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Calculate string value in javascript, not using eval
(12 answers)
Closed 4 months ago.
When the text was '2+3+5+1', the logic was easy
Split('+') so the string is converted to an array.
loop over the array and calculate the sum.
check the code below
void main() {
const text = '2+3+5+1';
final array = text.split('+');
int res =0;
for (var i=0; i<= array.length -1; i++){
res+=int.parse(array[i]);;
}
print(array);
print(res);
}
Now this String "2+3-5+1" contains minus.
how to get the right response using split method?
I am using dart.
note: I don't want to use any library (math expression) to solve this exercice.
Use the .replace() method.
text = text.replace("-", "+-");
When you run through the loop, it will calculate (-).
You can split your string using regex text.split(/\+|\-/).
This of course will fail if any space is added to the string (not to mention *, / or even decimal values).
const text = '20+3-5+10';
const arr = text.split(/\+|\-/)
let tot = 0
for (const num of arr) {
const pos = text.indexOf(num)
if (pos === 0) {
tot = parseInt(num)
} else {
switch (text.substr(text.indexOf(num) - 1, 1)) {
case '+':
tot += parseInt(num)
break
case '-':
tot -= parseInt(num)
break
}
}
}
console.log(tot)
I see 2 maybe 3 options, definitely there are hundreds
You don't use split and you just iterate through the string and just add or subtract on the way. As an example
You have '2+3-5+1'. You iterate until the second operator (+ or -) on your case. When you find it you just do the operation that you have iterated through and then you just keep going. You can do it recursive or not, doesn't matter
"2+3-5+1" -> "5-5+1" -> "0+1" -> 1
You use split on + for instance and you get [ '2', '3-5', '1' ] then you go through them with a loop with 2 conditions like
if(isNaN(x)) res+= x since you know it's been divided with a +
if(!isNaN(x)) res+= x.split('-')[0] - x.split('-')[1]
isNaN -> is not a number
Ofc you can make it look nicer. If you have parenthesis though, none of this will work
You can also use regex like split(/[-+]/) or more complex, but you'll have to find a way to know what operation follows each digit. One easy approach would be to iterate through both arrays. One of numbers and one of operators
"2+3-5+1".split(/[-+]/) -> [ '2', '3', '5', '1' ]
"2+3-5+1".split(/[0-9]*/).filter(x => x) -> [ '+', '-', '+' ]
You could probably find better regex, but you get the idea
You can ofc use a map or a switch for multiple operators
I am using fold on an array which hasn't been assign to a variable and want to check whether the element is the last value. With a conventional for loop I can do this:
List<int> ints = [1, 2, 3];
int sum = 0;
for (int num in ints]) {
if (num != ints.last) {
sum = sum + num;
}
}
print(sum);
Is it possible to do this with fold instead?
int foldSum = [1, 2, 3].fold(0, (int prev, element) => prev + element);
print(foldSum);
I can't find any way of check when fold is at the last value. Note: this is a simplified example of my problem and the reason the list isn't assigned to a variable (allowing me to use .last) is because it is the result of a call to .map().
For completeness, below is the actual code (which won't obviously won't be runnable in isolation but will help illustrate my problem) I am trying to convert to use .map and .fold:
String get fieldsToSqlInsert {
String val = "";
for (Column column in columns) {
if (data.containsKey(column.name)) {
val = '$val "${data[column.name]}"';
} else {
val = "$val NULL";
}
if (column != columns.last) {
val = "$val,";
}
}
return val;
}
But it doesn't work because I don't know how to check when fold is at the final element:
String get fieldsToSqlInsert => columns
.map((column) =>
data.containsKey(column.name) ? data[column.name] : "NULL")
.fold("", (val, column) => column != columns.last ? "$val," : val);
If you simply want to exclude the last element from further calculation, you can just use take to do so:
String get fieldsToSqlInsert => columns.take(columns.length - 1)...
I have this code
String encrypt(String x) {
String out;
var _x = x.codeUnits;
List dict;
/* <dict_assignment> */
dict[0] = 'a';
dict[1] = 'b';
dict[2] = 'c';
dict[3] = 'd';
dict[4] = 'e';
dict[5] = 'f';
dict[6] = 'g';
dict[7] = 'h';
dict[8] = 'i';
dict[9] = 'j';
/* </dict_assignment> */
_x.toList().forEach((i) {
var _i = i.toString();
_i.split("").forEach((k) {
var _k = int.parse(k);
print(_k);
print(dict[_k]);
out += dict[_k];
});
});
return out;
}
(Yes I'm writing HTML tags as comments in Dart...sue me)
(Idk why my indentations are messed up)
For some reason when I use this same function with a random string like this
var x = encrypt("hmm interesting");
I keep getting this
Unhandled exception:
NoSuchMethodError: The method '[]=' was called on null.
Receiver: null
Tried calling: []=(0, "a")
#0 Object.noSuchMethod (dart:core-patch/object_patch.dart:54:5)
Please help me I'm actually confused why this is happening
You have not initialized your dict variable, so it contains null.
If you change List dict; to List dict = []; then that would start working.
You also haven't initialized out.
The remainder of the code is leaning towards being overly complicated, and can be optimized as well. Here is a suggestion:
String encrypt(String x) {
var out = StringBuffer();
const dict = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j'];
for (var i in x.codeUnits) { // x.codeUnits is a list. Use for-in to iterate it.
for (var k in i.toString().codeUnits) {
var _k = k ^ 0x30; // Best way to convert code unit for 0-9 into integer 0-9.
// print(_k);
// print(dict[_k]);
out.write(dict[_k]); // Use a StringBuffer instead of repeated concatenation.
}
}
return out.toString();
}
It does not appear to be a decryptable encryption. The string "77" and the string "ᖳ" (aka "\u15b3") both encrypt to "ffff".
Or, if you want to "code-golf" rather than be readable or close to the original, it can also be a one-liner:
String encrypt(String x) => [
for (var i in x.codeUnits)
for (var k in "$i".codeUnits) "abcdefghij"[k ^ 0x30]
].join("");
i'm developing an app related to social messanging and i want to convert big numbers to Human readable format (e.g. 1500 to 1.5k) and also i'm new to Dart.
Your help will be appreciated.
You can use the NumberFormat class of flutter which has some in built functions for results you want..
Check out this link for NumberFormat class of flutter
Example:
This is one way if you want to use currency..
var _formattedNumber = NumberFormat.compactCurrency(
decimalDigits: 2,
symbol: '', // if you want to add currency symbol then pass that in this else leave it empty.
).format(numberToFormat);
print('Formatted Number is: $_formattedNumber');
Example:
This example is with locale.
var _formattedNumber = NumberFormat.compactCurrency(
decimalDigits: 2,
locale: 'en_IN'
symbol: '',
).format(numberToFormat);
print('Formatted Number is: $_formattedNumber');
The output of this is code would be:
If 1000 is entered then 1K is the output
Another way is by just using NumberFormat.compact() which gives the desired output...
// In this you won't have to worry about the symbol of the currency.
var _formattedNumber = NumberFormat.compact().format(numberToFormat);
print('Formatted Number is: $_formattedNumber');
The output of above example will also be:
If 1000 is entered then 1K is the output
I tried this and is working...
Make a class and used its static method every where.
class NumberFormatter{
static String formatter(String currentBalance) {
try{
// suffix = {' ', 'k', 'M', 'B', 'T', 'P', 'E'};
double value = double.parse(currentBalance);
if(value < 1000000){ // less than a million
return value.toStringAsFixed(2);
}else if(value >= 1000000 && value < (1000000*10*100)){ // less than 100 million
double result = value/1000000;
return result.toStringAsFixed(2)+"M";
}else if(value >= (1000000*10*100) && value < (1000000*10*100*100)){ // less than 100 billion
double result = value/(1000000*10*100);
return result.toStringAsFixed(2)+"B";
}else if(value >= (1000000*10*100*100) && value < (1000000*10*100*100*100)){ // less than 100 trillion
double result = value/(1000000*10*100*100);
return result.toStringAsFixed(2)+"T";
}
}catch(e){
print(e);
}
}
}
I have a string
Ex: "We prefer questions that can be answered; not just discussed "
now i want to split this string from ";"
like
We prefer questions that can be answered
and
not just discussed
is this possible in DXL.
i am learning DXL, so i don't have any idea whether we can split or not.
Note : This is not a home work.
I'm sorry for necroing this post. Being new to DXL I spent some time with the same challenge. I noticed that the implementations available on the have different specifications of "splitting" a string. Loving the Ruby language, I missed an implementation which comes at least close to the Ruby version of String#split.
Maybe my findings will be helpful to anybody.
Here's a functional comparison of
Variant A: niol's implementation (which at a first glance, appears to be the same implementation which is usually found at Capri Soft,
Variant B: PJT's implementation,
Variant C: Brett's implementation and
Variant D: my implementation (which provides the correct functionality imo).
To eliminate structural difference, all implementations were implemented in functions, returning a Skip list or an Array.
Splitting results
Note that all implementations return different results, depending on their definition of "splitting":
string mellow yellow; delimiter ello
splitVariantA returns 1 elements: ["mellow yellow" ]
splitVariantB returns 2 elements: ["m" "llow yellow" ]
splitVariantC returns 3 elements: ["w" "w y" "" ]
splitVariantD returns 3 elements: ["m" "w y" "w" ]
string now's the time; delimiter
splitVariantA returns 3 elements: ["now's" "the" "time" ]
splitVariantB returns 2 elements: ["" "now's the time" ]
splitVariantC returns 5 elements: ["time" "the" "" "now's" "" ]
splitVariantD returns 3 elements: ["now's" "the" "time" ]
string 1,2,,3,4,,; delimiter ,
splitVariantA returns 4 elements: ["1" "2" "3" "4" ]
splitVariantB returns 2 elements: ["1" "2,,3,4,," ]
splitVariantC returns 7 elements: ["" "" "4" "3" "" "2" "" ]
splitVariantD returns 7 elements: ["1" "2" "" "3" "4" "" "" ]
Timing
Splitting the string 1,2,,3,4,, with the pattern , for 10000 times on my machine gives these timings:
splitVariantA() : 406 ms
splitVariantB() : 46 ms
splitVariantC() : 749 ms
splitVariantD() : 1077 ms
Unfortunately, my implementation D is the slowest. Surprisingly, the regular expressions implementation C is pretty fast.
Source code
// niol, modified
Array splitVariantA(string splitter, string str){
Array tokens = create(1, 1);
Buffer buf = create;
int str_index;
buf = "";
for(str_index = 0; str_index < length(str); str_index++){
if( str[str_index:str_index] == splitter ){
array_push_str(tokens, stringOf(buf));
buf = "";
}
else
buf += str[str_index:str_index];
}
array_push_str(tokens, stringOf(buf));
delete buf;
return tokens;
}
// PJT, modified
Skip splitVariantB(string s, string delimiter) {
int offset
int len
Skip skp = create
if ( findPlainText(s, delimiter, offset, len, false)) {
put(skp, 0, s[0 : offset -1])
put(skp, 1, s[offset +1 :])
}
return skp
}
// Brett, modified
Skip splitVariantC (string s, string delim) {
Skip skp = create
int i = 0
Regexp split = regexp "^(.*)" delim "(.*)$"
while (split s) {
string temp_s = s[match 1]
put(skp, i++, s[match 2])
s = temp_s
}
put(skp, i++, s[match 2])
return skp
}
Skip splitVariantD(string str, string pattern) {
if (null(pattern) || 0 == length(pattern))
pattern = " ";
if (pattern == " ")
str = stringStrip(stringSqueeze(str, ' '));
Skip result = create;
int i = 0; // index for searching in str
int j = 0; // index counter for result array
bool found = true;
while (found) {
// find pattern
int pos = 0;
int len = 0;
found = findPlainText(str[i:], pattern, pos, len, true);
if (found) {
// insert into result
put(result, j++, str[i:i+pos-1]);
i += pos + len;
}
}
// append the rest after last found pattern
put(result, j, str[i:]);
return result;
}
Quick join&split I could come up with. Seams to work okay.
int array_size(Array a){
int size = 0;
while( !null(get(a, size, 0) ) )
size++;
return size;
}
void array_push_str(Array a, string str){
int array_index = array_size(a);
put(a, str, array_index, 0);
}
string array_get_str(Array a, int index){
return (string get(a, index, 0));
}
string str_join(string joiner, Array str_array){
Buffer joined = create;
int array_index = 0;
joined += "";
for(array_index = 0; array_index < array_size(str_array); array_index++){
joined += array_get_str(str_array, array_index);
if( array_index + 1 < array_size(str_array) )
joined += joiner;
}
return stringOf(joined)
}
Array str_split(string splitter, string str){
Array tokens = create(1, 1);
Buffer buf = create;
int str_index;
buf = "";
for(str_index = 0; str_index < length(str); str_index++){
if( str[str_index:str_index] == splitter ){
array_push_str(tokens, stringOf(buf));
buf = "";
}else{
buf += str[str_index:str_index];
}
}
array_push_str(tokens, stringOf(buf));
delete buf;
return tokens;
}
If you only split the string once this is how I would do it:
string s = "We prefer questions that can be answered; not just discussed"
string sub = ";"
int offset
int len
if ( findPlainText(s, sub, offset, len, false)) {
/* the reason why I subtract one and add one is to remove the delimiter from the out put.
First print is to print the prefix and then second is the suffix.*/
print s[0 : offset -1]
print s[offset +1 :]
} else {
// no delimiter found
print "Failed to match"
}
You could also use regular expressions refer to the DXL reference manual. It would be better to use regular expressions if you want to split up the string by multiple delimiters such as str = "this ; is an;example"
ACTUALLY WORKS:
This solution will split as many times as needed, or none, if the delimiter doesn't exist in the string.
This is what I have used instead of a traditional "split" command.
It actually skips the creation of an array, and just loops through each string that would be in the array and calls "someFunction" on each of those strings.
string s = "We prefer questions that can be answered; not just discussed"
// for this example, ";" is used as the delimiter
Regexp split = regexp "^(.*);(.*)$"
// while a ";" exists in s
while (split s) {
// save the text before the last ";"
string temp_s = s[match 1]
// call someFunction on the text after the last ";"
someFunction(s[match 2])
// remove the text after the last ";" (including ";")
s = temp_s
}
// call someFunction again for the last (or only) string
someFunction(s)
Sorry for necroing an old post; I just didn't find the other answers useful.
Perhaps someone would find handy this fused solution as well. It splits string in Skip, based on delimiter, which can actually have length more then one.
Skip splitString(string s1, string delimit)
{
int offset, len
Skip splited = create
while(findPlainText(s1, delimit, offset, len, false))
{
put(splited, s1[0:offset-1], s1[0:offset-1])
s1 = s1[offset+length(delimit):length(s1)-1]
}
if(length(s1)>0)
{
put (splited, s1, s1)
}
return splited
}
I tried this out and worked out for me...
string s = "We prefer questions that can be answered,not just discussed,hiyas"
string sub = ","
int offset
int len
string s1=s
while(length(s1)>0){
if ( findPlainText(s1, sub, offset, len, false)) {
print s1[0 : offset -1]"\n"
s1= s1[offset+1:length(s1)]
}
else
{
print s1
s1=""
}
}
Here is a better implementation. This is a recursive split of the string by searching a keyword.
pragma runLim, 10000
string s = "We prefer questions that can be answered,not just discussed,hiyas;
Next Line,Var1,Nemesis;
Next Line,Var2,Nemesis1;
Next Line,Var3,Nemesis2;
New,Var4,Nemesis3;
Next Line,Var5,Nemesis4;
New,Var5,Nemesis5;"
string sub = ","
int offset
int len
string searchkey=null
string curr=s
string nxt=s
string searchline=null
string Modulename=""
string Attributename=""
string Attributevalue=""
while(findPlainText(curr,"Next Line", offset,len,false))
{
int intlen=offset
searchkey=curr[offset:length(curr)]
if(findPlainText(searchkey,"Next Line",offset,len,false))
{
curr=searchkey[offset+1:length(searchkey)]
}
if(findPlainText(searchkey,";",offset,len,false))
{
searchline=searchkey[0:offset]
}
int counter=0
while(length(searchline)>0)
{
if (findPlainText(searchline, sub, offset, len, false))
{
if(counter==0)
{
Modulename=searchline[0 : offset -1]
counter++
}
else if(counter==1)
{
Attributename=searchline[0 : offset -1]
counter++
}
searchline= searchline[offset+1:length(searchline)]
}
else
{
if(counter==2)
{
Attributevalue=searchline[0:length(searchline)-2]
counter++
}
searchline=""
}
}
print "Modulename="Modulename " Attributename=" Attributename " Attributevalue= "Attributevalue "\n"
}