(Translated from German to Englisch)
I need help in this exercise :
Thread: String processing The user can make simple changes to an input sentence.
conditions
The program displays a menu for the user to select the following action. This is also displayed again after the action has been completed until the user terminates the program (a loop is therefore required).
The menu contains the following items, which should be executed when the specified letter is entered:
A. Enter the sentence
B. Determine the number of words
C. Determine the number of characters that are less than their sequence character
D. Replace all the words in the sentence with their uppercase initials
X. end
If the user enters a different letter, nothing happens or the menu is output again.
If the menu item A is selected, a prompt is issued to enter a set which is read into a string variable. This variable can not be changed by the actions of menu items B, C and D! Possibly. A copy of the set has to be prepared beforehand in another string variable.
In menu point B the number of all words in the block is to be counted. For simplicity, you can assume that there is always one space between two words. At the beginning and end of the sentence there are no spaces. The number of words is output after the calculation (e.g., "The set is 4 words").
If the user executes menu item C, the set is traversed character-by-character, and for each character it is checked whether it is smaller than its trailing character. Here is a simple character comparison (you can also write directly something like '1' <'d'). The number of characters so found is then output (e.g., "13 characters found in the sentence less than the trailing character").
In menu item D, the sentence is traversed and every word contained in it is replaced by its upper-case initial character. The capitalization is of course only made if the first character is a letter, otherwise the character remains unchanged. You can assume that the sentence never starts or ends with a space. Between two words there is always exactly one space and so it should be between the initial letters. For example, from "123 good mood" becomes "1 G L".
It is not permissible here to build up a completely new string piece by piece! Instead, you should work in a loop on a copy of the original sentence with pos, copy, length, delete and insert! It is also forbidden to "gather" the initial characters all at the beginning or end of the string; These should be inserted directly into the string at the position of the corresponding word!
Furthermore, a string can not be accessed at menu point D, because the work with string routines is to be practised explicitly here. Menu items B, C and D may only be selectable if a record has already been entered. Otherwise nothing happens or a fault message is entered when entering B, C or D in the menu and the menu is output again.
Each call to the menu items B, C or D will always work on the original set entered by the user and not on a set that has already been altered by previously executed menu items!
By entering the menu item A again, the entered block can be overwritten by a new one.
With an 'X' the user can terminate the program.
Use wherever it is the predefined string functions and do not write it yourself with difficulty loops, etc.! However, the use of the strreplace or reverseString functions is forbidden!
Here's my work till now, I only have problems with part D:
program Project2;
{$APPTYPE CONSOLE}
{$R *.res}
uses
System.SysUtils;
const
lz = ' ';
var
Satz: string;
Buchstabe: char;
i, p, j, zaehler2, index, count: integer;
writeln('A) Write Sentence');
readln(Satz);
'D':
begin
index := 2;
insert(lz, Satz, length(Satz)+1);
count := (pos(lz,Satz));
repeat
delete(Satz, index,(count - index));
index := index + 2;
count := pos(lz,copy(Satz,index,(length(Satz)-index)))+index-1;
until ;
writeln(uppercase(Satz));
end
I'm glad you've found your own solution, well done!
While you've been doing that, I've been writing the answer below and, as I
have finished it, I thought I'd post it here to show you another way
to go about the problem of extracting words from a string. There are
dozens of ways of doing that task, but I hope the one I've used is fairly
easy to follow.
Maybe the reason you were having a problem with this is that your string-indexing
expressions are a bit too complicated. I bet if you come back to your code in 6 months
it will take you a while to figure out what it is supposed to be doing and longer
to tell whether it is actually doing it. The key to avoiding problems like that
is to break your code up into chucks which are easier to follow and easier
to test. So instead of just telling you what your repeat condition should be,
I'll show you a way which is easier to follow.
The first thing to do is to extract a singe word from the input. So, first thing
I've written is a function, ExtractFirstWord which returns the first word in
the input string, whether or not the input includes spaces, and also returns
a Remainder string which is what is left or the input string after the first
word (and any spaces immediately following it have been removed). This is done
using some simple while loops which are coded to skip over the leading spaces
and then build a string from the non-space characters which follow.
Code
const
lz = ' ';
var
Satz: string;
FirstWord : String;
Remainder : String;
function ExtractFirstWord(const InputStr : String; var Remainder : String) : String;
var
P : Integer;
WordStart : Integer;
begin
Result := '';
P := 1;
// The following skips over any spaces at the start of InputStr
while (P <= Length(InputStr)) and (InputStr[P] = lz) do
Inc(P);
// Now we know where the first word starts
WordStart := P;
// Now we can get the first word, if there is one
while (P <= Length(InputStr)) and (InputStr[P] <> lz) do
Inc(P);
Result := Copy(InputStr, WordStart, P - WordStart);
Remainder := Copy(InputStr, P, Length(InputStr));
// the following is one way to remove spaces at the start of Remainder
while (Length(Remainder) > 0) and (Remainder[1] = lz) do
Delete(Remainder, 1, Length(lz));
// instead you could do something simlar to the first `while` loop above
end;
begin
Satz := ' cat dog ';
repeat
FirstWord := ExtractFirstWord(Satz, Remainder);
FirstWord := UpperCase(FirstWord);
Satz := Remainder;
writeln('First word: ', FirstWord, ' remainder: ', Remainder);
until Remainder = '';
readln;
end.
This particular way of doing it is not an ideal fit with the other requirements
specified in your task but should be easily adaptable to them. E.g, the upper-casing of words could be done "in place" on the input string by upper-casing the current character of it in the second While loop.
Btw, if you are using Delphi or Free Pascal/Lazarus, there is a much simpler
way of extracting the words from a string. It uses a TStringList. Try
looking it up in the online help and have a thing about how you might use it
to do the task.
Related
If I create a subprogram of type function that for instance orders you to type a string of a particular length and you type Overflow, it's supposed to type the last half of the string, so in this case it would be flow. But on the other end if I type an odd number of characters like Stack it's supposed to type the last half of the string + the middle letter, so in this case it would be "ack".
Let me make it clearer (text in bold is user input):
Type a string that's not longer than 7 characters: Candy
The other half of the string is: ndy
with Ada.Text_IO; use Ada.Text_IO;
with Ada.Integer_Text_IO; use Ada.Integer_Text_IO;
function Split_String (S : in String) return String is
begin
Mid := 1 + (S'Length / 2);
return S(Mid .. S'Last);
end Split_String;
S : String(1 .. 7);
I : Integer;
begin
Put("Type a string that's no longer than 7 characters: ");
Get_Line(S, I);
Put(Split_String(S));
end Split;
Let me tell you how I've been thinking. So I do a Get_Line to see how many characters the string contains. I then put I in my subprogram to determine if its evenly dividable by two or not. If it's dividable by two, the rest should be 0, thus it'll mean that typing out the other half of the string + THE MIDDLE CHARACTER is not needed. If in all the other cases, it's not dividable by two I have to type out the other half of the string + the middle character. But now I stumbled upon a big problem in my main program. I don't know how type out the other half of a string. If a string contains 4 words I can just type out Put(S(3 .. 4); but the thing is that I don't know a general formula for this. Help is appreciated! :) Have a good day!
You need a more general approach to your problem. Also, try to understand how Get_Line works for you.
For example, if you declare an input string with a large size such as
Input : String (1..1024);
You will have a string large enough to work with any likely input values.
Next, you need a variable to indicate how many characters were actually read by Get_Line.
Length : Natural;
The data returned by Get_Line will then be in the slice of the input string designated as
Input (1 .. Length);
Pass that slice to your function to return the second half of the string.
function last_half(S : string) return string;
last_half(Input(1..Length));
Now all you need is to calculate the last half of the string passed to the function last_half. The function will output a slice of the string passed to it. To find the first index of the last half of the input string you must perform the calculation
mid : Positive := 1 + (S'length / 2);
Then simply return the string S(mid .. S'Last).
It appears that the goal of this exercise is to learn how to use array slices. Concentrate on how slices work for you in the problem and the solution will be very simple.
One possible solution is
with Ada.Text_IO; use Ada.Text_IO;
procedure Main is
Input : String (1 .. 1_024);
Length : Natural;
function last_half (S : in String) return String is
Mid : Positive := 1 + (S'Length / 2);
begin
return S (Mid .. S'Last);
end last_half;
begin
Put ("Enter a string: ");
Get_Line (Input, Length);
Put_Line (Input (1 .. Length) & " : " & last_half (Input (1 .. Length)));
end Main;
Study how the solution uses array slices on the return value of Get_Line and on the parameter for the function last_half and on its return statement. It is also important to remember that the type String is defined as an unbounded array of character. This means that every slice of a string is also a string.
type String is array ( Positive range <> ) of Character;
Aside from being an untidy mess, your latest code edit (as of 20:11 GMT on 15 Nov 2021) doesn’t even compile. Please don’t show us code like this! (unless, of course, that’s the problem).
I’d like to strongly suggest this alternate way of inputting strings:
declare
S : constant String := Get_Line;
begin
-- do things with S, which is exactly as long as
-- the input you typed: no undefined characters at
-- the end to confuse the result, no need to worry
-- about overrunning an input buffer
end;
With this change, and obvious syntactic changes, your current code will do what you want.
I need a code on AHK
I have a variable look like this:
CYOMYACHOAYJGUGRYYQNYB
I need to get this:
YMAHAJURYNB
I meen, i need every second char from a variable. Thank in advance
Var := "CYOMYACHOAYJGUGRYYQNYB"
Loop, Parse, Var ; retrieves each character from the variable, one at a time
{
If (Mod(A_Index, 2) = 0) ; If A_Index is even (the remainder after division by 2 is 0)
NewVar .= A_LoopField ; add the retrieved character to the new variable
}
MsgBox %NewVar%
This works for me. I am using bit wise to determine if the index of the array of letters, given to me by StrSplit(TestString), is even or odd as I loop through them. I used this forum post for the bitwise logic. Then I concatenate if the line is even. So if index&1=0 will be true when the number is even, thus giving me every other letter to concatenate into NewString with this line NewString=%NewString%%letter%. Feel free to uncomment out the message box lines by removing the ; to better see how the loop parses the array.
TestString := "ABCD"
word_array := StrSplit(TestString)
NewString:=""
For index, letter in word_array{
if index&1=0
{
;MsgBox, %letter% added
NewString=%NewString%%letter%
;Msgbox, %NewString%
}
}
MsgBox, %NewString%
As you don't specify any language, I'll answer in pseudocode:
set counter to 1
set result to empty string
for every char in string:
if counter is even:
append char to result
increment counter by 1
user3419297 beat me to it, but mine is even easier:
InputBox, x, What Variable?, Enter Variable:
loop, % StrLen(x)
If mod(A_Index,2)=0
y.=substr(x,A_Index,1)
msgbox %y%
Clipboard := y
You input the variable in a dialog, and the result is shown, and put in clipboard. Hth,
EDIT: I like the bitwise logic from Zack Tarr! Substitute for the "if" above:
If A_Index&1=0
The rest is the same.
I work at a doctors office doing billing, and I've run into a problem. Because of an issue with the way the EMR (electronic medical record) program works, immunization codes don't always make it into the chart. In order to compensate for this, I have AHK also search for the name of the immunization (dangerous, I know). Just until recently, this was working fine, until one of the doctors canceled an order for an immunization. This threw a false positive because of the failsafe I included for the charts without the code (IMM94, and IMM97). I need to remove, from a very large string, everything after the word "CANCELED: " to the first line break that occurs after that word. I created a rather pretty string clipping function to do most of this kind of stuff, but it's taking issue with finding the line break because there are so many. The test case I'm working with is:
Assessment and Plan:
Need for prophylactic vaccination and inoculation against influenza (tab) CANCELED: Influenza, High Dose (65+) (newline)
Prediabetes Work on weight loss and exercise
Osteopenia Exercise and vit d and b 12
There's a lot of stuff both above this, and below this, but I want to extract ALL instances from my string of the word "CANCELED: " to the first line break after it.
This is what I'm trying, along with my function. It's not working, and in fact, it's duplicating the line beginning with "1. ":
^L::
lString := ClipBoard
If lString Contains Canceled
fString := ClipString(lString, "CANCELED: ", "`n")
Clipboard := fString
Return
ClipString(lString, aMarker, bMarker, Only := 0)
{
If (Only = 1)
Return SubStr(SubStr(lString, 1, InStr(lString, bMarker)+StrLen(bMarker)-1), InStr(lString, aMarker))
Else
Return SubStr(lString, 1, InStr(lString, aMarker)-StrLen(aMarker)) . SubStr(lString, InStr(lString, bMarker)+StrLen(bMarker))
}
Strips "cancelled:" and any text following it from each line in clipboard
Program:
Clipboard := RegExReplace(Clipboard, "im)\s*CANCELLED:.*\R?", "`r`n")
Sample Input:
alpha
beta cancelled: ABC asldkfalsd
delta
gamma omega CANCELLed: zeta
theta
Sample Output:
alpha
beta
delta
gamma omega
theta
Got it figured out. I'll comment it to help anyone who has this issue. I love how compact you can get functions usually, but this piece really needed a couple more lines than I'd like. If anyone can get it down to less, and have it still work, let me know!
^L::
ClipBoard := ClipString(ClipBoard, "Canceled: ", "`n",2) ;This is the function call here.
;The first parameter is the string to do stuff to, the second is what to cut out of the
;string, and the third parameter is how far after that piece to cut, so if you wrote
;Canceled: thenawholebunchofstuffthena newline, it'd remove all of it except for the
;newline. The third parameter is which mode to put ClipString to.
Return
ClipString(lString, aMarker, bMarker, Mode := 0) ;already explained above.
;Mode defaults to zero because I call the zero mode a lot.
{
If (Mode = 0) ;Mode Zero returns the ONLY the section from the original string
;between aMarker and bMarker.
Return SubStr(lString, 1, InStr(lString, aMarker)-StrLen(aMarker)) . SubStr(lString, InStr(lString, bMarker)+StrLen(bMarker))
Else If (Mode = 1) ;Mode One returns the original string with the section between aMarker
;and bMarker removed.
Return SubStr(SubStr(lString, 1, InStr(lString, bMarker)+StrLen(bMarker)-1), InStr(lString, aMarker))
Else If (Mode = 2) ;Mode Two returns the original string with all instances of aMarker
;to bMarker removed. I.E. Every occurrence of aMarker will be removed to bMarker.
{
aString := lString
StringReplace, aString, aString, %aMarker%, %aMarker%, UseErrorLevel ;Count how many
;instances of aMarker are in the original string.
CanCnt := Errorlevel ;actual count of # of aMarkers
Loop, %CanCnt% ;loop as many times as there are aMarkers
{
aString := SubStr(aString, 1, InStr(aString, aMarker)-1) . SubStr(SubStr(aString, InStr(aString, aMarker)), InStr(SubStr(aString, InStr(aString, aMarker)), bMarker)-1)
;this is a tad complicated. The first part before the concatenate takes the string, and
;keeps from the start point to the first aMarker. The concatenate adds the substring of the
;substring between aMarker and the end of the string, and the location of the substring in
;aString from the occurrence of aMarker to one position before the start of bMarker.
;An easier way to read this would be to replace "SubStr(aString, InStr(aString, aMarker))
;with a variable like bString. It shortens it up quite a bit.
}
Return aString
}
}
Extracting a specific word and a number of tokens on each side of it from each string in a column in SAS EG ?
For example,
row1: the sun is nice
row2: the sun looks great
row3: the sun left me
Is there a code that would produce the following result column (2 words where sun is the first):
SUN IS
SUN LOOKS
SUN LEFT
and possibly a second column with COUNT in case of duplicate matches.
So if there was 20 SUN LOOKS then it they would be grouped and have a count of 20.
Thanks
I think you can use functions findw() and scan() to do want you want. Both of those functions operate on the concept of word boundaries. findw() returns the position of the word in the string. Once you know the position, you can use scan() in a loop to get the next word or words following it.
Here is a simple example to show you the concept. It is by no means a finished or polished solution, but intended you point you in the right direction. The input data set (text) contains the sentences you provided in your question with slight modifications. The data step finds the word "sun" in the sentence and creates a variable named fragment that contains 3 words ("sun" + the next 2 words).
data text2;
set text;
length fragment $15;
word = 'sun'; * search term;
fragment_len = 3; * number of words in target output;
word_pos = findw(sentence, word, ' ', 'e');
if word_pos then do;
do i = 0 to fragmen_len-1;
fragment = catx(' ', fragment, scan(sentence, word_pos+i));
end;
end;
run;
Here is a partial print of the output data set.
You can use a combination of the INDEX, SUBSTR and SCAN functions to achieve this functionality.
INDEX - takes two arguments and returns the position at which a given substring appears in a string. You might use:
INDEX(str,'sun')
SUBSTR - simply returns a substring of the provided string, taking a second numeric argument referring to the starting position of the substring. Combine this with your INDEX function:
SUBSTR(str,INDEX(str,'sun'))
This returns the substring of str from the point where the word 'sun' first appears.
SCAN - returns the 'words' from a string, taking the string as the first argument, followed by a number referring to the 'word'. There is also a third argument that specifies the delimiter, but this defaults to space, so you wouldn't need it in your example.
To pick out the word after 'sun' you might do this:
SCAN(SUBSTR(str,INDEX(str,'sun')),2)
Now all that's left to do is build a new string containing the words of interest. That can be achieved with concatenation operators. To see how to concatenate two strings, run this illustrative example:
data _NULL_;
a = 'Hello';
b = 'World';
c = a||' - '||b;
put c;
run;
The log should contain this line:
Hello - World
As a result of displaying the value of the c variable using the put statement. There are a number of functions that can be used to concatenate strings, look in the documentation at CAT,CATX,CATS for some examples.
Hopefully there is enough here to help you.
I need to find out how many time one word appears in the string, but the catch is that the word you need to find can have spaces in between, for example you want to see how many times word text appears in *tOeOxOt" and it would give you output 1, or for example in textt it would give you output 2, I have written this procedure in pascal for this
procedure search(x:integer; i:integer);
var
x2:integer;
begin
x2:=x+1;
while (x2<=n) and (x2>0) do begin
if myarray[x2]=mystring[i+1] then
if i=length(mystring)-1 then
final:=final+1
else
search(x2,i+1);
x2:=x2+1;
end;
end;
and it checks number of time it appears from one letter, for example if I have ttext it would only give me one because I only check from the first t so I call the function every time I find a t in the string, but this method is too slow for 2D arrays with many characters, like 1000x1000 so I am looking for a faster solution.
You could check the array twice, on the first run trough it remove all spaces.
On the second one use a compare function like this(x is the array in which you search, y is the sub-string you are searching for and i is the current element you are checking):
function compare(var x,y:myarray; i:integer):boolean;
var l:integer;
Begin
compare:=false;
for l:=1 to length(y) do Begin
if x[i+l] <> y[l] then Exit;
End;
compare:=true;
End;
on each element of your array.