isdigit('a1 3 b.') = [0 1 0 1 0 0 0]
A(isdigit(A)) will produce a substring from A using only those values corresponding to a 1 in the logical array above.
A(isdigit(A)) = 13
that still returns a string, so you need to convert it into a number using str2num(). that, however, outputs a double precision number. so finally to get it to be an integer you can use int32()
I'm trying to simulate an algoritham in cryptography and I need to convert a string of 0s and 1s back into a word. Example:
I have: 01011110010101101000001101100001101
I have split it into an array of strings:
0101111, 0010101, ...
each member has 7 characters. I want to get a letter that 0101111 represents in UTF8? How do I do this?
I try CType("0010101", Byte), but it fails. I can pass max 111 this way.
Help :/
UTF-8 is 8 bit, those are only 7 bits. Do you mean 7 bit ASCII?
In that case here you go:
Function BinToStr(binStr As String) As String
Dim i As Long
For i = 0 To (Len(binStr) / 7) - 1
[A1] = CLng(Mid(binStr, i * 7 + 1, 7))
BinToStr = BinToStr & Chr([BIN2DEC(A1)])
Next
End Function
If that's not what you're looking for, let me know.
I need to count how often a number is present in a string. it should count EVERY occurence with a whitespace in front, except those followed by a =.
For example:
If i need to know how many "1" there are in this string: this is a 1 ramdnom string with 2 numbers 1 with 1=something it should return 2, as the third one is followed by an =
To find the occurrences I am using this: occurences = mystring.Split(" 1").Length - 1
But how to exclude those followed by a =?
Thanks
Something like,
Dim occurrences = Regex.Matches(yourString, "\W[0-9]([^=]|$)").Count
If you'd like to do replacements, use a Regex.Replace overload.
Breaking it down, this expression matches
\W // any whitespace character
[0-9] // any deciaml digit
( // either
[^=] // not =
| // or
$ // the end of the string
)