Replacing values from one set with values from another, if the value of set1 = the key of set2?

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box;
}

EDIT: Changed dictionaries to Sets, as I had not realised {} denoted a set. And fixed to say that sets are included in a tuple.

I want to iterate through each set in cardTuple, and for each value, replace it with the corresponding value(face) from imageDict. I assume we match value with with index, and somehow output the face. Maybe I need a third set or list to store results in before outputting?

imageDict = dict() # Contains index:face and looks like 1 👽 2 🐕 3 🐱 4 🚑 5 🚓 6 🐼 7 🐶 8 🐸 9 🐴 10 🐰 11 🐭 12 🐬 13 🐢 14 🐝



cardTuple = ({7, 42, 15, 47, 20, 52, 25, 30}, {3, 39, 14, 47, 55, 22, 23, 31})

My current approach:

newList = 

newList2 = 

for i in cardTuple:

      for j in i:

           if i == 1: ## maybe this needs to be 0?

               newList.append(imageDict[j])

           elif i == 2: ##  maybe 1?

               newList2.append(imageDict[j])

Any advice?

edited Nov 16 '18 at 14:42

asked Nov 16 '18 at 14:07

Karim

316

4

Are you aware that dict1 is not a dictionary? What is the expected output?

– Austin
Nov 16 '18 at 14:08

My advice is to take a look at How to Ask and understand how to provide a Minimal, Complete, and Verifiable example of your last attempt.

– Idlehands
Nov 16 '18 at 14:11

Sorry. My mistake. I fixed it to reference sets instead. I'll also try to follow IdleHands and add more details in a little while, providing more snippets from my code. Thanks!

– Karim
Nov 16 '18 at 14:31

If you use sets, the order of result might change as sets are unordered. I believe that's fine.

– Austin
Nov 16 '18 at 14:33

2

@Karim: To be clear, the meaning of curly braces ({}) depends on their content. If they're empty, or they store key-value pairs separated by colons (:), they're a dict. If they're non-empty, but don't have a colon in them, they're a set.

– ShadowRanger
Nov 16 '18 at 14:48

|
show 8 more comments

EDIT: Changed dictionaries to Sets, as I had not realised {} denoted a set. And fixed to say that sets are included in a tuple.

imageDict = dict() # Contains index:face and looks like 1 👽 2 🐕 3 🐱 4 🚑 5 🚓 6 🐼 7 🐶 8 🐸 9 🐴 10 🐰 11 🐭 12 🐬 13 🐢 14 🐝



cardTuple = ({7, 42, 15, 47, 20, 52, 25, 30}, {3, 39, 14, 47, 55, 22, 23, 31})

My current approach:

newList = 

newList2 = 

for i in cardTuple:

      for j in i:

           if i == 1: ## maybe this needs to be 0?

               newList.append(imageDict[j])

           elif i == 2: ##  maybe 1?

               newList2.append(imageDict[j])

Any advice?

edited Nov 16 '18 at 14:42

asked Nov 16 '18 at 14:07

Karim

316

4

Are you aware that dict1 is not a dictionary? What is the expected output?

– Austin
Nov 16 '18 at 14:08

My advice is to take a look at How to Ask and understand how to provide a Minimal, Complete, and Verifiable example of your last attempt.

– Idlehands
Nov 16 '18 at 14:11

Sorry. My mistake. I fixed it to reference sets instead. I'll also try to follow IdleHands and add more details in a little while, providing more snippets from my code. Thanks!

– Karim
Nov 16 '18 at 14:31

If you use sets, the order of result might change as sets are unordered. I believe that's fine.

– Austin
Nov 16 '18 at 14:33

2

@Karim: To be clear, the meaning of curly braces ({}) depends on their content. If they're empty, or they store key-value pairs separated by colons (:), they're a dict. If they're non-empty, but don't have a colon in them, they're a set.

– ShadowRanger
Nov 16 '18 at 14:48

|
show 8 more comments

EDIT: Changed dictionaries to Sets, as I had not realised {} denoted a set. And fixed to say that sets are included in a tuple.

imageDict = dict() # Contains index:face and looks like 1 👽 2 🐕 3 🐱 4 🚑 5 🚓 6 🐼 7 🐶 8 🐸 9 🐴 10 🐰 11 🐭 12 🐬 13 🐢 14 🐝



cardTuple = ({7, 42, 15, 47, 20, 52, 25, 30}, {3, 39, 14, 47, 55, 22, 23, 31})

My current approach:

newList = 

newList2 = 

for i in cardTuple:

      for j in i:

           if i == 1: ## maybe this needs to be 0?

               newList.append(imageDict[j])

           elif i == 2: ##  maybe 1?

               newList2.append(imageDict[j])

Any advice?

edited Nov 16 '18 at 14:42

asked Nov 16 '18 at 14:07

Karim

316

EDIT: Changed dictionaries to Sets, as I had not realised {} denoted a set. And fixed to say that sets are included in a tuple.

imageDict = dict() # Contains index:face and looks like 1 👽 2 🐕 3 🐱 4 🚑 5 🚓 6 🐼 7 🐶 8 🐸 9 🐴 10 🐰 11 🐭 12 🐬 13 🐢 14 🐝



cardTuple = ({7, 42, 15, 47, 20, 52, 25, 30}, {3, 39, 14, 47, 55, 22, 23, 31})

My current approach:

newList = 

newList2 = 

for i in cardTuple:

      for j in i:

           if i == 1: ## maybe this needs to be 0?

               newList.append(imageDict[j])

           elif i == 2: ##  maybe 1?

               newList2.append(imageDict[j])

Any advice?

python python-3.x dictionary replace output

edited Nov 16 '18 at 14:42

asked Nov 16 '18 at 14:07

Karim

316

edited Nov 16 '18 at 14:42

asked Nov 16 '18 at 14:07

Karim

316

edited Nov 16 '18 at 14:42

asked Nov 16 '18 at 14:07

Karim

316

asked Nov 16 '18 at 14:07

Karim

316

asked Nov 16 '18 at 14:07

Karim

316

4

Are you aware that dict1 is not a dictionary? What is the expected output?

– Austin
Nov 16 '18 at 14:08

My advice is to take a look at How to Ask and understand how to provide a Minimal, Complete, and Verifiable example of your last attempt.

– Idlehands
Nov 16 '18 at 14:11

Sorry. My mistake. I fixed it to reference sets instead. I'll also try to follow IdleHands and add more details in a little while, providing more snippets from my code. Thanks!

– Karim
Nov 16 '18 at 14:31

If you use sets, the order of result might change as sets are unordered. I believe that's fine.

– Austin
Nov 16 '18 at 14:33

2

@Karim: To be clear, the meaning of curly braces ({}) depends on their content. If they're empty, or they store key-value pairs separated by colons (:), they're a dict. If they're non-empty, but don't have a colon in them, they're a set.

– ShadowRanger
Nov 16 '18 at 14:48

|
show 8 more comments

4

Are you aware that dict1 is not a dictionary? What is the expected output?

– Austin
Nov 16 '18 at 14:08

My advice is to take a look at How to Ask and understand how to provide a Minimal, Complete, and Verifiable example of your last attempt.

– Idlehands
Nov 16 '18 at 14:11

Sorry. My mistake. I fixed it to reference sets instead. I'll also try to follow IdleHands and add more details in a little while, providing more snippets from my code. Thanks!

– Karim
Nov 16 '18 at 14:31

If you use sets, the order of result might change as sets are unordered. I believe that's fine.

– Austin
Nov 16 '18 at 14:33

2

@Karim: To be clear, the meaning of curly braces ({}) depends on their content. If they're empty, or they store key-value pairs separated by colons (:), they're a dict. If they're non-empty, but don't have a colon in them, they're a set.

– ShadowRanger
Nov 16 '18 at 14:48

Are you aware that dict1 is not a dictionary? What is the expected output?

– Austin
Nov 16 '18 at 14:08

My advice is to take a look at How to Ask and understand how to provide a Minimal, Complete, and Verifiable example of your last attempt.

– Idlehands
Nov 16 '18 at 14:11

Sorry. My mistake. I fixed it to reference sets instead. I'll also try to follow IdleHands and add more details in a little while, providing more snippets from my code. Thanks!

– Karim
Nov 16 '18 at 14:31

If you use sets, the order of result might change as sets are unordered. I believe that's fine.

– Austin
Nov 16 '18 at 14:33

@Karim: To be clear, the meaning of curly braces ({}) depends on their content. If they're empty, or they store key-value pairs separated by colons (:), they're a dict. If they're non-empty, but don't have a colon in them, they're a set.

– ShadowRanger
Nov 16 '18 at 14:48

|
show 8 more comments

1 Answer
1

active

oldest

votes

Firstly, as mentioned in the comments, dict1 is not a dict. It is a set. But you probably meant for it to be a list.
Here's a simple way to get what you want using list comprehensions:

mylist = [1,5,7,10,13]

mydict = {

          1:face1,2:face2,3:face3,4:face4,5:face5,6:face6,7:face7,

          8:face8,9:face9,10:face10,11:face11,12:face12,13:face13

         }



output = [mydict[key] for key in mylist]



>>> [face1, face5, face7, face10, face13]

answered Nov 16 '18 at 14:30

berkelem

9632721

So with output, you're creating a new list but I don't understand how the comparison works with "for key in mylist". Maybe you could explain what mydict[key] is doing? Is it comparing mydict key with key in mylist? If so, is it automatic that it outputs the value?

– Karim
Nov 16 '18 at 15:11

1

mydict[key] is extracting the value (face1,face2...) based on the key provided. In this case, the item in mylist. In this example it's essentially creating a list of mydict[1], mydict[5], mydict[7]... based on mylist = [1,5,7...]

– Idlehands
Nov 16 '18 at 15:33

1

dict objects are a collection of key: value pairs. By saying mydict[key] I am accessing the value associated with that key. By doing this in a for loop in a list comprehension, I am creating an output list type where each entry is the value associated with each key in mylist. You might want to look up list comprehensions - they are very useful Python constructions.

– berkelem
Nov 16 '18 at 15:54

Thank you! Much appreciated for the explanation. I'm going to try this solution than what I was working on now.

– Karim
Nov 16 '18 at 20:59

If this answered your question feel free to accept the answer :)

– berkelem
Nov 17 '18 at 8:37

add a comment |

Your Answer

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: "1"
};
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: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
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
});

}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53339438%2freplacing-values-from-one-set-with-values-from-another-if-the-value-of-set1-t%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

Firstly, as mentioned in the comments, dict1 is not a dict. It is a set. But you probably meant for it to be a list.
Here's a simple way to get what you want using list comprehensions:

mylist = [1,5,7,10,13]

mydict = {

          1:face1,2:face2,3:face3,4:face4,5:face5,6:face6,7:face7,

          8:face8,9:face9,10:face10,11:face11,12:face12,13:face13

         }



output = [mydict[key] for key in mylist]



>>> [face1, face5, face7, face10, face13]

answered Nov 16 '18 at 14:30

berkelem

9632721

So with output, you're creating a new list but I don't understand how the comparison works with "for key in mylist". Maybe you could explain what mydict[key] is doing? Is it comparing mydict key with key in mylist? If so, is it automatic that it outputs the value?

– Karim
Nov 16 '18 at 15:11

1

mydict[key] is extracting the value (face1,face2...) based on the key provided. In this case, the item in mylist. In this example it's essentially creating a list of mydict[1], mydict[5], mydict[7]... based on mylist = [1,5,7...]

– Idlehands
Nov 16 '18 at 15:33

1

dict objects are a collection of key: value pairs. By saying mydict[key] I am accessing the value associated with that key. By doing this in a for loop in a list comprehension, I am creating an output list type where each entry is the value associated with each key in mylist. You might want to look up list comprehensions - they are very useful Python constructions.

– berkelem
Nov 16 '18 at 15:54

Thank you! Much appreciated for the explanation. I'm going to try this solution than what I was working on now.

– Karim
Nov 16 '18 at 20:59

If this answered your question feel free to accept the answer :)

– berkelem
Nov 17 '18 at 8:37

add a comment |

Firstly, as mentioned in the comments, dict1 is not a dict. It is a set. But you probably meant for it to be a list.
Here's a simple way to get what you want using list comprehensions:

mylist = [1,5,7,10,13]

mydict = {

          1:face1,2:face2,3:face3,4:face4,5:face5,6:face6,7:face7,

          8:face8,9:face9,10:face10,11:face11,12:face12,13:face13

         }



output = [mydict[key] for key in mylist]



>>> [face1, face5, face7, face10, face13]

answered Nov 16 '18 at 14:30

berkelem

9632721

So with output, you're creating a new list but I don't understand how the comparison works with "for key in mylist". Maybe you could explain what mydict[key] is doing? Is it comparing mydict key with key in mylist? If so, is it automatic that it outputs the value?

– Karim
Nov 16 '18 at 15:11

1

mydict[key] is extracting the value (face1,face2...) based on the key provided. In this case, the item in mylist. In this example it's essentially creating a list of mydict[1], mydict[5], mydict[7]... based on mylist = [1,5,7...]

– Idlehands
Nov 16 '18 at 15:33

1

dict objects are a collection of key: value pairs. By saying mydict[key] I am accessing the value associated with that key. By doing this in a for loop in a list comprehension, I am creating an output list type where each entry is the value associated with each key in mylist. You might want to look up list comprehensions - they are very useful Python constructions.

– berkelem
Nov 16 '18 at 15:54

Thank you! Much appreciated for the explanation. I'm going to try this solution than what I was working on now.

– Karim
Nov 16 '18 at 20:59

If this answered your question feel free to accept the answer :)

– berkelem
Nov 17 '18 at 8:37

add a comment |

Firstly, as mentioned in the comments, dict1 is not a dict. It is a set. But you probably meant for it to be a list.
Here's a simple way to get what you want using list comprehensions:

mylist = [1,5,7,10,13]

mydict = {

          1:face1,2:face2,3:face3,4:face4,5:face5,6:face6,7:face7,

          8:face8,9:face9,10:face10,11:face11,12:face12,13:face13

         }



output = [mydict[key] for key in mylist]



>>> [face1, face5, face7, face10, face13]

answered Nov 16 '18 at 14:30

berkelem

9632721

Firstly, as mentioned in the comments, dict1 is not a dict. It is a set. But you probably meant for it to be a list.
Here's a simple way to get what you want using list comprehensions:

mylist = [1,5,7,10,13]

mydict = {

          1:face1,2:face2,3:face3,4:face4,5:face5,6:face6,7:face7,

          8:face8,9:face9,10:face10,11:face11,12:face12,13:face13

         }



output = [mydict[key] for key in mylist]



>>> [face1, face5, face7, face10, face13]

answered Nov 16 '18 at 14:30

berkelem

9632721

answered Nov 16 '18 at 14:30

berkelem

9632721

answered Nov 16 '18 at 14:30

berkelem

9632721

answered Nov 16 '18 at 14:30

berkelem

9632721

So with output, you're creating a new list but I don't understand how the comparison works with "for key in mylist". Maybe you could explain what mydict[key] is doing? Is it comparing mydict key with key in mylist? If so, is it automatic that it outputs the value?

– Karim
Nov 16 '18 at 15:11

1

mydict[key] is extracting the value (face1,face2...) based on the key provided. In this case, the item in mylist. In this example it's essentially creating a list of mydict[1], mydict[5], mydict[7]... based on mylist = [1,5,7...]

– Idlehands
Nov 16 '18 at 15:33

1

dict objects are a collection of key: value pairs. By saying mydict[key] I am accessing the value associated with that key. By doing this in a for loop in a list comprehension, I am creating an output list type where each entry is the value associated with each key in mylist. You might want to look up list comprehensions - they are very useful Python constructions.

– berkelem
Nov 16 '18 at 15:54

Thank you! Much appreciated for the explanation. I'm going to try this solution than what I was working on now.

– Karim
Nov 16 '18 at 20:59

If this answered your question feel free to accept the answer :)

– berkelem
Nov 17 '18 at 8:37

add a comment |

So with output, you're creating a new list but I don't understand how the comparison works with "for key in mylist". Maybe you could explain what mydict[key] is doing? Is it comparing mydict key with key in mylist? If so, is it automatic that it outputs the value?

– Karim
Nov 16 '18 at 15:11

1

mydict[key] is extracting the value (face1,face2...) based on the key provided. In this case, the item in mylist. In this example it's essentially creating a list of mydict[1], mydict[5], mydict[7]... based on mylist = [1,5,7...]

– Idlehands
Nov 16 '18 at 15:33

1

dict objects are a collection of key: value pairs. By saying mydict[key] I am accessing the value associated with that key. By doing this in a for loop in a list comprehension, I am creating an output list type where each entry is the value associated with each key in mylist. You might want to look up list comprehensions - they are very useful Python constructions.

– berkelem
Nov 16 '18 at 15:54

Thank you! Much appreciated for the explanation. I'm going to try this solution than what I was working on now.

– Karim
Nov 16 '18 at 20:59

If this answered your question feel free to accept the answer :)

– berkelem
Nov 17 '18 at 8:37

So with output, you're creating a new list but I don't understand how the comparison works with "for key in mylist". Maybe you could explain what mydict[key] is doing? Is it comparing mydict key with key in mylist? If so, is it automatic that it outputs the value?

– Karim
Nov 16 '18 at 15:11

mydict[key] is extracting the value (face1,face2...) based on the key provided. In this case, the item in mylist. In this example it's essentially creating a list of mydict[1], mydict[5], mydict[7]... based on mylist = [1,5,7...]

– Idlehands
Nov 16 '18 at 15:33

dict objects are a collection of key: value pairs. By saying mydict[key] I am accessing the value associated with that key. By doing this in a for loop in a list comprehension, I am creating an output list type where each entry is the value associated with each key in mylist. You might want to look up list comprehensions - they are very useful Python constructions.

– berkelem
Nov 16 '18 at 15:54

Thank you! Much appreciated for the explanation. I'm going to try this solution than what I was working on now.

– Karim
Nov 16 '18 at 20:59

If this answered your question feel free to accept the answer :)

– berkelem
Nov 17 '18 at 8:37

add a comment |

draft saved

draft discarded

Thanks for contributing an answer to Stack Overflow!

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.

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');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

搜尋此網誌

Ndtyjky