Can I make this Dictionary sort (smallest to biggest) algorithm more concise? Swift
0
$begingroup$
I have the bellow sorting algorithm which takes an array of dictionary values as declared below.
guard var imageUrlString = anyImage.value as? [String:AnyObject] else { return }
I then loop through it adding the values with the smallest Int key to a new array to be used afterward. Thus the values in this array of AnyObjects would go from 1 to n.
I was wondering if I could make it more concise.
var values = [AnyObject]()
var keys = [String]()
var Done = false
var j = 1
while !Done {
for i in imageUrlString {
let key = Int(String(i.key.last!))
if j == key {
values.append(i.value)
keys.append(i.key)
print(i, " This is teh i for in if ")
if imageUrlString.count == j {
print("Done yet: yes", values[0], " ", values[3])
Done = true
break;
}
j+=1
} else {
print("No,,.")
}
}
}
swift ios dictionary
asked 4 mins ago
OutsiderOutsider
11
New contributor
Outsider is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
0
$begingroup$
I have the bellow sorting algorithm which takes an array of dictionary values as declared below.
guard var imageUrlString = anyImage.value as? [String:AnyObject] else { return }
I then loop through it adding the values with the smallest Int key to a new array to be used afterward. Thus the values in this array of AnyObjects would go from 1 to n.
I was wondering if I could make it more concise.
var values = [AnyObject]()
var keys = [String]()
var Done = false
var j = 1
while !Done {
for i in imageUrlString {
let key = Int(String(i.key.last!))
if j == key {
values.append(i.value)
keys.append(i.key)
print(i, " This is teh i for in if ")
if imageUrlString.count == j {
print("Done yet: yes", values[0], " ", values[3])
Done = true
break;
}
j+=1
} else {
print("No,,.")
}
}
}
swift ios dictionary
asked 4 mins ago
OutsiderOutsider
11
New contributor
Outsider is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
0
0
0
1
$begingroup$
I have the bellow sorting algorithm which takes an array of dictionary values as declared below.
guard var imageUrlString = anyImage.value as? [String:AnyObject] else { return }
I then loop through it adding the values with the smallest Int key to a new array to be used afterward. Thus the values in this array of AnyObjects would go from 1 to n.
I was wondering if I could make it more concise.
var values = [AnyObject]()
var keys = [String]()
var Done = false
var j = 1
while !Done {
for i in imageUrlString {
let key = Int(String(i.key.last!))
if j == key {
values.append(i.value)
keys.append(i.key)
print(i, " This is teh i for in if ")
if imageUrlString.count == j {
print("Done yet: yes", values[0], " ", values[3])
Done = true
break;
}
j+=1
} else {
print("No,,.")
}
}
}
swift ios dictionary
asked 4 mins ago
OutsiderOutsider
11
New contributor
Outsider is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I have the bellow sorting algorithm which takes an array of dictionary values as declared below.
guard var imageUrlString = anyImage.value as? [String:AnyObject] else { return }
I then loop through it adding the values with the smallest Int key to a new array to be used afterward. Thus the values in this array of AnyObjects would go from 1 to n.
I was wondering if I could make it more concise.
var values = [AnyObject]()
var keys = [String]()
var Done = false
var j = 1
while !Done {
for i in imageUrlString {
let key = Int(String(i.key.last!))
if j == key {
values.append(i.value)
keys.append(i.key)
print(i, " This is teh i for in if ")
if imageUrlString.count == j {
print("Done yet: yes", values[0], " ", values[3])
Done = true
break;
}
j+=1
} else {
print("No,,.")
}
}
}
swift ios dictionary
swift ios dictionary
asked 4 mins ago
OutsiderOutsider
11
New contributor
Outsider is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 mins ago
OutsiderOutsider
11
New contributor
Outsider is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 mins ago
OutsiderOutsider
11
New contributor
Outsider is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 mins ago
OutsiderOutsider
11
asked 4 mins ago
OutsiderOutsider
11
11
New contributor
Outsider is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Outsider is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Outsider is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
});
});
}, "mathjax-editing");
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "196"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Outsider is a new contributor. Be nice, and check out our Code of Conduct.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodereview.stackexchange.com%2fquestions%2f215392%2fcan-i-make-this-dictionary-sort-smallest-to-biggest-algorithm-more-concise-sw%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Outsider is a new contributor. Be nice, and check out our Code of Conduct.
draft saved
draft discarded
Outsider is a new contributor. Be nice, and check out our Code of Conduct.
Outsider is a new contributor. Be nice, and check out our Code of Conduct.
Outsider is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Code Review Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodereview.stackexchange.com%2fquestions%2f215392%2fcan-i-make-this-dictionary-sort-smallest-to-biggest-algorithm-more-concise-sw%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
