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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]);
Related
I have column in a df with a size range with different sizeformats.
artikelkleurnummer size
6725 0161810ZWA B080
6726 0161810ZWA B085
6727 0161810ZWA B090
6728 0161810ZWA B095
6729 0161810ZWA B100
in the sizerange are also these other size formats like XS - XXL, 36-50 , 36/38 - 52/54, ONE, XS/S - XL/XXL, 363-545
I have tried to get the prefix '0' out of all sizes with start with a letter in range (A:K). For exemple: Want to change B080 into B80. B100 stays B100.
steps:
1 look for items in column ['size'] with first letter of string in range (A:K),
2 if True change second position in string into ''
for range I use:
from string import ascii_letters
def range_alpha(start_letter, end_letter):
return ascii_letters[ascii_letters.index(start_letter):ascii_letters.index(end_letter) + 1]
then I've tried a for loop
for items in df['size']:
if df.loc[df['size'].str[0] in range_alpha('A','K'):
df.loc[df['size'].str[1] == ''
message
SyntaxError: unexpected EOF while parsing
what's wrong?
You can do it with regex and the pd.Series.str.replace -
df = pd.DataFrame([['0161810ZWA']*5, ['B080', 'B085', 'B090', 'B095', 'B100']]).T
df.columns = "artikelkleurnummer size".split()
replacement = lambda mpat: ''.join(g for g in mpat.groups() if mpat.groups().index(g) != 1)
df['size_cleaned'] = df['size'].str.replace(r'([a-kA-K])(0*)(\d+)', replacement)
Output
artikelkleurnummer size size_cleaned
0 0161810ZWA B080 B80
1 0161810ZWA B085 B85
2 0161810ZWA B090 B90
3 0161810ZWA B095 B95
4 0161810ZWA B100 B100
TL;DR
Find a pattern "LetterZeroDigits" and change it to "LetterDigits" using a regular expression.
Slightly longer explanation
Regexes are very handy but also hard. In the solution above, we are trying to find the pattern of interest and then replace it. In our case, the pattern of interest is made of 3 parts -
A letter in from A-K
Zero or more 0's
Some more digits
In regex terms - this can be written as r'([a-kA-K])(0*)(\d+)'. Note that the 3 brackets make up the 3 parts - they are called groups. It might make a little or no sense depending on how exposed you have been to regexes in the past - but you can get it from any introduction to regexes online.
Once we have the parts, what we want to do is retain everything else except part-2, which is the 0s.
The pd.Series.str.replace documentation has the details on the replacement portion. In essence replacement is a function that takes all the matching groups as the input and produces an output.
In the first part - where we identified three groups or parts. These groups are accessed with the mpat.groups() function - which returns a tuple containing the match for each group. We want to reconstruct a string with the middle part excluded, which is what the replacement function does
sizes = [{"size": "B080"},{"size": "B085"},{"size": "B090"},{"size": "B095"},{"size": "B100"}]
def range_char(start, stop):
return (chr(n) for n in range(ord(start), ord(stop) + 1))
for s in sizes:
if s['size'][0].upper() in range_char("A", "K"):
s['size'] = s['size'][0]+s['size'][1:].lstrip('0')
print(sizes)
Using a List/Dict here for example.
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 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)
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
)
I'm trying to emulate a function in SQL that a client has produced in Excel. In effect, they have a unique, 10-digit numeric value (VARCHAR) as the primary key in one of their enterprise database systems. Within another database, they require a unique, 5-digit alphanumeric identifier. They want that 5-digit alphanumeric value to be a representation of the 10-digit number. So what they did in excel was to split the 10-digit number into pairs, then convert each of those pairs into a hexadecimal value, then stitch them back together.
The EXCEL equation is:
=IF(VALUE(MID(A2,1,4))>0,DEC2HEX(VALUE(MID(A2,3,2)))&DEC2HEX(VALUE(MID(A2,5,2)))&DEC2HEX(VALUE(MID(A2,7,2)))&DEC2HEX(VALUE(MID(A2,9,2))),DEC2HEX(VALUE(MID(A2,5,2)))&DEC2HEX(VALUE(MID(A2,7,2)))&DEC2HEX((VALUE(MID(A2,9,2)))))
I need the SQL equivalent of this. Of course, should someone out there know a better way to accomplish their goal of "a 5-digit alphanumeric identifier" based off the 10-digit number, I'm all ears.
ADDED 8/2/2011
First of all, thank you to everyone for the replies. Nice to see folks willing to help and even enjoying it! Based on all the responses, I'm apt to tell my client they're intent is sound, only their method is off kilter. I'd also like to recommend a solution. So the challenge remains, just modified slightly:
CHALLENGE: Within SQL, take a 10 digit, unique NUMERIC string and represent it ALPHANUMERICALLY in as few characters as possible. The resulting string must also be unique.
Note that the first 3-4 characters in the 10-digit string are likely to be zeros, and that they could be stripped to shorten the resulting alphanumeric string. Not required, but perhaps helpful.
This problem is inherently impossible. You have a 10 digit numeric value that you want to convert to a 5 digit alphanumeric value. Since there are 10 numeric characters, this means that there are 10^10 = 10 000 000 000 unique values for your 10 digit number. Since there are 36 alphanumeric characters (26 letters + 10 numbers), there are 36^5 = 60 466 176 unique values for your 5 digit number. You cannot map a set of 10 billion elements into a set with around 60 million.
Now, lets take a closer look at what your client's code is doing:
So what they did in excel was to split the 10-digit number into pairs, then convert each of those pairs into a hexadecimal value, then stitch them back together.
This isn't 100% accurate. The excel code never uses the first 2 digits, but performs this operation on the remaining 8. There are two main problems with this algorithm which may not be intuitively obvious:
Two 10 digit numbers can map to the same 5 digit number. Consider the numbers 1000000117 and 1000001701. The last four digits of 1000000117 get mapped to 1 11, where the last four digits of 1000001701 get mapped to 11 1. This causes both to map to 00111.
The 5 digit number may not even end up being 5 digits! For example, 1000001616 gets mapped to 001010.
So, what is a possible solution? Well, if you don't care if that 5 digit number is unique or not, in MySQL you can use something like:
hex(<NUMERIC VALUE> % 0xFFFFF)
The log of 10^10 base 2 is 33.219280948874
> return math.log(10 ^ 10) / math.log(2)
33.219280948874
> = 2 ^ 33.21928
9999993422.9114
So, it takes 34 bits to represent this number. In hex this will take 34/4 = 8.5 characters, much more than 5.
> return math.log(10 ^ 10) / math.log(16)
8.3048202372184
The Excel macro is ignoring the first 4 (or 6) characters of the 10 character string.
You could try encoding in base 36 instead of 16. This will get you to 7 characters or less.
> return math.log(10 ^ 10) / math.log(36)
6.4254860446923
The popular base 64 encoding will get you to 6 characters
> return math.log(10 ^ 10) / math.log(64)
5.5365468248123
Even Ascii85 encoding won't get you down to 5.
> return math.log(10 ^ 10) / math.log(85)
5.1829075929158
You need base 100 to get to 5 characters
> return math.log(10 ^ 10) / math.log(100)
5
There aren't 100 printable ASCII characters, so this is not going to work, as zkhr explained as well, unless you're willing to go beyond ASCII.
I found your question interesting (although I don't claim to know the answer) - I googled a bit for you out of interest and found this which may help you http://dpatrickcaldwell.blogspot.com/2009/05/converting-decimal-to-hexadecimal-with.html