Mixing
Function to determine the direction of an arrow
I have a function in a project that I feel could be written better. The purpose of the function is to return the direction of an arrow based on two parameters.
positiveDirection: The direction for a positive result. The values could beincreasingordecreasing.
changeType: Whether the result waspositive,negativeorno-change.
I guess it's the if statement that is bothering me the most; it looks like it could be reduced based on some logic gate that I don't know of.
getArrowDirection = (positiveDirection, changeType) => {
let direction = null
if(changeType === 'no-change') {
return direction
}
if(
(changeType === 'positive' && positiveDirection === 'increasing') ||
(changeType === 'negative' && positiveDirection === 'decreasing')
) {
direction = 'up-arrow'
} else if(
(changeType === 'positive' && positiveDirection === 'decreasing') ||
(changeType === 'negative' && positiveDirection === 'increasing')
) {
direction = 'down-arrow'
}
return direction
}javascript ecmascript-6
edited Nov 21 '18 at 16:38
Toby Speight
23.4k639111
asked Nov 21 '18 at 12:40
Richard Healy
364
add a comment |
I have a function in a project that I feel could be written better. The purpose of the function is to return the direction of an arrow based on two parameters.
positiveDirection: The direction for a positive result. The values could beincreasingordecreasing.
changeType: Whether the result waspositive,negativeorno-change.
I guess it's the if statement that is bothering me the most; it looks like it could be reduced based on some logic gate that I don't know of.
getArrowDirection = (positiveDirection, changeType) => {
let direction = null
if(changeType === 'no-change') {
return direction
}
if(
(changeType === 'positive' && positiveDirection === 'increasing') ||
(changeType === 'negative' && positiveDirection === 'decreasing')
) {
direction = 'up-arrow'
} else if(
(changeType === 'positive' && positiveDirection === 'decreasing') ||
(changeType === 'negative' && positiveDirection === 'increasing')
) {
direction = 'down-arrow'
}
return direction
}javascript ecmascript-6
edited Nov 21 '18 at 16:38
Toby Speight
23.4k639111
asked Nov 21 '18 at 12:40
Richard Healy
364
What's possible values forchangeValueandpositiveDirection?
– Calak
Nov 21 '18 at 14:16
We might be able to improve the code further if we knew the context. Who calls this code, and where do the parameter values come from?
– 200_success
Nov 21 '18 at 15:23
add a comment |
I have a function in a project that I feel could be written better. The purpose of the function is to return the direction of an arrow based on two parameters.
positiveDirection: The direction for a positive result. The values could beincreasingordecreasing.
changeType: Whether the result waspositive,negativeorno-change.
I guess it's the if statement that is bothering me the most; it looks like it could be reduced based on some logic gate that I don't know of.
getArrowDirection = (positiveDirection, changeType) => {
let direction = null
if(changeType === 'no-change') {
return direction
}
if(
(changeType === 'positive' && positiveDirection === 'increasing') ||
(changeType === 'negative' && positiveDirection === 'decreasing')
) {
direction = 'up-arrow'
} else if(
(changeType === 'positive' && positiveDirection === 'decreasing') ||
(changeType === 'negative' && positiveDirection === 'increasing')
) {
direction = 'down-arrow'
}
return direction
}javascript ecmascript-6
edited Nov 21 '18 at 16:38
Toby Speight
23.4k639111
asked Nov 21 '18 at 12:40
Richard Healy
364
I have a function in a project that I feel could be written better. The purpose of the function is to return the direction of an arrow based on two parameters.
positiveDirection: The direction for a positive result. The values could beincreasingordecreasing.
changeType: Whether the result waspositive,negativeorno-change.
I guess it's the if statement that is bothering me the most; it looks like it could be reduced based on some logic gate that I don't know of.
getArrowDirection = (positiveDirection, changeType) => {
let direction = null
if(changeType === 'no-change') {
return direction
}
if(
(changeType === 'positive' && positiveDirection === 'increasing') ||
(changeType === 'negative' && positiveDirection === 'decreasing')
) {
direction = 'up-arrow'
} else if(
(changeType === 'positive' && positiveDirection === 'decreasing') ||
(changeType === 'negative' && positiveDirection === 'increasing')
) {
direction = 'down-arrow'
}
return direction
} getArrowDirection = (positiveDirection, changeType) => {
let direction = null
if(changeType === 'no-change') {
return direction
}
if(
(changeType === 'positive' && positiveDirection === 'increasing') ||
(changeType === 'negative' && positiveDirection === 'decreasing')
) {
direction = 'up-arrow'
} else if(
(changeType === 'positive' && positiveDirection === 'decreasing') ||
(changeType === 'negative' && positiveDirection === 'increasing')
) {
direction = 'down-arrow'
}
return direction
} getArrowDirection = (positiveDirection, changeType) => {
let direction = null
if(changeType === 'no-change') {
return direction
}
if(
(changeType === 'positive' && positiveDirection === 'increasing') ||
(changeType === 'negative' && positiveDirection === 'decreasing')
) {
direction = 'up-arrow'
} else if(
(changeType === 'positive' && positiveDirection === 'decreasing') ||
(changeType === 'negative' && positiveDirection === 'increasing')
) {
direction = 'down-arrow'
}
return direction
}javascript ecmascript-6
javascript ecmascript-6
edited Nov 21 '18 at 16:38
Toby Speight
23.4k639111
asked Nov 21 '18 at 12:40
Richard Healy
364
edited Nov 21 '18 at 16:38
Toby Speight
23.4k639111
asked Nov 21 '18 at 12:40
Richard Healy
364
edited Nov 21 '18 at 16:38
Toby Speight
23.4k639111
edited Nov 21 '18 at 16:38
Toby Speight
23.4k639111
edited Nov 21 '18 at 16:38
Toby Speight
23.4k639111
23.4k639111
asked Nov 21 '18 at 12:40
Richard Healy
364
asked Nov 21 '18 at 12:40
Richard Healy
364
asked Nov 21 '18 at 12:40
Richard Healy
364
364
What's possible values forchangeValueandpositiveDirection?
– Calak
Nov 21 '18 at 14:16
We might be able to improve the code further if we knew the context. Who calls this code, and where do the parameter values come from?
– 200_success
Nov 21 '18 at 15:23
add a comment |
What's possible values forchangeValueandpositiveDirection?
– Calak
Nov 21 '18 at 14:16
We might be able to improve the code further if we knew the context. Who calls this code, and where do the parameter values come from?
– 200_success
Nov 21 '18 at 15:23
What's possible values for
changeValue and positiveDirection?– Calak
Nov 21 '18 at 14:16
What's possible values for
changeValue and positiveDirection?– Calak
Nov 21 '18 at 14:16
We might be able to improve the code further if we knew the context. Who calls this code, and where do the parameter values come from?
– 200_success
Nov 21 '18 at 15:23
We might be able to improve the code further if we knew the context. Who calls this code, and where do the parameter values come from?
– 200_success
Nov 21 '18 at 15:23
add a comment |
4 Answers
4
active
oldest
votes
If changeValue can only be (after your early check for 'no-change') equal to 'positive' or 'negative' and positiveDirection can only be equal to 'increasing' or 'decreasing, your if should cover all possible ways.
But you can simplify it a lot making use of the xor operator:
direction = (changeType === 'positive' ^ positiveDirection === 'decreasing')
? 'up-arrow'
: 'down-arrow';
Or make the whole function a one-liner:
return (changeType !== 'no-change')
? ((changeType === 'positive' ^ positiveDirection === 'decreasing') ? 'up-arrow' : 'down-arrow')
: null;
answered Nov 21 '18 at 14:43
Calak
2,102318
This is slick. Definitely a better solution than my original code. The ^ syntax was something I didn't know about.
– Richard Healy
Nov 21 '18 at 15:21
GivenA^B, ifAandBare either bothtrueorfalse, returnsfalse. Otherwise, returnstrue. If fact, you can also replace it with!=but I found the logic xor prettier/smarter :)
– Calak
Nov 21 '18 at 15:32
add a comment |
The test for if(changeType === 'no-change') { is redundant and not needed as it will fall through if the other tests fail and return null anyways.
You could also break the statement up returning the result as needed and falling through if they fail for null.
const getArrowDirection = (dir, type) => {
if (type === "positive") {
if (dir === "increasing") { return "up-arrow" }
if (dir === "decreasing") { return "down-arrow" }
}else if (type === "negative") {
if (dir === "increasing") { return "down-arrow" }
if (dir === "decreasing") { return "up-arrow" }
}
return null;
}
Assuming that you have given all possible values you can make assumptions and reduce the code further
const getArrowDirection = (dir, type) => {
if (type !== 'no-change') {
if (type === "positive") { return dir === "increasing" ? "up-arrow" : "down-arrow" }
return dir === "increasing" ? "down-arrow" : "up-arrow";
}
return null; // returning null is not the best. Returning undefined would
// be better and would not need this line
}
You can use an object as a lookup using the combined strings.
const getArrowDirection = (() => {
const directions = {
positiveincreasing: "up-arrow",
negativedecreasing: "up-arrow",
positivedecreasing: "down-arrow",
negativeincreasing: "down-arrow",
};
return (dir, type) => directions[type + dir] ? directions[type + dir] : null;
})();
answered Nov 21 '18 at 13:11
Blindman67
7,2011521
add a comment |
We don't need the direction variable - we can simply return at the appropriate point.
Once you know that changeType is one of 'positive' or 'negative' and that positiveDirection is either 'increasing' or 'decreasing', you can test whether the two equality tests match:
// Winging it, because this isn't my language!
getArrowDirection = (positiveDirection, changeType) => {
if (changeType != 'positive' && changeType != 'negative') {
return null;
}
if (positiveDirection != 'increasing' && positiveDirection != 'decreasing')
return null;
}
if ((changeType === 'positive') == (positiveDirection === 'increasing')) {
return 'up-arrow';
else
return 'down-arrow';
}
}
answered Nov 21 '18 at 14:17
Toby Speight
23.4k639111
add a comment |
You could could extract your conditions into functions with very clear condition description names. Would cut down the code because many of those conditions are repeated.
answered Nov 21 '18 at 13:09
Dave
1
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
If changeValue can only be (after your early check for 'no-change') equal to 'positive' or 'negative' and positiveDirection can only be equal to 'increasing' or 'decreasing, your if should cover all possible ways.
But you can simplify it a lot making use of the xor operator:
direction = (changeType === 'positive' ^ positiveDirection === 'decreasing')
? 'up-arrow'
: 'down-arrow';
Or make the whole function a one-liner:
return (changeType !== 'no-change')
? ((changeType === 'positive' ^ positiveDirection === 'decreasing') ? 'up-arrow' : 'down-arrow')
: null;
answered Nov 21 '18 at 14:43
Calak
2,102318
This is slick. Definitely a better solution than my original code. The ^ syntax was something I didn't know about.
– Richard Healy
Nov 21 '18 at 15:21
GivenA^B, ifAandBare either bothtrueorfalse, returnsfalse. Otherwise, returnstrue. If fact, you can also replace it with!=but I found the logic xor prettier/smarter :)
– Calak
Nov 21 '18 at 15:32
add a comment |
If changeValue can only be (after your early check for 'no-change') equal to 'positive' or 'negative' and positiveDirection can only be equal to 'increasing' or 'decreasing, your if should cover all possible ways.
But you can simplify it a lot making use of the xor operator:
direction = (changeType === 'positive' ^ positiveDirection === 'decreasing')
? 'up-arrow'
: 'down-arrow';
Or make the whole function a one-liner:
return (changeType !== 'no-change')
? ((changeType === 'positive' ^ positiveDirection === 'decreasing') ? 'up-arrow' : 'down-arrow')
: null;
answered Nov 21 '18 at 14:43
Calak
2,102318
This is slick. Definitely a better solution than my original code. The ^ syntax was something I didn't know about.
– Richard Healy
Nov 21 '18 at 15:21
GivenA^B, ifAandBare either bothtrueorfalse, returnsfalse. Otherwise, returnstrue. If fact, you can also replace it with!=but I found the logic xor prettier/smarter :)
– Calak
Nov 21 '18 at 15:32
add a comment |
If changeValue can only be (after your early check for 'no-change') equal to 'positive' or 'negative' and positiveDirection can only be equal to 'increasing' or 'decreasing, your if should cover all possible ways.
But you can simplify it a lot making use of the xor operator:
direction = (changeType === 'positive' ^ positiveDirection === 'decreasing')
? 'up-arrow'
: 'down-arrow';
Or make the whole function a one-liner:
return (changeType !== 'no-change')
? ((changeType === 'positive' ^ positiveDirection === 'decreasing') ? 'up-arrow' : 'down-arrow')
: null;
answered Nov 21 '18 at 14:43
Calak
2,102318
If changeValue can only be (after your early check for 'no-change') equal to 'positive' or 'negative' and positiveDirection can only be equal to 'increasing' or 'decreasing, your if should cover all possible ways.
But you can simplify it a lot making use of the xor operator:
direction = (changeType === 'positive' ^ positiveDirection === 'decreasing')
? 'up-arrow'
: 'down-arrow';
Or make the whole function a one-liner:
return (changeType !== 'no-change')
? ((changeType === 'positive' ^ positiveDirection === 'decreasing') ? 'up-arrow' : 'down-arrow')
: null;
answered Nov 21 '18 at 14:43
Calak
2,102318
answered Nov 21 '18 at 14:43
Calak
2,102318
answered Nov 21 '18 at 14:43
Calak
2,102318
answered Nov 21 '18 at 14:43
Calak
2,102318
2,102318
This is slick. Definitely a better solution than my original code. The ^ syntax was something I didn't know about.
– Richard Healy
Nov 21 '18 at 15:21
GivenA^B, ifAandBare either bothtrueorfalse, returnsfalse. Otherwise, returnstrue. If fact, you can also replace it with!=but I found the logic xor prettier/smarter :)
– Calak
Nov 21 '18 at 15:32
add a comment |
This is slick. Definitely a better solution than my original code. The ^ syntax was something I didn't know about.
– Richard Healy
Nov 21 '18 at 15:21
GivenA^B, ifAandBare either bothtrueorfalse, returnsfalse. Otherwise, returnstrue. If fact, you can also replace it with!=but I found the logic xor prettier/smarter :)
– Calak
Nov 21 '18 at 15:32
This is slick. Definitely a better solution than my original code. The ^ syntax was something I didn't know about.
– Richard Healy
Nov 21 '18 at 15:21
This is slick. Definitely a better solution than my original code. The ^ syntax was something I didn't know about.
– Richard Healy
Nov 21 '18 at 15:21
Given
A^B, if A and B are either both true or false, returns false. Otherwise, returns true. If fact, you can also replace it with != but I found the logic xor prettier/smarter :)– Calak
Nov 21 '18 at 15:32
Given
A^B, if A and B are either both true or false, returns false. Otherwise, returns true. If fact, you can also replace it with != but I found the logic xor prettier/smarter :)– Calak
Nov 21 '18 at 15:32
add a comment |
The test for if(changeType === 'no-change') { is redundant and not needed as it will fall through if the other tests fail and return null anyways.
You could also break the statement up returning the result as needed and falling through if they fail for null.
const getArrowDirection = (dir, type) => {
if (type === "positive") {
if (dir === "increasing") { return "up-arrow" }
if (dir === "decreasing") { return "down-arrow" }
}else if (type === "negative") {
if (dir === "increasing") { return "down-arrow" }
if (dir === "decreasing") { return "up-arrow" }
}
return null;
}
Assuming that you have given all possible values you can make assumptions and reduce the code further
const getArrowDirection = (dir, type) => {
if (type !== 'no-change') {
if (type === "positive") { return dir === "increasing" ? "up-arrow" : "down-arrow" }
return dir === "increasing" ? "down-arrow" : "up-arrow";
}
return null; // returning null is not the best. Returning undefined would
// be better and would not need this line
}
You can use an object as a lookup using the combined strings.
const getArrowDirection = (() => {
const directions = {
positiveincreasing: "up-arrow",
negativedecreasing: "up-arrow",
positivedecreasing: "down-arrow",
negativeincreasing: "down-arrow",
};
return (dir, type) => directions[type + dir] ? directions[type + dir] : null;
})();
answered Nov 21 '18 at 13:11
Blindman67
7,2011521
add a comment |
The test for if(changeType === 'no-change') { is redundant and not needed as it will fall through if the other tests fail and return null anyways.
You could also break the statement up returning the result as needed and falling through if they fail for null.
const getArrowDirection = (dir, type) => {
if (type === "positive") {
if (dir === "increasing") { return "up-arrow" }
if (dir === "decreasing") { return "down-arrow" }
}else if (type === "negative") {
if (dir === "increasing") { return "down-arrow" }
if (dir === "decreasing") { return "up-arrow" }
}
return null;
}
Assuming that you have given all possible values you can make assumptions and reduce the code further
const getArrowDirection = (dir, type) => {
if (type !== 'no-change') {
if (type === "positive") { return dir === "increasing" ? "up-arrow" : "down-arrow" }
return dir === "increasing" ? "down-arrow" : "up-arrow";
}
return null; // returning null is not the best. Returning undefined would
// be better and would not need this line
}
You can use an object as a lookup using the combined strings.
const getArrowDirection = (() => {
const directions = {
positiveincreasing: "up-arrow",
negativedecreasing: "up-arrow",
positivedecreasing: "down-arrow",
negativeincreasing: "down-arrow",
};
return (dir, type) => directions[type + dir] ? directions[type + dir] : null;
})();
answered Nov 21 '18 at 13:11
Blindman67
7,2011521
add a comment |
The test for if(changeType === 'no-change') { is redundant and not needed as it will fall through if the other tests fail and return null anyways.
You could also break the statement up returning the result as needed and falling through if they fail for null.
const getArrowDirection = (dir, type) => {
if (type === "positive") {
if (dir === "increasing") { return "up-arrow" }
if (dir === "decreasing") { return "down-arrow" }
}else if (type === "negative") {
if (dir === "increasing") { return "down-arrow" }
if (dir === "decreasing") { return "up-arrow" }
}
return null;
}
Assuming that you have given all possible values you can make assumptions and reduce the code further
const getArrowDirection = (dir, type) => {
if (type !== 'no-change') {
if (type === "positive") { return dir === "increasing" ? "up-arrow" : "down-arrow" }
return dir === "increasing" ? "down-arrow" : "up-arrow";
}
return null; // returning null is not the best. Returning undefined would
// be better and would not need this line
}
You can use an object as a lookup using the combined strings.
const getArrowDirection = (() => {
const directions = {
positiveincreasing: "up-arrow",
negativedecreasing: "up-arrow",
positivedecreasing: "down-arrow",
negativeincreasing: "down-arrow",
};
return (dir, type) => directions[type + dir] ? directions[type + dir] : null;
})();
answered Nov 21 '18 at 13:11
Blindman67
7,2011521
The test for if(changeType === 'no-change') { is redundant and not needed as it will fall through if the other tests fail and return null anyways.
You could also break the statement up returning the result as needed and falling through if they fail for null.
const getArrowDirection = (dir, type) => {
if (type === "positive") {
if (dir === "increasing") { return "up-arrow" }
if (dir === "decreasing") { return "down-arrow" }
}else if (type === "negative") {
if (dir === "increasing") { return "down-arrow" }
if (dir === "decreasing") { return "up-arrow" }
}
return null;
}
Assuming that you have given all possible values you can make assumptions and reduce the code further
const getArrowDirection = (dir, type) => {
if (type !== 'no-change') {
if (type === "positive") { return dir === "increasing" ? "up-arrow" : "down-arrow" }
return dir === "increasing" ? "down-arrow" : "up-arrow";
}
return null; // returning null is not the best. Returning undefined would
// be better and would not need this line
}
You can use an object as a lookup using the combined strings.
const getArrowDirection = (() => {
const directions = {
positiveincreasing: "up-arrow",
negativedecreasing: "up-arrow",
positivedecreasing: "down-arrow",
negativeincreasing: "down-arrow",
};
return (dir, type) => directions[type + dir] ? directions[type + dir] : null;
})();
answered Nov 21 '18 at 13:11
Blindman67
7,2011521
edited Nov 21 '18 at 15:18
answered Nov 21 '18 at 13:11
Blindman67
7,2011521
answered Nov 21 '18 at 13:11
Blindman67
7,2011521
answered Nov 21 '18 at 13:11
Blindman67
7,2011521
7,2011521
add a comment |
add a comment |
We don't need the direction variable - we can simply return at the appropriate point.
Once you know that changeType is one of 'positive' or 'negative' and that positiveDirection is either 'increasing' or 'decreasing', you can test whether the two equality tests match:
// Winging it, because this isn't my language!
getArrowDirection = (positiveDirection, changeType) => {
if (changeType != 'positive' && changeType != 'negative') {
return null;
}
if (positiveDirection != 'increasing' && positiveDirection != 'decreasing')
return null;
}
if ((changeType === 'positive') == (positiveDirection === 'increasing')) {
return 'up-arrow';
else
return 'down-arrow';
}
}
answered Nov 21 '18 at 14:17
Toby Speight
23.4k639111
add a comment |
We don't need the direction variable - we can simply return at the appropriate point.
Once you know that changeType is one of 'positive' or 'negative' and that positiveDirection is either 'increasing' or 'decreasing', you can test whether the two equality tests match:
// Winging it, because this isn't my language!
getArrowDirection = (positiveDirection, changeType) => {
if (changeType != 'positive' && changeType != 'negative') {
return null;
}
if (positiveDirection != 'increasing' && positiveDirection != 'decreasing')
return null;
}
if ((changeType === 'positive') == (positiveDirection === 'increasing')) {
return 'up-arrow';
else
return 'down-arrow';
}
}
answered Nov 21 '18 at 14:17
Toby Speight
23.4k639111
add a comment |
We don't need the direction variable - we can simply return at the appropriate point.
Once you know that changeType is one of 'positive' or 'negative' and that positiveDirection is either 'increasing' or 'decreasing', you can test whether the two equality tests match:
// Winging it, because this isn't my language!
getArrowDirection = (positiveDirection, changeType) => {
if (changeType != 'positive' && changeType != 'negative') {
return null;
}
if (positiveDirection != 'increasing' && positiveDirection != 'decreasing')
return null;
}
if ((changeType === 'positive') == (positiveDirection === 'increasing')) {
return 'up-arrow';
else
return 'down-arrow';
}
}
answered Nov 21 '18 at 14:17
Toby Speight
23.4k639111
We don't need the direction variable - we can simply return at the appropriate point.
Once you know that changeType is one of 'positive' or 'negative' and that positiveDirection is either 'increasing' or 'decreasing', you can test whether the two equality tests match:
// Winging it, because this isn't my language!
getArrowDirection = (positiveDirection, changeType) => {
if (changeType != 'positive' && changeType != 'negative') {
return null;
}
if (positiveDirection != 'increasing' && positiveDirection != 'decreasing')
return null;
}
if ((changeType === 'positive') == (positiveDirection === 'increasing')) {
return 'up-arrow';
else
return 'down-arrow';
}
}
answered Nov 21 '18 at 14:17
Toby Speight
23.4k639111
answered Nov 21 '18 at 14:17
Toby Speight
23.4k639111
answered Nov 21 '18 at 14:17
Toby Speight
23.4k639111
answered Nov 21 '18 at 14:17
Toby Speight
23.4k639111
23.4k639111
add a comment |
add a comment |
You could could extract your conditions into functions with very clear condition description names. Would cut down the code because many of those conditions are repeated.
answered Nov 21 '18 at 13:09
Dave
1
add a comment |
You could could extract your conditions into functions with very clear condition description names. Would cut down the code because many of those conditions are repeated.
answered Nov 21 '18 at 13:09
Dave
1
add a comment |
You could could extract your conditions into functions with very clear condition description names. Would cut down the code because many of those conditions are repeated.
answered Nov 21 '18 at 13:09
Dave
1
You could could extract your conditions into functions with very clear condition description names. Would cut down the code because many of those conditions are repeated.
answered Nov 21 '18 at 13:09
Dave
1
answered Nov 21 '18 at 13:09
Dave
1
answered Nov 21 '18 at 13:09
Dave
1
answered Nov 21 '18 at 13:09
Dave
1
1
add a comment |
add a comment |
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What's possible values for
changeValueandpositiveDirection?– Calak
Nov 21 '18 at 14:16
We might be able to improve the code further if we knew the context. Who calls this code, and where do the parameter values come from?
– 200_success
Nov 21 '18 at 15:23