switching keys and values in a dictionary in python

Say I have a dictionary like so:

my_dict = {2:3, 5:6, 8:9}

Is there a way that I can switch the keys and values to get:

{3:2, 6:5, 9:8}

edited Nov 24 '18 at 11:06

ViAik

5019

asked Nov 29 '11 at 3:36

me45

3193714

1

Are the values unique? Are the values hashable?

– wim
Nov 29 '11 at 3:45

add a comment |

Say I have a dictionary like so:

my_dict = {2:3, 5:6, 8:9}

Is there a way that I can switch the keys and values to get:

{3:2, 6:5, 9:8}

edited Nov 24 '18 at 11:06

ViAik

5019

asked Nov 29 '11 at 3:36

me45

3193714

1

Are the values unique? Are the values hashable?

– wim
Nov 29 '11 at 3:45

add a comment |

Say I have a dictionary like so:

my_dict = {2:3, 5:6, 8:9}

Is there a way that I can switch the keys and values to get:

{3:2, 6:5, 9:8}

edited Nov 24 '18 at 11:06

ViAik

5019

asked Nov 29 '11 at 3:36

me45

3193714

Say I have a dictionary like so:

my_dict = {2:3, 5:6, 8:9}

Is there a way that I can switch the keys and values to get:

{3:2, 6:5, 9:8}

python dictionary key

edited Nov 24 '18 at 11:06

ViAik

5019

asked Nov 29 '11 at 3:36

me45

3193714

edited Nov 24 '18 at 11:06

ViAik

5019

asked Nov 29 '11 at 3:36

me45

3193714

edited Nov 24 '18 at 11:06

ViAik

5019

edited Nov 24 '18 at 11:06

ViAik

5019

edited Nov 24 '18 at 11:06

ViAik

5019

asked Nov 29 '11 at 3:36

me45

3193714

asked Nov 29 '11 at 3:36

me45

3193714

asked Nov 29 '11 at 3:36

me45

3193714

1

Are the values unique? Are the values hashable?

– wim
Nov 29 '11 at 3:45

add a comment |

1

Are the values unique? Are the values hashable?

– wim
Nov 29 '11 at 3:45

Are the values unique? Are the values hashable?

– wim
Nov 29 '11 at 3:45

add a comment |

8 Answers
8

active

oldest

votes

my_dict2 = dict((y,x) for x,y in my_dict.iteritems())

If you are using python 2.7 or 3.x you can use a dictionary comprehension instead:

my_dict2 = {y:x for x,y in my_dict.iteritems()}

Edit

As noted in the comments by JBernardo, for python 3.x you need to use items instead of iteritems

edited Nov 29 '11 at 3:48

answered Nov 29 '11 at 3:39

GWW

33k68796

3

.iteritems() is not valid on Python 3.x...

– JBernardo
Nov 29 '11 at 3:47

@JBernardo: Fixed thanks I forgot about that

– GWW
Nov 29 '11 at 3:48

1

another correction - doing the dictionary comprehension you suggest {(y, x) for x, y in dic_cols_w.items()} returns a set, not a dictionary. You would still need to wrap this in a dict(...) to get a dictionary

– tsando
Apr 5 '17 at 13:06

3

@tsando: I think you have read my answer incorrectly. I have {y:x ...

– GWW
Apr 5 '17 at 22:34

How would you do this if the values were a list of items and each item in the list should be a key?

– Woody Pride
Feb 6 at 14:49

|
show 1 more comment

Use this code (trivially modified) from the accepted answer at Python reverse / invert a mapping:

dict((v,k) for k, v in my_dict.iteritems())

Note that this assumes that the values in the original dictionary are unique. Otherwise you'd end up with duplicate keys in the resulting dictionary, and that is not allowed.

And, as @wim points out, it also assumes the values are hashable. See the Python glossary if you're not sure what is and isn't hashable.

edited May 23 '17 at 12:10

Community♦

answered Nov 29 '11 at 3:41

Trott

37.9k19104152

add a comment |

Try this:

my_dict = {2:3, 5:6, 8:9}



new_dict = {}

for k, v in my_dict.items():

    new_dict[v] = k

answered Nov 29 '11 at 3:40

Óscar López

178k25227323

add a comment |

maybe:

flipped_dict = dict(zip(my_dict.values(), my_dict.keys()))

answered Nov 29 '11 at 3:41

aaron

835812

2

this is probably not safe as it relies on key and value ordering, although it happened to work for me. I like JBernardo/GWW's answer(s).

– aaron
Nov 29 '11 at 3:43

It's probably better to use iterkeys and itervalues

– GWW
Nov 29 '11 at 3:44

That's safe if the dict isn't changed during the operation. I liked because you did it without a comprehension.

– JBernardo
Nov 29 '11 at 3:51

add a comment |

Sometimes, the condition that the values are all unique will not hold, in which case, the answers above will destroy any duplicate values.

The following rolls the values that might be duplicates up into a list:

from itertools import count

dict([(a,[list(d.keys())[i] for i,j in zip(count(), d.values())if j==a in set(d.values())])

I'm sure there's a better (non-list-comp) method, but I had a problem with the earlier answers, so thought I'd provide my solution in case others have a similar use-case.

P.S. Don't expect the dict to remain neatly arranged after any changes to the original! This method is a one-time use only on a static dict - you have been warned!

answered Jul 29 '16 at 20:48

Thomas Kimber

3,36921323

add a comment |

my_dict = { my_dict[k]:k for k in my_dict}

edited Jul 19 '16 at 11:39

Ingo Karkat

132k14148199

answered Jul 19 '16 at 11:21

VladEg

111

add a comment |

First of all it is not guaranteed that this is possible, since the values of a dictionary can be unhashable.

In case these are not, we can use a functional approach with:

reversed_dict = dict(map(reversed, original_dict.items()))

answered Jan 19 '18 at 10:25

Willem Van Onsem

148k16142232

add a comment |

If the values are not unique this will collide the key space in conversion.
Best is to keep the keys in list when switching places

below handles this -

RvsD = dict()

for k,v in MyDict.iteritems():

    RsvD.setdefault(v, ).append(k)

answered Dec 23 '18 at 3:56

mungayree

1118

add a comment |

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8 Answers
8

active

oldest

votes

8 Answers
8

active

oldest

votes

my_dict2 = dict((y,x) for x,y in my_dict.iteritems())

If you are using python 2.7 or 3.x you can use a dictionary comprehension instead:

my_dict2 = {y:x for x,y in my_dict.iteritems()}

Edit

As noted in the comments by JBernardo, for python 3.x you need to use items instead of iteritems

edited Nov 29 '11 at 3:48

answered Nov 29 '11 at 3:39

GWW

33k68796

3

.iteritems() is not valid on Python 3.x...

– JBernardo
Nov 29 '11 at 3:47

@JBernardo: Fixed thanks I forgot about that

– GWW
Nov 29 '11 at 3:48

1

another correction - doing the dictionary comprehension you suggest {(y, x) for x, y in dic_cols_w.items()} returns a set, not a dictionary. You would still need to wrap this in a dict(...) to get a dictionary

– tsando
Apr 5 '17 at 13:06

3

@tsando: I think you have read my answer incorrectly. I have {y:x ...

– GWW
Apr 5 '17 at 22:34

How would you do this if the values were a list of items and each item in the list should be a key?

– Woody Pride
Feb 6 at 14:49

|
show 1 more comment

my_dict2 = dict((y,x) for x,y in my_dict.iteritems())

If you are using python 2.7 or 3.x you can use a dictionary comprehension instead:

my_dict2 = {y:x for x,y in my_dict.iteritems()}

Edit

As noted in the comments by JBernardo, for python 3.x you need to use items instead of iteritems

edited Nov 29 '11 at 3:48

answered Nov 29 '11 at 3:39

GWW

33k68796

3

.iteritems() is not valid on Python 3.x...

– JBernardo
Nov 29 '11 at 3:47

@JBernardo: Fixed thanks I forgot about that

– GWW
Nov 29 '11 at 3:48

1

another correction - doing the dictionary comprehension you suggest {(y, x) for x, y in dic_cols_w.items()} returns a set, not a dictionary. You would still need to wrap this in a dict(...) to get a dictionary

– tsando
Apr 5 '17 at 13:06

3

@tsando: I think you have read my answer incorrectly. I have {y:x ...

– GWW
Apr 5 '17 at 22:34

How would you do this if the values were a list of items and each item in the list should be a key?

– Woody Pride
Feb 6 at 14:49

|
show 1 more comment

my_dict2 = dict((y,x) for x,y in my_dict.iteritems())

If you are using python 2.7 or 3.x you can use a dictionary comprehension instead:

my_dict2 = {y:x for x,y in my_dict.iteritems()}

Edit

As noted in the comments by JBernardo, for python 3.x you need to use items instead of iteritems

edited Nov 29 '11 at 3:48

answered Nov 29 '11 at 3:39

GWW

33k68796

my_dict2 = dict((y,x) for x,y in my_dict.iteritems())

If you are using python 2.7 or 3.x you can use a dictionary comprehension instead:

my_dict2 = {y:x for x,y in my_dict.iteritems()}

Edit

As noted in the comments by JBernardo, for python 3.x you need to use items instead of iteritems

edited Nov 29 '11 at 3:48

answered Nov 29 '11 at 3:39

GWW

33k68796

edited Nov 29 '11 at 3:48

answered Nov 29 '11 at 3:39

GWW

33k68796

answered Nov 29 '11 at 3:39

GWW

33k68796

answered Nov 29 '11 at 3:39

GWW

33k68796

3

.iteritems() is not valid on Python 3.x...

– JBernardo
Nov 29 '11 at 3:47

@JBernardo: Fixed thanks I forgot about that

– GWW
Nov 29 '11 at 3:48

1

another correction - doing the dictionary comprehension you suggest {(y, x) for x, y in dic_cols_w.items()} returns a set, not a dictionary. You would still need to wrap this in a dict(...) to get a dictionary

– tsando
Apr 5 '17 at 13:06

3

@tsando: I think you have read my answer incorrectly. I have {y:x ...

– GWW
Apr 5 '17 at 22:34

How would you do this if the values were a list of items and each item in the list should be a key?

– Woody Pride
Feb 6 at 14:49

|
show 1 more comment

3

.iteritems() is not valid on Python 3.x...

– JBernardo
Nov 29 '11 at 3:47

@JBernardo: Fixed thanks I forgot about that

– GWW
Nov 29 '11 at 3:48

1

another correction - doing the dictionary comprehension you suggest {(y, x) for x, y in dic_cols_w.items()} returns a set, not a dictionary. You would still need to wrap this in a dict(...) to get a dictionary

– tsando
Apr 5 '17 at 13:06

3

@tsando: I think you have read my answer incorrectly. I have {y:x ...

– GWW
Apr 5 '17 at 22:34

How would you do this if the values were a list of items and each item in the list should be a key?

– Woody Pride
Feb 6 at 14:49

.iteritems() is not valid on Python 3.x...

– JBernardo
Nov 29 '11 at 3:47

@JBernardo: Fixed thanks I forgot about that

– GWW
Nov 29 '11 at 3:48

another correction - doing the dictionary comprehension you suggest {(y, x) for x, y in dic_cols_w.items()} returns a set, not a dictionary. You would still need to wrap this in a dict(...) to get a dictionary

– tsando
Apr 5 '17 at 13:06

@tsando: I think you have read my answer incorrectly. I have {y:x ...

– GWW
Apr 5 '17 at 22:34

How would you do this if the values were a list of items and each item in the list should be a key?

– Woody Pride
Feb 6 at 14:49

|
show 1 more comment

Use this code (trivially modified) from the accepted answer at Python reverse / invert a mapping:

dict((v,k) for k, v in my_dict.iteritems())

Note that this assumes that the values in the original dictionary are unique. Otherwise you'd end up with duplicate keys in the resulting dictionary, and that is not allowed.

And, as @wim points out, it also assumes the values are hashable. See the Python glossary if you're not sure what is and isn't hashable.

edited May 23 '17 at 12:10

Community♦

answered Nov 29 '11 at 3:41

Trott

37.9k19104152

add a comment |

Use this code (trivially modified) from the accepted answer at Python reverse / invert a mapping:

dict((v,k) for k, v in my_dict.iteritems())

Note that this assumes that the values in the original dictionary are unique. Otherwise you'd end up with duplicate keys in the resulting dictionary, and that is not allowed.

And, as @wim points out, it also assumes the values are hashable. See the Python glossary if you're not sure what is and isn't hashable.

edited May 23 '17 at 12:10

Community♦

answered Nov 29 '11 at 3:41

Trott

37.9k19104152

add a comment |

Use this code (trivially modified) from the accepted answer at Python reverse / invert a mapping:

dict((v,k) for k, v in my_dict.iteritems())

Note that this assumes that the values in the original dictionary are unique. Otherwise you'd end up with duplicate keys in the resulting dictionary, and that is not allowed.

And, as @wim points out, it also assumes the values are hashable. See the Python glossary if you're not sure what is and isn't hashable.

edited May 23 '17 at 12:10

Community♦

answered Nov 29 '11 at 3:41

Trott

37.9k19104152

Use this code (trivially modified) from the accepted answer at Python reverse / invert a mapping:

dict((v,k) for k, v in my_dict.iteritems())

Note that this assumes that the values in the original dictionary are unique. Otherwise you'd end up with duplicate keys in the resulting dictionary, and that is not allowed.

And, as @wim points out, it also assumes the values are hashable. See the Python glossary if you're not sure what is and isn't hashable.

edited May 23 '17 at 12:10

Community♦

answered Nov 29 '11 at 3:41

Trott

37.9k19104152

edited May 23 '17 at 12:10

Community♦

edited May 23 '17 at 12:10

Community♦

edited May 23 '17 at 12:10

Community♦

answered Nov 29 '11 at 3:41

Trott

37.9k19104152

answered Nov 29 '11 at 3:41

Trott

37.9k19104152

answered Nov 29 '11 at 3:41

Trott

37.9k19104152

add a comment |

Try this:

my_dict = {2:3, 5:6, 8:9}



new_dict = {}

for k, v in my_dict.items():

    new_dict[v] = k

answered Nov 29 '11 at 3:40

Óscar López

178k25227323

add a comment |

Try this:

my_dict = {2:3, 5:6, 8:9}



new_dict = {}

for k, v in my_dict.items():

    new_dict[v] = k

answered Nov 29 '11 at 3:40

Óscar López

178k25227323

add a comment |

Try this:

my_dict = {2:3, 5:6, 8:9}



new_dict = {}

for k, v in my_dict.items():

    new_dict[v] = k

answered Nov 29 '11 at 3:40

Óscar López

178k25227323

Try this:

my_dict = {2:3, 5:6, 8:9}



new_dict = {}

for k, v in my_dict.items():

    new_dict[v] = k

answered Nov 29 '11 at 3:40

Óscar López

178k25227323

answered Nov 29 '11 at 3:40

Óscar López

178k25227323

answered Nov 29 '11 at 3:40

Óscar López

178k25227323

answered Nov 29 '11 at 3:40

Óscar López

178k25227323

add a comment |

maybe:

flipped_dict = dict(zip(my_dict.values(), my_dict.keys()))

answered Nov 29 '11 at 3:41

aaron

835812

2

this is probably not safe as it relies on key and value ordering, although it happened to work for me. I like JBernardo/GWW's answer(s).

– aaron
Nov 29 '11 at 3:43

It's probably better to use iterkeys and itervalues

– GWW
Nov 29 '11 at 3:44

That's safe if the dict isn't changed during the operation. I liked because you did it without a comprehension.

– JBernardo
Nov 29 '11 at 3:51

add a comment |

maybe:

flipped_dict = dict(zip(my_dict.values(), my_dict.keys()))

answered Nov 29 '11 at 3:41

aaron

835812

2

this is probably not safe as it relies on key and value ordering, although it happened to work for me. I like JBernardo/GWW's answer(s).

– aaron
Nov 29 '11 at 3:43

It's probably better to use iterkeys and itervalues

– GWW
Nov 29 '11 at 3:44

That's safe if the dict isn't changed during the operation. I liked because you did it without a comprehension.

– JBernardo
Nov 29 '11 at 3:51

add a comment |

maybe:

flipped_dict = dict(zip(my_dict.values(), my_dict.keys()))

answered Nov 29 '11 at 3:41

aaron

835812

maybe:

flipped_dict = dict(zip(my_dict.values(), my_dict.keys()))

answered Nov 29 '11 at 3:41

aaron

835812

answered Nov 29 '11 at 3:41

aaron

835812

answered Nov 29 '11 at 3:41

aaron

835812

answered Nov 29 '11 at 3:41

aaron

835812

2

this is probably not safe as it relies on key and value ordering, although it happened to work for me. I like JBernardo/GWW's answer(s).

– aaron
Nov 29 '11 at 3:43

It's probably better to use iterkeys and itervalues

– GWW
Nov 29 '11 at 3:44

That's safe if the dict isn't changed during the operation. I liked because you did it without a comprehension.

– JBernardo
Nov 29 '11 at 3:51

add a comment |

2

this is probably not safe as it relies on key and value ordering, although it happened to work for me. I like JBernardo/GWW's answer(s).

– aaron
Nov 29 '11 at 3:43

It's probably better to use iterkeys and itervalues

– GWW
Nov 29 '11 at 3:44

That's safe if the dict isn't changed during the operation. I liked because you did it without a comprehension.

– JBernardo
Nov 29 '11 at 3:51

this is probably not safe as it relies on key and value ordering, although it happened to work for me. I like JBernardo/GWW's answer(s).

– aaron
Nov 29 '11 at 3:43

It's probably better to use iterkeys and itervalues

– GWW
Nov 29 '11 at 3:44

That's safe if the dict isn't changed during the operation. I liked because you did it without a comprehension.

– JBernardo
Nov 29 '11 at 3:51

add a comment |

Sometimes, the condition that the values are all unique will not hold, in which case, the answers above will destroy any duplicate values.

The following rolls the values that might be duplicates up into a list:

from itertools import count

dict([(a,[list(d.keys())[i] for i,j in zip(count(), d.values())if j==a in set(d.values())])

I'm sure there's a better (non-list-comp) method, but I had a problem with the earlier answers, so thought I'd provide my solution in case others have a similar use-case.

P.S. Don't expect the dict to remain neatly arranged after any changes to the original! This method is a one-time use only on a static dict - you have been warned!

answered Jul 29 '16 at 20:48

Thomas Kimber

3,36921323

add a comment |

Sometimes, the condition that the values are all unique will not hold, in which case, the answers above will destroy any duplicate values.

The following rolls the values that might be duplicates up into a list:

from itertools import count

dict([(a,[list(d.keys())[i] for i,j in zip(count(), d.values())if j==a in set(d.values())])

I'm sure there's a better (non-list-comp) method, but I had a problem with the earlier answers, so thought I'd provide my solution in case others have a similar use-case.

P.S. Don't expect the dict to remain neatly arranged after any changes to the original! This method is a one-time use only on a static dict - you have been warned!

answered Jul 29 '16 at 20:48

Thomas Kimber

3,36921323

add a comment |

Sometimes, the condition that the values are all unique will not hold, in which case, the answers above will destroy any duplicate values.

The following rolls the values that might be duplicates up into a list:

from itertools import count

dict([(a,[list(d.keys())[i] for i,j in zip(count(), d.values())if j==a in set(d.values())])

I'm sure there's a better (non-list-comp) method, but I had a problem with the earlier answers, so thought I'd provide my solution in case others have a similar use-case.

P.S. Don't expect the dict to remain neatly arranged after any changes to the original! This method is a one-time use only on a static dict - you have been warned!

answered Jul 29 '16 at 20:48

Thomas Kimber

3,36921323

Sometimes, the condition that the values are all unique will not hold, in which case, the answers above will destroy any duplicate values.

The following rolls the values that might be duplicates up into a list:

from itertools import count

dict([(a,[list(d.keys())[i] for i,j in zip(count(), d.values())if j==a in set(d.values())])

I'm sure there's a better (non-list-comp) method, but I had a problem with the earlier answers, so thought I'd provide my solution in case others have a similar use-case.

P.S. Don't expect the dict to remain neatly arranged after any changes to the original! This method is a one-time use only on a static dict - you have been warned!

answered Jul 29 '16 at 20:48

Thomas Kimber

3,36921323

answered Jul 29 '16 at 20:48

Thomas Kimber

3,36921323

answered Jul 29 '16 at 20:48

Thomas Kimber

3,36921323

answered Jul 29 '16 at 20:48

Thomas Kimber

3,36921323

add a comment |

my_dict = { my_dict[k]:k for k in my_dict}

edited Jul 19 '16 at 11:39

Ingo Karkat

132k14148199

answered Jul 19 '16 at 11:21

VladEg

111

add a comment |

my_dict = { my_dict[k]:k for k in my_dict}

edited Jul 19 '16 at 11:39

Ingo Karkat

132k14148199

answered Jul 19 '16 at 11:21

VladEg

111

add a comment |

my_dict = { my_dict[k]:k for k in my_dict}

edited Jul 19 '16 at 11:39

Ingo Karkat

132k14148199

answered Jul 19 '16 at 11:21

VladEg

111

my_dict = { my_dict[k]:k for k in my_dict}

edited Jul 19 '16 at 11:39

Ingo Karkat

132k14148199

answered Jul 19 '16 at 11:21

VladEg

111

edited Jul 19 '16 at 11:39

Ingo Karkat

132k14148199

edited Jul 19 '16 at 11:39

Ingo Karkat

132k14148199

edited Jul 19 '16 at 11:39

Ingo Karkat

132k14148199

answered Jul 19 '16 at 11:21

VladEg

111

answered Jul 19 '16 at 11:21

VladEg

111

answered Jul 19 '16 at 11:21

VladEg

111

add a comment |

First of all it is not guaranteed that this is possible, since the values of a dictionary can be unhashable.

In case these are not, we can use a functional approach with:

reversed_dict = dict(map(reversed, original_dict.items()))

answered Jan 19 '18 at 10:25

Willem Van Onsem

148k16142232

add a comment |

First of all it is not guaranteed that this is possible, since the values of a dictionary can be unhashable.

In case these are not, we can use a functional approach with:

reversed_dict = dict(map(reversed, original_dict.items()))

answered Jan 19 '18 at 10:25

Willem Van Onsem

148k16142232

add a comment |

First of all it is not guaranteed that this is possible, since the values of a dictionary can be unhashable.

In case these are not, we can use a functional approach with:

reversed_dict = dict(map(reversed, original_dict.items()))

answered Jan 19 '18 at 10:25

Willem Van Onsem

148k16142232

First of all it is not guaranteed that this is possible, since the values of a dictionary can be unhashable.

In case these are not, we can use a functional approach with:

reversed_dict = dict(map(reversed, original_dict.items()))

answered Jan 19 '18 at 10:25

Willem Van Onsem

148k16142232

answered Jan 19 '18 at 10:25

Willem Van Onsem

148k16142232

answered Jan 19 '18 at 10:25

Willem Van Onsem

148k16142232

answered Jan 19 '18 at 10:25

Willem Van Onsem

148k16142232

add a comment |

If the values are not unique this will collide the key space in conversion.
Best is to keep the keys in list when switching places

below handles this -

RvsD = dict()

for k,v in MyDict.iteritems():

    RsvD.setdefault(v, ).append(k)

answered Dec 23 '18 at 3:56

mungayree

1118

add a comment |

If the values are not unique this will collide the key space in conversion.
Best is to keep the keys in list when switching places

below handles this -

RvsD = dict()

for k,v in MyDict.iteritems():

    RsvD.setdefault(v, ).append(k)

answered Dec 23 '18 at 3:56

mungayree

1118

add a comment |

If the values are not unique this will collide the key space in conversion.
Best is to keep the keys in list when switching places

below handles this -

RvsD = dict()

for k,v in MyDict.iteritems():

    RsvD.setdefault(v, ).append(k)

answered Dec 23 '18 at 3:56

mungayree

1118

If the values are not unique this will collide the key space in conversion.
Best is to keep the keys in list when switching places

below handles this -

RvsD = dict()

for k,v in MyDict.iteritems():

    RsvD.setdefault(v, ).append(k)

answered Dec 23 '18 at 3:56

mungayree

1118

answered Dec 23 '18 at 3:56

mungayree

1118

answered Dec 23 '18 at 3:56

mungayree

1118

answered Dec 23 '18 at 3:56

mungayree

1118

add a comment |

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