Reverse dictionary where values are lists

I'm working on the following problem: I've got a dictionary like this one:

dic={0:[0,1,2],1:[3,4,5]}

And I want to reverse it so it looks like this:

dic2={0:0,1:0,2:0,3:1,4:1,5:1}

I managed to make it, but doing this:

dic2={}

for i in dic:

    for j in dic[i]:

        dic2[j]=i

I know about list and dict comprehensions and this code reeks of it, but I'm not good at them when there are nested for and dicts. How would you make it more efficiently?

edited 5 hours ago

Sᴀᴍ Onᴇᴌᴀ

10.1k62167

asked 6 hours ago

Juan C

1235

New contributor

add a comment |

I'm working on the following problem: I've got a dictionary like this one:

dic={0:[0,1,2],1:[3,4,5]}

And I want to reverse it so it looks like this:

dic2={0:0,1:0,2:0,3:1,4:1,5:1}

I managed to make it, but doing this:

dic2={}

for i in dic:

    for j in dic[i]:

        dic2[j]=i

I know about list and dict comprehensions and this code reeks of it, but I'm not good at them when there are nested for and dicts. How would you make it more efficiently?

edited 5 hours ago

Sᴀᴍ Onᴇᴌᴀ

10.1k62167

asked 6 hours ago

Juan C

1235

New contributor

add a comment |

I'm working on the following problem: I've got a dictionary like this one:

dic={0:[0,1,2],1:[3,4,5]}

And I want to reverse it so it looks like this:

dic2={0:0,1:0,2:0,3:1,4:1,5:1}

I managed to make it, but doing this:

dic2={}

for i in dic:

    for j in dic[i]:

        dic2[j]=i

I know about list and dict comprehensions and this code reeks of it, but I'm not good at them when there are nested for and dicts. How would you make it more efficiently?

edited 5 hours ago

Sᴀᴍ Onᴇᴌᴀ

10.1k62167

asked 6 hours ago

Juan C

1235

New contributor

I'm working on the following problem: I've got a dictionary like this one:

dic={0:[0,1,2],1:[3,4,5]}

And I want to reverse it so it looks like this:

dic2={0:0,1:0,2:0,3:1,4:1,5:1}

I managed to make it, but doing this:

dic2={}

for i in dic:

    for j in dic[i]:

        dic2[j]=i

I know about list and dict comprehensions and this code reeks of it, but I'm not good at them when there are nested for and dicts. How would you make it more efficiently?

python dictionary

edited 5 hours ago

Sᴀᴍ Onᴇᴌᴀ

10.1k62167

asked 6 hours ago

Juan C

1235

New contributor

edited 5 hours ago

Sᴀᴍ Onᴇᴌᴀ

10.1k62167

asked 6 hours ago

Juan C

1235

New contributor

edited 5 hours ago

Sᴀᴍ Onᴇᴌᴀ

10.1k62167

edited 5 hours ago

Sᴀᴍ Onᴇᴌᴀ

10.1k62167

edited 5 hours ago

Sᴀᴍ Onᴇᴌᴀ

10.1k62167

asked 6 hours ago

Juan C

1235

New contributor

asked 6 hours ago

Juan C

1235

asked 6 hours ago

Juan C

1235

New contributor

Juan C 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 |

1 Answer
1

active

oldest

votes

There's really not much to say about this code, it is straightforward.

Style

These are only nitpicks.

Generic dictionary keys are typically named k instead of i or j, but in a specific application a more descriptive name would be even better.

Collections should be named by their purpose in the application, not their type.

By convention, assignments and other binary operators should be surrounded by spaces: a = b, not a=b. See PEP 8 for reference, if you're not already aware of it. Following this style guide makes your code easier to read for others.

Code improvements

When iterating over the keys and values of a dictionary at the same time, you can use

for k, v in dic.items():

    # ... use k, v ...

instead of

for k in dic:

    v = dic[k]

    # ...

The nested loop can be transformed to a dictionary comprehension like this:

dic2 = {v: k for k, values in dic.items() for v in values}

You can remember that the order of for clauses in the comprehension is the same as the order of corresponding nested for loops.

Potential pitfalls

You should be aware that this transformation from a dictionary of lists to a reverse dictionary only works if all items in the original lists are unique. Counterexample:

>>> dic = {0: [1, 2], 1: [2, 3]}

>>> {v: k for k, values in dic.items() for v in values}

{1: 0, 2: 1, 3: 1}  # missing 2: 0

To correct this, the output dictionary should have the same format as the input, namely mapping each of the items in the input lists to a list of corresponding keys.

If the input represents a directed graph (mapping nodes to lists of neighbours), this corresponds to computing the transposed or reversed graph. It can be done by using a collections.defaultdict. I don't see an easy way to write it as a comprehension in this case.

from collections import defaultdict



graph = {0: [1, 2], 1: [2, 3]}



transposed_graph = defaultdict(list)

for node, neighbours in graph.items():

    for neighbour in neighbours:

        transposed_graph[neighbour].append(node)



# {1: [0], 2: [0, 1], 3: [1]}

answered 5 hours ago

mkrieger1

1,4231824

add a comment |

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

There's really not much to say about this code, it is straightforward.

Style

These are only nitpicks.

Generic dictionary keys are typically named k instead of i or j, but in a specific application a more descriptive name would be even better.

Collections should be named by their purpose in the application, not their type.

By convention, assignments and other binary operators should be surrounded by spaces: a = b, not a=b. See PEP 8 for reference, if you're not already aware of it. Following this style guide makes your code easier to read for others.

Code improvements

When iterating over the keys and values of a dictionary at the same time, you can use

for k, v in dic.items():

    # ... use k, v ...

instead of

for k in dic:

    v = dic[k]

    # ...

The nested loop can be transformed to a dictionary comprehension like this:

dic2 = {v: k for k, values in dic.items() for v in values}

You can remember that the order of for clauses in the comprehension is the same as the order of corresponding nested for loops.

Potential pitfalls

You should be aware that this transformation from a dictionary of lists to a reverse dictionary only works if all items in the original lists are unique. Counterexample:

>>> dic = {0: [1, 2], 1: [2, 3]}

>>> {v: k for k, values in dic.items() for v in values}

{1: 0, 2: 1, 3: 1}  # missing 2: 0

To correct this, the output dictionary should have the same format as the input, namely mapping each of the items in the input lists to a list of corresponding keys.

from collections import defaultdict



graph = {0: [1, 2], 1: [2, 3]}



transposed_graph = defaultdict(list)

for node, neighbours in graph.items():

    for neighbour in neighbours:

        transposed_graph[neighbour].append(node)



# {1: [0], 2: [0, 1], 3: [1]}

answered 5 hours ago

mkrieger1

1,4231824

add a comment |

There's really not much to say about this code, it is straightforward.

Style

These are only nitpicks.

Generic dictionary keys are typically named k instead of i or j, but in a specific application a more descriptive name would be even better.

Collections should be named by their purpose in the application, not their type.

By convention, assignments and other binary operators should be surrounded by spaces: a = b, not a=b. See PEP 8 for reference, if you're not already aware of it. Following this style guide makes your code easier to read for others.

Code improvements

When iterating over the keys and values of a dictionary at the same time, you can use

for k, v in dic.items():

    # ... use k, v ...

instead of

for k in dic:

    v = dic[k]

    # ...

The nested loop can be transformed to a dictionary comprehension like this:

dic2 = {v: k for k, values in dic.items() for v in values}

You can remember that the order of for clauses in the comprehension is the same as the order of corresponding nested for loops.

Potential pitfalls

You should be aware that this transformation from a dictionary of lists to a reverse dictionary only works if all items in the original lists are unique. Counterexample:

>>> dic = {0: [1, 2], 1: [2, 3]}

>>> {v: k for k, values in dic.items() for v in values}

{1: 0, 2: 1, 3: 1}  # missing 2: 0

To correct this, the output dictionary should have the same format as the input, namely mapping each of the items in the input lists to a list of corresponding keys.

from collections import defaultdict



graph = {0: [1, 2], 1: [2, 3]}



transposed_graph = defaultdict(list)

for node, neighbours in graph.items():

    for neighbour in neighbours:

        transposed_graph[neighbour].append(node)



# {1: [0], 2: [0, 1], 3: [1]}

answered 5 hours ago

mkrieger1

1,4231824

add a comment |

There's really not much to say about this code, it is straightforward.

Style

These are only nitpicks.

Generic dictionary keys are typically named k instead of i or j, but in a specific application a more descriptive name would be even better.

Collections should be named by their purpose in the application, not their type.

By convention, assignments and other binary operators should be surrounded by spaces: a = b, not a=b. See PEP 8 for reference, if you're not already aware of it. Following this style guide makes your code easier to read for others.

Code improvements

When iterating over the keys and values of a dictionary at the same time, you can use

for k, v in dic.items():

    # ... use k, v ...

instead of

for k in dic:

    v = dic[k]

    # ...

The nested loop can be transformed to a dictionary comprehension like this:

dic2 = {v: k for k, values in dic.items() for v in values}

You can remember that the order of for clauses in the comprehension is the same as the order of corresponding nested for loops.

Potential pitfalls

You should be aware that this transformation from a dictionary of lists to a reverse dictionary only works if all items in the original lists are unique. Counterexample:

>>> dic = {0: [1, 2], 1: [2, 3]}

>>> {v: k for k, values in dic.items() for v in values}

{1: 0, 2: 1, 3: 1}  # missing 2: 0

To correct this, the output dictionary should have the same format as the input, namely mapping each of the items in the input lists to a list of corresponding keys.

from collections import defaultdict



graph = {0: [1, 2], 1: [2, 3]}



transposed_graph = defaultdict(list)

for node, neighbours in graph.items():

    for neighbour in neighbours:

        transposed_graph[neighbour].append(node)



# {1: [0], 2: [0, 1], 3: [1]}

answered 5 hours ago

mkrieger1

1,4231824

There's really not much to say about this code, it is straightforward.

Style

These are only nitpicks.

Generic dictionary keys are typically named k instead of i or j, but in a specific application a more descriptive name would be even better.

Collections should be named by their purpose in the application, not their type.

By convention, assignments and other binary operators should be surrounded by spaces: a = b, not a=b. See PEP 8 for reference, if you're not already aware of it. Following this style guide makes your code easier to read for others.

Code improvements

When iterating over the keys and values of a dictionary at the same time, you can use

for k, v in dic.items():

    # ... use k, v ...

instead of

for k in dic:

    v = dic[k]

    # ...

The nested loop can be transformed to a dictionary comprehension like this:

dic2 = {v: k for k, values in dic.items() for v in values}

You can remember that the order of for clauses in the comprehension is the same as the order of corresponding nested for loops.

Potential pitfalls

You should be aware that this transformation from a dictionary of lists to a reverse dictionary only works if all items in the original lists are unique. Counterexample:

>>> dic = {0: [1, 2], 1: [2, 3]}

>>> {v: k for k, values in dic.items() for v in values}

{1: 0, 2: 1, 3: 1}  # missing 2: 0

To correct this, the output dictionary should have the same format as the input, namely mapping each of the items in the input lists to a list of corresponding keys.

from collections import defaultdict



graph = {0: [1, 2], 1: [2, 3]}



transposed_graph = defaultdict(list)

for node, neighbours in graph.items():

    for neighbour in neighbours:

        transposed_graph[neighbour].append(node)



# {1: [0], 2: [0, 1], 3: [1]}

answered 5 hours ago

mkrieger1

1,4231824

answered 5 hours ago

mkrieger1

1,4231824

answered 5 hours ago

mkrieger1

1,4231824

answered 5 hours ago

mkrieger1

1,4231824

add a comment |

Juan C is a new contributor. Be nice, and check out our Code of Conduct.

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