How to remove item from locals()? [duplicate]

This question already has an answer here:

Any way to modify locals dictionary?

5 answers

I'm working on a script that requires manipulation of locals(), and I'm having trouble removing values from it. I've tried locals().pop(key) and del locals()[key], but neither works.

As an example:

def locals_test():

    a, b = 1, 2

    locals().pop('a')

    del locals()['b']

    return locals()



def dict_test():

    test = {'a':1, 'b':2}

    test.pop('a')

    del test['b']

    return test



print(locals_test()) # --> {'a':1, 'b':2}

print(dict_test())   # --> {}

I'm trying to replicate the behavior of dict_test() in locals_test(), but I've yet to find a solution. Does anyone know how to solve this?

edited Jan 2 at 1:06

martineau

68.4k1090183

asked Jan 1 at 22:59

aedificatori

557

marked as duplicate by juanpa.arrivillaga python
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Jan 1 at 23:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

3

Changes to the dictionary returned by locals will not affect the local namespace. Did you read the documentation for locals? Why do you believe you even need to manipulate locals???

– juanpa.arrivillaga
Jan 1 at 23:03

2

locals() is a one-way street. You can't remove local variables by deleting from the locals() dictionary.

– Martijn Pieters♦
Jan 1 at 23:09

I missed that part of the documentation, looks like. It didn't occur to me that the returned dictionary would be by reference as opposed to the actual objects. As for why I'm even doing this, I'm working on a Python interface written in Python, and I'm pulling locals between snippets of code.

– aedificatori
Jan 1 at 23:13

1

"by reference as opposed to the actual objects" What? No, that isn't what's going on. locals simply builds a dict on each invocation from the local namespace. It returns that dict, but local namespaces aren't affected by modifying that dict. But that dict contains the actual objects being referenced in the local namespace. Anyway, that does seem like a good reason to use locals.

– juanpa.arrivillaga
Jan 1 at 23:20

Ahh, sorry, I used the wrong words for that. Build by invocation makes a lot of sense, though, thank you for clarifying. I appreciate it.

– aedificatori
Jan 1 at 23:28

add a comment |

This question already has an answer here:

Any way to modify locals dictionary?

5 answers

I'm working on a script that requires manipulation of locals(), and I'm having trouble removing values from it. I've tried locals().pop(key) and del locals()[key], but neither works.

As an example:

def locals_test():

    a, b = 1, 2

    locals().pop('a')

    del locals()['b']

    return locals()



def dict_test():

    test = {'a':1, 'b':2}

    test.pop('a')

    del test['b']

    return test



print(locals_test()) # --> {'a':1, 'b':2}

print(dict_test())   # --> {}

I'm trying to replicate the behavior of dict_test() in locals_test(), but I've yet to find a solution. Does anyone know how to solve this?

edited Jan 2 at 1:06

martineau

68.4k1090183

asked Jan 1 at 22:59

aedificatori

557

marked as duplicate by juanpa.arrivillaga python
Users with the python badge can single-handedly close python questions as duplicates and reopen them as needed.

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Jan 1 at 23:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

3

Changes to the dictionary returned by locals will not affect the local namespace. Did you read the documentation for locals? Why do you believe you even need to manipulate locals???

– juanpa.arrivillaga
Jan 1 at 23:03

2

locals() is a one-way street. You can't remove local variables by deleting from the locals() dictionary.

– Martijn Pieters♦
Jan 1 at 23:09

I missed that part of the documentation, looks like. It didn't occur to me that the returned dictionary would be by reference as opposed to the actual objects. As for why I'm even doing this, I'm working on a Python interface written in Python, and I'm pulling locals between snippets of code.

– aedificatori
Jan 1 at 23:13

1

"by reference as opposed to the actual objects" What? No, that isn't what's going on. locals simply builds a dict on each invocation from the local namespace. It returns that dict, but local namespaces aren't affected by modifying that dict. But that dict contains the actual objects being referenced in the local namespace. Anyway, that does seem like a good reason to use locals.

– juanpa.arrivillaga
Jan 1 at 23:20

Ahh, sorry, I used the wrong words for that. Build by invocation makes a lot of sense, though, thank you for clarifying. I appreciate it.

– aedificatori
Jan 1 at 23:28

add a comment |

This question already has an answer here:

Any way to modify locals dictionary?

5 answers

I'm working on a script that requires manipulation of locals(), and I'm having trouble removing values from it. I've tried locals().pop(key) and del locals()[key], but neither works.

As an example:

def locals_test():

    a, b = 1, 2

    locals().pop('a')

    del locals()['b']

    return locals()



def dict_test():

    test = {'a':1, 'b':2}

    test.pop('a')

    del test['b']

    return test



print(locals_test()) # --> {'a':1, 'b':2}

print(dict_test())   # --> {}

I'm trying to replicate the behavior of dict_test() in locals_test(), but I've yet to find a solution. Does anyone know how to solve this?

edited Jan 2 at 1:06

martineau

68.4k1090183

asked Jan 1 at 22:59

aedificatori

557

This question already has an answer here:

Any way to modify locals dictionary?

5 answers

I'm working on a script that requires manipulation of locals(), and I'm having trouble removing values from it. I've tried locals().pop(key) and del locals()[key], but neither works.

As an example:

def locals_test():

    a, b = 1, 2

    locals().pop('a')

    del locals()['b']

    return locals()



def dict_test():

    test = {'a':1, 'b':2}

    test.pop('a')

    del test['b']

    return test



print(locals_test()) # --> {'a':1, 'b':2}

print(dict_test())   # --> {}

I'm trying to replicate the behavior of dict_test() in locals_test(), but I've yet to find a solution. Does anyone know how to solve this?

This question already has an answer here:

Any way to modify locals dictionary?

5 answers

python dictionary local-variables

edited Jan 2 at 1:06

martineau

68.4k1090183

asked Jan 1 at 22:59

aedificatori

557

edited Jan 2 at 1:06

martineau

68.4k1090183

asked Jan 1 at 22:59

aedificatori

557

edited Jan 2 at 1:06

martineau

68.4k1090183

edited Jan 2 at 1:06

martineau

68.4k1090183

edited Jan 2 at 1:06

martineau

68.4k1090183

asked Jan 1 at 22:59

aedificatori

557

asked Jan 1 at 22:59

aedificatori

557

asked Jan 1 at 22:59

aedificatori

557

marked as duplicate by juanpa.arrivillaga python
Users with the python badge can single-handedly close python questions as duplicates and reopen them as needed.

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Jan 1 at 23:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by juanpa.arrivillaga python
Users with the python badge can single-handedly close python questions as duplicates and reopen them as needed.

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Jan 1 at 23:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

3

Changes to the dictionary returned by locals will not affect the local namespace. Did you read the documentation for locals? Why do you believe you even need to manipulate locals???

– juanpa.arrivillaga
Jan 1 at 23:03

2

locals() is a one-way street. You can't remove local variables by deleting from the locals() dictionary.

– Martijn Pieters♦
Jan 1 at 23:09

I missed that part of the documentation, looks like. It didn't occur to me that the returned dictionary would be by reference as opposed to the actual objects. As for why I'm even doing this, I'm working on a Python interface written in Python, and I'm pulling locals between snippets of code.

– aedificatori
Jan 1 at 23:13

1

"by reference as opposed to the actual objects" What? No, that isn't what's going on. locals simply builds a dict on each invocation from the local namespace. It returns that dict, but local namespaces aren't affected by modifying that dict. But that dict contains the actual objects being referenced in the local namespace. Anyway, that does seem like a good reason to use locals.

– juanpa.arrivillaga
Jan 1 at 23:20

Ahh, sorry, I used the wrong words for that. Build by invocation makes a lot of sense, though, thank you for clarifying. I appreciate it.

– aedificatori
Jan 1 at 23:28

add a comment |

3

Changes to the dictionary returned by locals will not affect the local namespace. Did you read the documentation for locals? Why do you believe you even need to manipulate locals???

– juanpa.arrivillaga
Jan 1 at 23:03

2

locals() is a one-way street. You can't remove local variables by deleting from the locals() dictionary.

– Martijn Pieters♦
Jan 1 at 23:09

I missed that part of the documentation, looks like. It didn't occur to me that the returned dictionary would be by reference as opposed to the actual objects. As for why I'm even doing this, I'm working on a Python interface written in Python, and I'm pulling locals between snippets of code.

– aedificatori
Jan 1 at 23:13

1

"by reference as opposed to the actual objects" What? No, that isn't what's going on. locals simply builds a dict on each invocation from the local namespace. It returns that dict, but local namespaces aren't affected by modifying that dict. But that dict contains the actual objects being referenced in the local namespace. Anyway, that does seem like a good reason to use locals.

– juanpa.arrivillaga
Jan 1 at 23:20

Ahh, sorry, I used the wrong words for that. Build by invocation makes a lot of sense, though, thank you for clarifying. I appreciate it.

– aedificatori
Jan 1 at 23:28

Changes to the dictionary returned by locals will not affect the local namespace. Did you read the documentation for locals? Why do you believe you even need to manipulate locals???

– juanpa.arrivillaga
Jan 1 at 23:03

locals() is a one-way street. You can't remove local variables by deleting from the locals() dictionary.

– Martijn Pieters♦
Jan 1 at 23:09

I missed that part of the documentation, looks like. It didn't occur to me that the returned dictionary would be by reference as opposed to the actual objects. As for why I'm even doing this, I'm working on a Python interface written in Python, and I'm pulling locals between snippets of code.

– aedificatori
Jan 1 at 23:13

"by reference as opposed to the actual objects" What? No, that isn't what's going on. locals simply builds a dict on each invocation from the local namespace. It returns that dict, but local namespaces aren't affected by modifying that dict. But that dict contains the actual objects being referenced in the local namespace. Anyway, that does seem like a good reason to use locals.

– juanpa.arrivillaga
Jan 1 at 23:20

Ahh, sorry, I used the wrong words for that. Build by invocation makes a lot of sense, though, thank you for clarifying. I appreciate it.

– aedificatori
Jan 1 at 23:28

add a comment |

1 Answer
1

active

oldest

votes

locals() returns a dictionary which is a copy of the local variables. Changing a copy won't remove the variable. The way that works:

def locals_test():

    a, b = 1, 2

    del a

    del b

    return locals()

General advice: playing with locals & globals like this is often the symptom of a XY problem. You'd be better off with a dictionary of values so you're not polluted by the rest of the local variables.

answered Jan 1 at 23:05

Jean-François Fabre

105k955112

Got it, thanks. I'd missed the part of the documentation where it described that the output is a copy of the local variables. (And goodness, do I wish it were an XY problem.)

– aedificatori
Jan 1 at 23:13

add a comment |

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

locals() returns a dictionary which is a copy of the local variables. Changing a copy won't remove the variable. The way that works:

def locals_test():

    a, b = 1, 2

    del a

    del b

    return locals()

answered Jan 1 at 23:05

Jean-François Fabre

105k955112

Got it, thanks. I'd missed the part of the documentation where it described that the output is a copy of the local variables. (And goodness, do I wish it were an XY problem.)

– aedificatori
Jan 1 at 23:13

add a comment |

locals() returns a dictionary which is a copy of the local variables. Changing a copy won't remove the variable. The way that works:

def locals_test():

    a, b = 1, 2

    del a

    del b

    return locals()

answered Jan 1 at 23:05

Jean-François Fabre

105k955112

Got it, thanks. I'd missed the part of the documentation where it described that the output is a copy of the local variables. (And goodness, do I wish it were an XY problem.)

– aedificatori
Jan 1 at 23:13

add a comment |

locals() returns a dictionary which is a copy of the local variables. Changing a copy won't remove the variable. The way that works:

def locals_test():

    a, b = 1, 2

    del a

    del b

    return locals()

answered Jan 1 at 23:05

Jean-François Fabre

105k955112

locals() returns a dictionary which is a copy of the local variables. Changing a copy won't remove the variable. The way that works:

def locals_test():

    a, b = 1, 2

    del a

    del b

    return locals()

answered Jan 1 at 23:05

Jean-François Fabre

105k955112

answered Jan 1 at 23:05

Jean-François Fabre

105k955112

answered Jan 1 at 23:05

Jean-François Fabre

105k955112

answered Jan 1 at 23:05

Jean-François Fabre

105k955112

Got it, thanks. I'd missed the part of the documentation where it described that the output is a copy of the local variables. (And goodness, do I wish it were an XY problem.)

– aedificatori
Jan 1 at 23:13

add a comment |

Got it, thanks. I'd missed the part of the documentation where it described that the output is a copy of the local variables. (And goodness, do I wish it were an XY problem.)

– aedificatori
Jan 1 at 23:13

Got it, thanks. I'd missed the part of the documentation where it described that the output is a copy of the local variables. (And goodness, do I wish it were an XY problem.)

– aedificatori
Jan 1 at 23:13

add a comment |

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