How can I read each character in a String? For example, I want to read each character in String "a7m4d0". After that I want to verify that each character is a character or a number. Any tips or ideas?
DATA: smth TYPE string VALUE `qwert1yua22sd123bnm,`,
index TYPE i,
length TYPE i,
char TYPE c,
num TYPE i.
length = STRLEN( smth ).
WHILE index < length.
char = smth+index(1).
TRY .
num = char.
WRITE: / num,'was a number'.
CATCH cx_sy_conversion_no_number.
WRITE: / char,'was no number'.
ENDTRY.
ADD 1 TO index.
ENDWHILE.
Here's your problem solved :P
A bit convoluted and on a recent 740 ABAP server. :)
DATA: lv_text TYPE string VALUE `a7m4d0`.
DO strlen( lv_text ) TIMES.
DATA(lv_single) = substring( val = lv_text off = sy-index - 1 len = 1 ) && ` is ` &&
COND string( WHEN substring( val = lv_text off = sy-index - 1 len = 1 ) CO '0123456789' THEN 'Numeric'
ELSE 'Character' ).
WRITE : / lv_single.
ENDDO.
Here is how you can access a single character within a string:
This example will extract out the character "t" into the variable "lv_char1".
DATA: lv_string TYPE char10,
lv_char TYPE char1.
lv_string = "Something";
lv_char1 = lv_string+4(1).
Appending "+4" to the string name specifies the offset from the start of the string (in this case 4), and "(1)" specifies the number of characters to pick up.
See the documentation here for more info:
http://help.sap.com/saphelp_nw04/Helpdata/EN/fc/eb341a358411d1829f0000e829fbfe/content.htm
If you want to look at each character in turn, you could get the length of the field using "strlen( )" and do a loop for each character.
One more approach
PERFORM analyze_string USING `qwert1yua22sd123bnm,`.
FORM analyze_string USING VALUE(p_string) TYPE string.
WHILE p_string IS NOT INITIAL.
IF p_string(1) CA '0123456798'.
WRITE: / p_string(1) , 'was a number'.
ELSE.
WRITE: / p_string(1) , 'was no number'.
ENDIF.
p_string = p_string+1.
ENDWHILE.
ENDFORM.
No DATA statements, string functions or explicit indexing required.
I know the post it's old but this might be useful, this is what use :)
DATA lv_counter TYPE i.
DO STRLEN( lv_word ) TIMES.
IF lv_word+lv_counter(1) CA '0123456789'
"It's a number
ENDIF.
lv_counter = lv_counter + 1.
ENDDO.
Related
if you look at the pseudo code, i am trying to make a new string without certain elements.
thesentence = 'i need help!*'
bettersentence = ''.join([char for char in thesentence if char != '!' and '*'])
print(bettersentence)
comparing a character with two strings at the same time doesnt work. But im wondering wether there isnt any easy approach to this?
You can turn your string into a list and then use a for loop to see which characters in the string equal to * or ! and replace them with an empty string
def bettersentence(text):
text = list(text)
for i, character in enumerate(text):
if character == '*' or character == '!':
text[i] = ''
return ('').join(text)
How can I, in ABAP, split a string into n parts AND determine which one is the biggest element? In my solution I would need to know how many elements there are, but I want to solve it for WHATEVER NUMBER of elements.
I tried the below code. And i searched the web.
DATA: string TYPE string VALUE 'this is a string'.
DATA: part1 TYPE c LENGTH 20.
DATA: part2 TYPE c LENGTH 20.
DATA: part3 TYPE c LENGTH 20.
DATA: part4 TYPE c LENGTH 20.
DATA: del TYPE c VALUE ' '.
DATA: bigger TYPE c LENGTH 20.
split: string AT del INTO part1 part2 part3 part4.
bigger = part1.
IF bigger > part2.
bigger = part1.
ELSEIF bigger > part3.
bigger = part2.
ELSE.
bigger = part4.
ENDIF.
WRITE: bigger.
Expected result: Works with any number of elements in a string and determines which one is biggest.
Actual result: I need to know how many elements there are
Here is one way to solve it:
DATA: string TYPE string VALUE 'this is a string'.
TYPES: BEGIN OF ty_words,
word TYPE string,
length TYPE i,
END OF ty_words.
DATA: ls_words TYPE ty_words.
DATA: gt_words TYPE STANDARD TABLE OF ty_words.
START-OF-SELECTION.
WHILE string IS NOT INITIAL.
SPLIT string AT space INTO ls_words-word string.
ls_words-length = strlen( ls_words-word ).
APPEND ls_words TO gt_words.
ENDWHILE.
SORT gt_words BY length DESCENDING.
READ TABLE gt_words
ASSIGNING FIELD-SYMBOL(<ls_longest_word>)
INDEX 1.
IF sy-subrc EQ 0.
WRITE: 'The longest word is:', <ls_longest_word>-word.
ENDIF.
Please note, it does not cover the case if there are more longest words with the same length, it will just show one of them.
You don't need to know the number of splitted parts if you split the string into an array. Then you LOOP over the array and check the string length to find the longest one.
While József Szikszai's solution works, it may be too complex for the functionality you need. This would work just as well: (also with the same limitation that it willl only output the first longest word and no other ones of the same length)
DATA string TYPE string VALUE 'this is a string'.
DATA parts TYPE STANDARD TABLE OF string.
DATA biggest TYPE string.
FIELD-SYMBOLS <part> TYPE string.
SPLIT string AT space INTO TABLE parts.
LOOP AT parts ASSIGNING <part>.
IF STRLEN( <part> ) > STRLEN( biggest ).
biggest = <part>.
ENDIF.
ENDLOOP.
WRITE biggest.
Edit: I assumed 'biggest' meant longest, but if you actually wanted the word that would be last in an alphabet, then you could sort the array descending and just output the first entry like this:
DATA string TYPE string VALUE 'this is a string'.
DATA parts TYPE STANDARD TABLE OF string.
DATA biggest TYPE string.
SPLIT string AT space INTO TABLE parts.
SORT parts DESCENDING.
READ TABLE parts INDEX 1 INTO biggest.
WRITE biggest.
With ABAP 740, you can also shorten it to:
SPLIT lv_s AT space INTO TABLE DATA(lt_word).
DATA(lv_longest) = REDUCE string( INIT longest = `` FOR <word> IN lt_word NEXT longest = COND #( WHEN strlen( <word> ) > strlen( longest ) THEN <word> ELSE longest ) ).
DATA(lv_alphabetic) = REDUCE string( INIT alph = `` FOR <word> IN lt_word NEXT alph = COND #( WHEN <word> > alph THEN <word> ELSE alph ) ).
If "biggest" means "longest" word here is the Regex way to do this:
FIND ALL OCCURRENCES OF REGEX '\w+' IN string RESULTS DATA(words).
SORT words BY length DESCENDING.
WRITE substring( val = string off = words[ 1 ]-offset len = words[ 1 ]-length ).
I currently have to a code in ABAP which contains a String that has multiple words that start with Capital letters/Uppercase and there is no space in-between.
I have to separate it into an internal table like this:
INPUT :
NameAgeAddress
OUTPUT :
Name
Age
Address
Here is the shortest code I could find, which uses a regular expression combined with SPLIT:
SPLIT replace( val = 'NameAgeAddress' regex = `(?!^.)\u` with = ` $0` occ = 0 )
AT ` `
INTO TABLE itab.
So, replace converts 'NameAgeAddress' into 'Name Age Address' and SPLIT puts the 3 words into an internal table.
Details:
(?!^.) to say the next character to find (\u) should not be the first character
\u being any upper case letter
$0 to replace the found string ($0) by itself preceded with a space character
occ = 0 to replace all occurrences
Unfortunately, the SPLIT statement in ABAP does not allow a regex as separator expression. Therefore, we have to use progressive matching, which is a bit awkward in ABAP:
report zz_test_split_capital.
parameters: p_input type string default 'NameAgeAddress' lower case.
data: output type stringtab,
off type i,
moff type i,
mlen type i.
while off < strlen( p_input ).
find regex '[A-Z][^A-Z]*'
in section offset off of p_input
match offset moff match length mlen.
if sy-subrc eq 0.
append substring( val = p_input off = moff len = mlen ) to output.
off = moff + mlen.
else.
exit.
endif.
endwhile.
cl_demo_output=>display_data( output ).
Just for comparison, the following statement would do the job in Perl:
my $input = "NameAgeAddress";
my #output = split /(?=[A-Z])/, $input;
# gives #output = ('Name','Age','Address')
It is easy with using regular expressions. The solution could look like this.
REPORT ZZZ.
DATA: g_string TYPE string VALUE `NameAgeAddress`.
DATA(gcl_regex) = NEW cl_abap_regex( pattern = `[A-Z]{1}[a-z]+` ).
DATA(gcl_matcher) = gcl_regex->create_matcher( text = g_string ).
WHILE gcl_matcher->find_next( ).
DATA(g_match_result) = gcl_matcher->get_match( ).
WRITE / g_string+g_match_result-offset(g_match_result-length).
ENDWHILE.
For when regular expressions are just overkill and plain old ABAP will do:
DATA(str) = 'NameAgeAddress'.
IF str CA sy-abcde.
DATA(off) = 0.
DO.
data(tailstart) = off + 1.
IF str+tailstart CA sy-abcde.
DATA(len) = sy-fdpos + 1.
WRITE: / str+off(len).
add len to off.
ELSE.
EXIT.
ENDIF.
ENDDO.
write / str+off.
ENDIF.
If you do not want to use or cannot use Regex, here another solution:
DATA: lf_input TYPE string VALUE 'NameAgeAddress',
lf_offset TYPE i,
lf_current_letter TYPE char1,
lf_letter_in_capital TYPE char1,
lf_word TYPE string,
lt_word LIKE TABLE OF lf_word.
DO strlen( lf_input ) TIMES.
lf_offset = sy-index - 1.
lf_current_letter = lf_input+lf_offset(1).
lf_letter_in_capital = to_upper( lf_current_letter ).
IF lf_current_letter = lf_letter_in_capital.
APPEND INITIAL LINE TO lt_word ASSIGNING FIELD-SYMBOL(<ls_word>).
ENDIF.
IF <ls_word> IS ASSIGNED. "if input string does not start with capital letter
<ls_word> = <ls_word> && lf_current_letter.
ENDIF.
ENDDO.
I want to write a code in MATLAB that converts a letter into NATO alphabet. Such as the word 'hello' would be re-written as Hotel-Echo-Lima-Lima-Oscar. I have been having some trouble with the code. So far I have the following:
function natoText = textToNato(plaintext)
plaintext = lower(plaintext);
r = zeros(1, length(plaintext))
%Define my NATO alphabet
natalph = ["Alpha","Bravo","Charlie","Delta","Echo","Foxtrot","Golf", ...
"Hotel","India","Juliet","Kilo","Lima","Mike","November","Oscar", ...
"Papa","Quebec","Romeo","Sierra","Tango","Uniform","Victor",...
"Whiskey","Xray","Yankee","Zulu"];
%Define the normal lower alphabet
noralpha = ['a' : 'z'];
%Now we need to make a loop for matlab to check for each letter
for i = 1:length(text)
for j = 1:26
n = r(i) == natalph(j);
if noralpha(j) == text(i) : n
else r(i) = r(i)
natoText = ''
end
end
end
for v = 1:length(plaintext)
natoText = natoText + r(v) + ''
natoText = natoText(:,-1)
end
end
I know the above code is a mess and I am a bit in doubt what really I have been doing. Is there anyone who knows a better way of doing this? Can I modify the above code so that it works?
It is because now when I run the code, I am getting an empty plot, which I don't know why because I have not asked for a plot in any lines.
You can actually do your conversion in one line. Given your string array natalph:
plaintext = 'hello'; % Your input; could also be "hello"
natoText = strjoin(natalph(char(lower(plaintext))-96), '-');
And the result:
natoText =
string
"Hotel-Echo-Lima-Lima-Oscar"
This uses a trick that character arrays can be treated as numeric arrays of their ASCII equivalent values. The code char(lower(plaintext))-96 converts plaintext to lowercase, then to a character array (if it isn't already) and implicitly converts it to a numeric vector of ASCII values by subtracting 96. Since 'a' is equal to 97, this creates an index vector containing the values 1 ('a') through 26 ('z'). This is used to index the string array natalph, and these are then joined together with hyphens.
If my string is this: 'this is a string', how can I produce all possible combinations by joining each word with its neighboring word?
What this output would look like:
this is a string
thisis a string
thisisa string
thisisastring
thisis astring
this isa string
this isastring
this is astring
What I have tried:
s = 'this is a string'.split()
for i, l in enumerate(s):
''.join(s[0:i])+' '.join(s[i:])
This produces:
'this is a string'
'thisis a string'
'thisisa string'
'thisisastring'
I realize I need to change the s[0:i] part because it's statically anchored at 0 but I don't know how to move to the next word is while still including this in the output.
A simpler (and 3x faster than the accepted answer) way to use itertools product:
s = 'this is a string'
s2 = s.replace('%', '%%').replace(' ', '%s')
for i in itertools.product((' ', ''), repeat=s.count(' ')):
print(s2 % i)
You can also use itertools.product():
import itertools
s = 'this is a string'
words = s.split()
for t in itertools.product(range(len('01')), repeat=len(words)-1):
print(''.join([words[i]+t[i]*' ' for i in range(len(t))])+words[-1])
Well, it took me a little longer than I expected... this is actually tricker than I thought :)
The main idea:
The number of spaces when you split the string is the length or the split array - 1. In our example there are 3 spaces:
'this is a string'
^ ^ ^
We'll take a binary representation of all the options to have/not have either one of the spaces, so in our case it'll be:
000
001
011
100
101
...
and for each option we'll generate the sentence respectively, where 111 represents all 3 spaces: 'this is a string' and 000 represents no-space at all: 'thisisastring'
def binaries(n):
res = []
for x in range(n ** 2 - 1):
tmp = bin(x)
res.append(tmp.replace('0b', '').zfill(n))
return res
def generate(arr, bins):
res = []
for bin in bins:
tmp = arr[0]
i = 1
for digit in list(bin):
if digit == '1':
tmp = tmp + " " + arr[i]
else:
tmp = tmp + arr[i]
i += 1
res.append(tmp)
return res
def combinations(string):
s = string.split(' ')
bins = binaries(len(s) - 1)
res = generate(s, bins)
return res
print combinations('this is a string')
# ['thisisastring', 'thisisa string', 'thisis astring', 'thisis a string', 'this isastring', 'this isa string', 'this is astring', 'this is a string']
UPDATE:
I now see that Amadan thought of the same idea - kudos for being quicker than me to think about! Great minds think alike ;)
The easiest is to do it recursively.
Terminating condition: Schrödinger join of a single element list is that word.
Recurring condition: say that L is the Schrödinger join of all the words but the first. Then the Schrödinger join of the list consists of all elements from L with the first word directly prepended, and all elements from L with the first word prepended with an intervening space.
(Assuming you are missing thisis astring by accident. If it is deliberately, I am sure I have no idea what the question is :P )
Another, non-recursive way you can do it is to enumerate all numbers from 0 to 2^(number of words - 1) - 1, then use the binary representation of each number as a selector whether or not a space needs to be present. So, for example, the abovementioned thisis astring corresponds to 0b010, for "nospace, space, nospace".