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Remove key:value pair from dictionary based on a condition?
Remove key-value pair based on grouping from dictionaries in pythonHow to remove an element from a list by index?How to remove items from a list while iterating?Delete an element from a dictionaryHow to remove a key from a Python dictionary?Select rows from a DataFrame based on values in a column in pandasPythonic way to delete an item completely from a dictionaryCreating a nested dictionary from a list of tuplesPython cant make dictionary from list of dictionariesRemove digits from string pythonRemove nan's from the nested dictionary in python
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
I have nested dictionary 'my_dict' as given below. I want to remove common keys from nested dictionary grouped by main key name format.
my_dict = 'abc_1': '00000000': 0.01555745891946835,
'facility': 0.04667237675840505,
'among': 0.01555745891946835,
'abc_2': '00000000': 0.01555745891946835,
'before': 0.04667237675840505,
'last': 0.01555745891946835,
'mno_1': 'hello': 0.01555745891946835,
'hola': 0.04667237675840505,
'0000150000': 0.01555745891946835,
'mno_2': 'hello': 0.01555745891946835,
'name': 0.04667237675840505,
'0000150000': 0.01555745891946835,
'oko_1': 'err': 0.01555745891946835,
'error': 0.04667237675840505,
'7812': 0.01555745891946835,
'oko_2': '9872': 0.01555745891946835,
'error': 0.04667237675840505,
'00': 0.01555745891946835
For example, common keys in nested dictionary for keys starting abc* is 00000000. So, I want to remove this key. Likewise, i want to do for all.
Expected result is given below:
Expected Result:
result_dict = 'abc_1': 'facility': 0.04667237675840505,
'among': 0.01555745891946835,
'abc_2': 'before': 0.04667237675840505,
'last': 0.01555745891946835,
'mno_1': 'hola': 0.04667237675840505,
'mno_2': 'name': 0.04667237675840505,
'oko_1': 'err': 0.01555745891946835,
'7812': 0.01555745891946835,
'oko_2': '9872': 0.01555745891946835,
'00': 0.01555745891946835
python python-3.x dictionary
edited Mar 25 at 20:32
martineau
73.4k10 gold badges101 silver badges191 bronze badges
asked Mar 25 at 19:31
user15051990user15051990
7567 silver badges19 bronze badges
add a comment |
I have nested dictionary 'my_dict' as given below. I want to remove common keys from nested dictionary grouped by main key name format.
my_dict = 'abc_1': '00000000': 0.01555745891946835,
'facility': 0.04667237675840505,
'among': 0.01555745891946835,
'abc_2': '00000000': 0.01555745891946835,
'before': 0.04667237675840505,
'last': 0.01555745891946835,
'mno_1': 'hello': 0.01555745891946835,
'hola': 0.04667237675840505,
'0000150000': 0.01555745891946835,
'mno_2': 'hello': 0.01555745891946835,
'name': 0.04667237675840505,
'0000150000': 0.01555745891946835,
'oko_1': 'err': 0.01555745891946835,
'error': 0.04667237675840505,
'7812': 0.01555745891946835,
'oko_2': '9872': 0.01555745891946835,
'error': 0.04667237675840505,
'00': 0.01555745891946835
For example, common keys in nested dictionary for keys starting abc* is 00000000. So, I want to remove this key. Likewise, i want to do for all.
Expected result is given below:
Expected Result:
result_dict = 'abc_1': 'facility': 0.04667237675840505,
'among': 0.01555745891946835,
'abc_2': 'before': 0.04667237675840505,
'last': 0.01555745891946835,
'mno_1': 'hola': 0.04667237675840505,
'mno_2': 'name': 0.04667237675840505,
'oko_1': 'err': 0.01555745891946835,
'7812': 0.01555745891946835,
'oko_2': '9872': 0.01555745891946835,
'00': 0.01555745891946835
python python-3.x dictionary
edited Mar 25 at 20:32
martineau
73.4k10 gold badges101 silver badges191 bronze badges
asked Mar 25 at 19:31
user15051990user15051990
7567 silver badges19 bronze badges
1
What have you tried so far?
– Stephen Newell
Mar 25 at 19:34
Is it the common key itself, or only if the common keys share a value?
– chepner
Mar 25 at 19:52
only if common key share a value. Delete then that particular key:value par in nested dictionary.
– user15051990
Mar 25 at 19:53
add a comment |
I have nested dictionary 'my_dict' as given below. I want to remove common keys from nested dictionary grouped by main key name format.
my_dict = 'abc_1': '00000000': 0.01555745891946835,
'facility': 0.04667237675840505,
'among': 0.01555745891946835,
'abc_2': '00000000': 0.01555745891946835,
'before': 0.04667237675840505,
'last': 0.01555745891946835,
'mno_1': 'hello': 0.01555745891946835,
'hola': 0.04667237675840505,
'0000150000': 0.01555745891946835,
'mno_2': 'hello': 0.01555745891946835,
'name': 0.04667237675840505,
'0000150000': 0.01555745891946835,
'oko_1': 'err': 0.01555745891946835,
'error': 0.04667237675840505,
'7812': 0.01555745891946835,
'oko_2': '9872': 0.01555745891946835,
'error': 0.04667237675840505,
'00': 0.01555745891946835
For example, common keys in nested dictionary for keys starting abc* is 00000000. So, I want to remove this key. Likewise, i want to do for all.
Expected result is given below:
Expected Result:
result_dict = 'abc_1': 'facility': 0.04667237675840505,
'among': 0.01555745891946835,
'abc_2': 'before': 0.04667237675840505,
'last': 0.01555745891946835,
'mno_1': 'hola': 0.04667237675840505,
'mno_2': 'name': 0.04667237675840505,
'oko_1': 'err': 0.01555745891946835,
'7812': 0.01555745891946835,
'oko_2': '9872': 0.01555745891946835,
'00': 0.01555745891946835
python python-3.x dictionary
edited Mar 25 at 20:32
martineau
73.4k10 gold badges101 silver badges191 bronze badges
asked Mar 25 at 19:31
user15051990user15051990
7567 silver badges19 bronze badges
I have nested dictionary 'my_dict' as given below. I want to remove common keys from nested dictionary grouped by main key name format.
my_dict = 'abc_1': '00000000': 0.01555745891946835,
'facility': 0.04667237675840505,
'among': 0.01555745891946835,
'abc_2': '00000000': 0.01555745891946835,
'before': 0.04667237675840505,
'last': 0.01555745891946835,
'mno_1': 'hello': 0.01555745891946835,
'hola': 0.04667237675840505,
'0000150000': 0.01555745891946835,
'mno_2': 'hello': 0.01555745891946835,
'name': 0.04667237675840505,
'0000150000': 0.01555745891946835,
'oko_1': 'err': 0.01555745891946835,
'error': 0.04667237675840505,
'7812': 0.01555745891946835,
'oko_2': '9872': 0.01555745891946835,
'error': 0.04667237675840505,
'00': 0.01555745891946835
For example, common keys in nested dictionary for keys starting abc* is 00000000. So, I want to remove this key. Likewise, i want to do for all.
Expected result is given below:
Expected Result:
result_dict = 'abc_1': 'facility': 0.04667237675840505,
'among': 0.01555745891946835,
'abc_2': 'before': 0.04667237675840505,
'last': 0.01555745891946835,
'mno_1': 'hola': 0.04667237675840505,
'mno_2': 'name': 0.04667237675840505,
'oko_1': 'err': 0.01555745891946835,
'7812': 0.01555745891946835,
'oko_2': '9872': 0.01555745891946835,
'00': 0.01555745891946835
python python-3.x dictionary
python python-3.x dictionary
edited Mar 25 at 20:32
martineau
73.4k10 gold badges101 silver badges191 bronze badges
asked Mar 25 at 19:31
user15051990user15051990
7567 silver badges19 bronze badges
edited Mar 25 at 20:32
martineau
73.4k10 gold badges101 silver badges191 bronze badges
asked Mar 25 at 19:31
user15051990user15051990
7567 silver badges19 bronze badges
edited Mar 25 at 20:32
martineau
73.4k10 gold badges101 silver badges191 bronze badges
edited Mar 25 at 20:32
martineau
73.4k10 gold badges101 silver badges191 bronze badges
edited Mar 25 at 20:32
martineau
73.4k10 gold badges101 silver badges191 bronze badges
73.4k10 gold badges101 silver badges191 bronze badges
asked Mar 25 at 19:31
user15051990user15051990
7567 silver badges19 bronze badges
asked Mar 25 at 19:31
user15051990user15051990
7567 silver badges19 bronze badges
asked Mar 25 at 19:31
user15051990user15051990
7567 silver badges19 bronze badges
7567 silver badges19 bronze badges
1
What have you tried so far?
– Stephen Newell
Mar 25 at 19:34
Is it the common key itself, or only if the common keys share a value?
– chepner
Mar 25 at 19:52
only if common key share a value. Delete then that particular key:value par in nested dictionary.
– user15051990
Mar 25 at 19:53
add a comment |
1
What have you tried so far?
– Stephen Newell
Mar 25 at 19:34
Is it the common key itself, or only if the common keys share a value?
– chepner
Mar 25 at 19:52
only if common key share a value. Delete then that particular key:value par in nested dictionary.
– user15051990
Mar 25 at 19:53
1
1
What have you tried so far?
– Stephen Newell
Mar 25 at 19:34
What have you tried so far?
– Stephen Newell
Mar 25 at 19:34
Is it the common key itself, or only if the common keys share a value?
– chepner
Mar 25 at 19:52
Is it the common key itself, or only if the common keys share a value?
– chepner
Mar 25 at 19:52
only if common key share a value. Delete then that particular key:value par in nested dictionary.
– user15051990
Mar 25 at 19:53
only if common key share a value. Delete then that particular key:value par in nested dictionary.
– user15051990
Mar 25 at 19:53
add a comment |
1 Answer
1
active
oldest
votes
First, get all the keys, then filter which keys you wish to keep. Then you can reconstruct the new dict with only the keys to keep:
all_keys = [n for k in my_dict.values() for n in k.keys()]
keys_to_keep = k for k in all_keys if all_keys.count(k) == 1
result_dict = k: kk: v[kk] for kk in keys_to_keep if kk in v for k, v in my_dict.items()
result:
'abc_1': 'facility': 0.04667237675840505, 'among': 0.01555745891946835, 'abc_2': 'before': 0.04667237675840505, 'last': 0.01555745891946835, 'mno_1': 'hola': 0.04667237675840505, 'mno_2': 'name': 0.04667237675840505, 'oko_1': 'err': 0.01555745891946835, '7812': 0.01555745891946835, 'oko_2': '9872': 0.01555745891946835, '00': 0.01555745891946835
answered Mar 25 at 20:00
Maxime ChéramyMaxime Chéramy
10.5k4 gold badges32 silver badges61 bronze badges
I have similar kind of problem - stackoverflow.com/questions/55895451/…. Can you please help me in solving it?
– user15051990
May 1 at 18:46
add a comment |
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1 Answer
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First, get all the keys, then filter which keys you wish to keep. Then you can reconstruct the new dict with only the keys to keep:
all_keys = [n for k in my_dict.values() for n in k.keys()]
keys_to_keep = k for k in all_keys if all_keys.count(k) == 1
result_dict = k: kk: v[kk] for kk in keys_to_keep if kk in v for k, v in my_dict.items()
result:
'abc_1': 'facility': 0.04667237675840505, 'among': 0.01555745891946835, 'abc_2': 'before': 0.04667237675840505, 'last': 0.01555745891946835, 'mno_1': 'hola': 0.04667237675840505, 'mno_2': 'name': 0.04667237675840505, 'oko_1': 'err': 0.01555745891946835, '7812': 0.01555745891946835, 'oko_2': '9872': 0.01555745891946835, '00': 0.01555745891946835
answered Mar 25 at 20:00
Maxime ChéramyMaxime Chéramy
10.5k4 gold badges32 silver badges61 bronze badges
I have similar kind of problem - stackoverflow.com/questions/55895451/…. Can you please help me in solving it?
– user15051990
May 1 at 18:46
add a comment |
First, get all the keys, then filter which keys you wish to keep. Then you can reconstruct the new dict with only the keys to keep:
all_keys = [n for k in my_dict.values() for n in k.keys()]
keys_to_keep = k for k in all_keys if all_keys.count(k) == 1
result_dict = k: kk: v[kk] for kk in keys_to_keep if kk in v for k, v in my_dict.items()
result:
'abc_1': 'facility': 0.04667237675840505, 'among': 0.01555745891946835, 'abc_2': 'before': 0.04667237675840505, 'last': 0.01555745891946835, 'mno_1': 'hola': 0.04667237675840505, 'mno_2': 'name': 0.04667237675840505, 'oko_1': 'err': 0.01555745891946835, '7812': 0.01555745891946835, 'oko_2': '9872': 0.01555745891946835, '00': 0.01555745891946835
answered Mar 25 at 20:00
Maxime ChéramyMaxime Chéramy
10.5k4 gold badges32 silver badges61 bronze badges
I have similar kind of problem - stackoverflow.com/questions/55895451/…. Can you please help me in solving it?
– user15051990
May 1 at 18:46
add a comment |
First, get all the keys, then filter which keys you wish to keep. Then you can reconstruct the new dict with only the keys to keep:
all_keys = [n for k in my_dict.values() for n in k.keys()]
keys_to_keep = k for k in all_keys if all_keys.count(k) == 1
result_dict = k: kk: v[kk] for kk in keys_to_keep if kk in v for k, v in my_dict.items()
result:
'abc_1': 'facility': 0.04667237675840505, 'among': 0.01555745891946835, 'abc_2': 'before': 0.04667237675840505, 'last': 0.01555745891946835, 'mno_1': 'hola': 0.04667237675840505, 'mno_2': 'name': 0.04667237675840505, 'oko_1': 'err': 0.01555745891946835, '7812': 0.01555745891946835, 'oko_2': '9872': 0.01555745891946835, '00': 0.01555745891946835
answered Mar 25 at 20:00
Maxime ChéramyMaxime Chéramy
10.5k4 gold badges32 silver badges61 bronze badges
First, get all the keys, then filter which keys you wish to keep. Then you can reconstruct the new dict with only the keys to keep:
all_keys = [n for k in my_dict.values() for n in k.keys()]
keys_to_keep = k for k in all_keys if all_keys.count(k) == 1
result_dict = k: kk: v[kk] for kk in keys_to_keep if kk in v for k, v in my_dict.items()
result:
'abc_1': 'facility': 0.04667237675840505, 'among': 0.01555745891946835, 'abc_2': 'before': 0.04667237675840505, 'last': 0.01555745891946835, 'mno_1': 'hola': 0.04667237675840505, 'mno_2': 'name': 0.04667237675840505, 'oko_1': 'err': 0.01555745891946835, '7812': 0.01555745891946835, 'oko_2': '9872': 0.01555745891946835, '00': 0.01555745891946835
answered Mar 25 at 20:00
Maxime ChéramyMaxime Chéramy
10.5k4 gold badges32 silver badges61 bronze badges
edited Mar 25 at 20:13
answered Mar 25 at 20:00
Maxime ChéramyMaxime Chéramy
10.5k4 gold badges32 silver badges61 bronze badges
answered Mar 25 at 20:00
Maxime ChéramyMaxime Chéramy
10.5k4 gold badges32 silver badges61 bronze badges
answered Mar 25 at 20:00
Maxime ChéramyMaxime Chéramy
10.5k4 gold badges32 silver badges61 bronze badges
10.5k4 gold badges32 silver badges61 bronze badges
I have similar kind of problem - stackoverflow.com/questions/55895451/…. Can you please help me in solving it?
– user15051990
May 1 at 18:46
add a comment |
I have similar kind of problem - stackoverflow.com/questions/55895451/…. Can you please help me in solving it?
– user15051990
May 1 at 18:46
I have similar kind of problem - stackoverflow.com/questions/55895451/…. Can you please help me in solving it?
– user15051990
May 1 at 18:46
I have similar kind of problem - stackoverflow.com/questions/55895451/…. Can you please help me in solving it?
– user15051990
May 1 at 18:46
add a comment |
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1
What have you tried so far?
– Stephen Newell
Mar 25 at 19:34
Is it the common key itself, or only if the common keys share a value?
– chepner
Mar 25 at 19:52
only if common key share a value. Delete then that particular key:value par in nested dictionary.
– user15051990
Mar 25 at 19:53