return the key whose values are repeated
2
I have dictionary whose values of the keys are lists. each list may contain some values repeated more than once like this
{'VSS': ['A2', 'A3', 'A1'], 'X_P1_1': ['A2', 'A1'], 'X_P2': ['A3', 'A2'], 'X_P1_3': ['A2', 'A1'], 'VDD': ['A3', 'A3', 'A3', 'A3', 'A3', 'A3', 'A3', 'B', 'A3'], 'X': ['B', 'A1', 'A1', 'A1', 'A1', 'A1', 'A1', 'A1', 'B', 'A1']}
I want to do loop that takes the keys whose values have frequencies more than 1
for example:
VDD: ['A3':8]
X : ['A1':8, 'B':2]
how this can be done?
python dictionary for-loop counter
edited Nov 12 '18 at 0:45
jpp
91.7k2052102
asked Nov 12 '18 at 0:36
M.salameh
597
add a comment |
2
I have dictionary whose values of the keys are lists. each list may contain some values repeated more than once like this
{'VSS': ['A2', 'A3', 'A1'], 'X_P1_1': ['A2', 'A1'], 'X_P2': ['A3', 'A2'], 'X_P1_3': ['A2', 'A1'], 'VDD': ['A3', 'A3', 'A3', 'A3', 'A3', 'A3', 'A3', 'B', 'A3'], 'X': ['B', 'A1', 'A1', 'A1', 'A1', 'A1', 'A1', 'A1', 'B', 'A1']}
I want to do loop that takes the keys whose values have frequencies more than 1
for example:
VDD: ['A3':8]
X : ['A1':8, 'B':2]
how this can be done?
python dictionary for-loop counter
edited Nov 12 '18 at 0:45
jpp
91.7k2052102
asked Nov 12 '18 at 0:36
M.salameh
597
would you accept pandas?
– W-B
Nov 12 '18 at 0:38
@W-B what is pandas?
– M.salameh
Nov 12 '18 at 0:40
add a comment |
2
2
2
I have dictionary whose values of the keys are lists. each list may contain some values repeated more than once like this
{'VSS': ['A2', 'A3', 'A1'], 'X_P1_1': ['A2', 'A1'], 'X_P2': ['A3', 'A2'], 'X_P1_3': ['A2', 'A1'], 'VDD': ['A3', 'A3', 'A3', 'A3', 'A3', 'A3', 'A3', 'B', 'A3'], 'X': ['B', 'A1', 'A1', 'A1', 'A1', 'A1', 'A1', 'A1', 'B', 'A1']}
I want to do loop that takes the keys whose values have frequencies more than 1
for example:
VDD: ['A3':8]
X : ['A1':8, 'B':2]
how this can be done?
python dictionary for-loop counter
edited Nov 12 '18 at 0:45
jpp
91.7k2052102
asked Nov 12 '18 at 0:36
M.salameh
597
I have dictionary whose values of the keys are lists. each list may contain some values repeated more than once like this
{'VSS': ['A2', 'A3', 'A1'], 'X_P1_1': ['A2', 'A1'], 'X_P2': ['A3', 'A2'], 'X_P1_3': ['A2', 'A1'], 'VDD': ['A3', 'A3', 'A3', 'A3', 'A3', 'A3', 'A3', 'B', 'A3'], 'X': ['B', 'A1', 'A1', 'A1', 'A1', 'A1', 'A1', 'A1', 'B', 'A1']}
I want to do loop that takes the keys whose values have frequencies more than 1
for example:
VDD: ['A3':8]
X : ['A1':8, 'B':2]
how this can be done?
python dictionary for-loop counter
python dictionary for-loop counter
edited Nov 12 '18 at 0:45
jpp
91.7k2052102
asked Nov 12 '18 at 0:36
M.salameh
597
edited Nov 12 '18 at 0:45
jpp
91.7k2052102
asked Nov 12 '18 at 0:36
M.salameh
597
edited Nov 12 '18 at 0:45
jpp
91.7k2052102
edited Nov 12 '18 at 0:45
jpp
91.7k2052102
edited Nov 12 '18 at 0:45
jpp
91.7k2052102
91.7k2052102
asked Nov 12 '18 at 0:36
M.salameh
597
asked Nov 12 '18 at 0:36
M.salameh
597
asked Nov 12 '18 at 0:36
M.salameh
597
597
would you accept pandas?
– W-B
Nov 12 '18 at 0:38
@W-B what is pandas?
– M.salameh
Nov 12 '18 at 0:40
add a comment |
would you accept pandas?
– W-B
Nov 12 '18 at 0:38
@W-B what is pandas?
– M.salameh
Nov 12 '18 at 0:40
would you accept pandas?
– W-B
Nov 12 '18 at 0:38
would you accept pandas?
– W-B
Nov 12 '18 at 0:38
@W-B what is pandas?
– M.salameh
Nov 12 '18 at 0:40
@W-B what is pandas?
– M.salameh
Nov 12 '18 at 0:40
add a comment |
3 Answers
3
active
oldest
votes
1
Using collections.Counter:
from collections import Counter
# count values in lists, only including counts greater than 1
c = {k: {val: count for val, count in Counter(v).items() if count > 1}
for k, v in d.items()}
# isolate only keys where Counter value is non-empty
res = {k: v for k, v in c.items() if v}
{'VDD': {'A3': 8},
'X': {'B': 2, 'A1': 8}}
answered Nov 12 '18 at 0:43
jpp
91.7k2052102
Line continuation isn't needed inside a parenthesised expression ;-)
– coldspeed
Nov 12 '18 at 0:56
@coldspeed, Yep I know :). But I find it more readable.. I can see myself seeing}at the end of one line and thinking the comprehension ends there!
– jpp
Nov 12 '18 at 0:57
add a comment |
2
You could use Counter:
from collections import Counter
data = {'VSS': ['A2', 'A3', 'A1'], 'X_P1_1': ['A2', 'A1'], 'X_P2': ['A3', 'A2'], 'X_P1_3': ['A2', 'A1'], 'VDD': ['A3', 'A3', 'A3', 'A3', 'A3', 'A3', 'A3', 'B', 'A3'], 'X': ['B', 'A1', 'A1', 'A1', 'A1', 'A1', 'A1', 'A1', 'B', 'A1']}
result = {key : { e: count for e, count in Counter(values).items() if count > 1} for key, values in data.items() if any(value > 1 for value in Counter(values).values())}
print(result)
Output
{'VDD': {'A3': 8}, 'X': {'B': 2, 'A1': 8}}
Or if you prefer the values as a list of tuples:
from collections import Counter
data = {'VSS': ['A2', 'A3', 'A1'], 'X_P1_1': ['A2', 'A1'], 'X_P2': ['A3', 'A2'], 'X_P1_3': ['A2', 'A1'],
'VDD': ['A3', 'A3', 'A3', 'A3', 'A3', 'A3', 'A3', 'B', 'A3'],
'X': ['B', 'A1', 'A1', 'A1', 'A1', 'A1', 'A1', 'A1', 'B', 'A1']}
result = {key: [(element, count) for element, count in counts.items() if count > 1] for key, counts in map(lambda x: (x[0], Counter(x[1])), data.items()) if
any(count > 1 for count in counts.values())}
print(result)
Output
{'VDD': [('A3', 8)], 'X': [('A1', 8), ('B', 2)]}
answered Nov 12 '18 at 0:42
Daniel Mesejo
13.4k11027
add a comment |
0
The first step is to reverse the mapping.
valuekeydict = {}
for k, vs in orig_dict.items():
for v in vs:
valuekeydict.setdefault(v, ).append(v)
Then grab a set of all values where len > 1
result = {v for vs in valuekeydict.values() for v in vs if len(vs) > 1}
answered Nov 12 '18 at 0:44
Adam Smith
33.2k53174
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
Using collections.Counter:
from collections import Counter
# count values in lists, only including counts greater than 1
c = {k: {val: count for val, count in Counter(v).items() if count > 1}
for k, v in d.items()}
# isolate only keys where Counter value is non-empty
res = {k: v for k, v in c.items() if v}
{'VDD': {'A3': 8},
'X': {'B': 2, 'A1': 8}}
answered Nov 12 '18 at 0:43
jpp
91.7k2052102
Line continuation isn't needed inside a parenthesised expression ;-)
– coldspeed
Nov 12 '18 at 0:56
@coldspeed, Yep I know :). But I find it more readable.. I can see myself seeing}at the end of one line and thinking the comprehension ends there!
– jpp
Nov 12 '18 at 0:57
add a comment |
1
Using collections.Counter:
from collections import Counter
# count values in lists, only including counts greater than 1
c = {k: {val: count for val, count in Counter(v).items() if count > 1}
for k, v in d.items()}
# isolate only keys where Counter value is non-empty
res = {k: v for k, v in c.items() if v}
{'VDD': {'A3': 8},
'X': {'B': 2, 'A1': 8}}
answered Nov 12 '18 at 0:43
jpp
91.7k2052102
Line continuation isn't needed inside a parenthesised expression ;-)
– coldspeed
Nov 12 '18 at 0:56
@coldspeed, Yep I know :). But I find it more readable.. I can see myself seeing}at the end of one line and thinking the comprehension ends there!
– jpp
Nov 12 '18 at 0:57
add a comment |
1
1
1
Using collections.Counter:
from collections import Counter
# count values in lists, only including counts greater than 1
c = {k: {val: count for val, count in Counter(v).items() if count > 1}
for k, v in d.items()}
# isolate only keys where Counter value is non-empty
res = {k: v for k, v in c.items() if v}
{'VDD': {'A3': 8},
'X': {'B': 2, 'A1': 8}}
answered Nov 12 '18 at 0:43
jpp
91.7k2052102
Using collections.Counter:
from collections import Counter
# count values in lists, only including counts greater than 1
c = {k: {val: count for val, count in Counter(v).items() if count > 1}
for k, v in d.items()}
# isolate only keys where Counter value is non-empty
res = {k: v for k, v in c.items() if v}
{'VDD': {'A3': 8},
'X': {'B': 2, 'A1': 8}}
answered Nov 12 '18 at 0:43
jpp
91.7k2052102
answered Nov 12 '18 at 0:43
jpp
91.7k2052102
answered Nov 12 '18 at 0:43
jpp
91.7k2052102
answered Nov 12 '18 at 0:43
jpp
91.7k2052102
91.7k2052102
Line continuation isn't needed inside a parenthesised expression ;-)
– coldspeed
Nov 12 '18 at 0:56
@coldspeed, Yep I know :). But I find it more readable.. I can see myself seeing}at the end of one line and thinking the comprehension ends there!
– jpp
Nov 12 '18 at 0:57
add a comment |
Line continuation isn't needed inside a parenthesised expression ;-)
– coldspeed
Nov 12 '18 at 0:56
@coldspeed, Yep I know :). But I find it more readable.. I can see myself seeing}at the end of one line and thinking the comprehension ends there!
– jpp
Nov 12 '18 at 0:57
Line continuation isn't needed inside a parenthesised expression ;-)
– coldspeed
Nov 12 '18 at 0:56
Line continuation isn't needed inside a parenthesised expression ;-)
– coldspeed
Nov 12 '18 at 0:56
@coldspeed, Yep I know :). But I find it more readable.. I can see myself seeing
} at the end of one line and thinking the comprehension ends there!– jpp
Nov 12 '18 at 0:57
@coldspeed, Yep I know :). But I find it more readable.. I can see myself seeing
} at the end of one line and thinking the comprehension ends there!– jpp
Nov 12 '18 at 0:57
add a comment |
2
You could use Counter:
from collections import Counter
data = {'VSS': ['A2', 'A3', 'A1'], 'X_P1_1': ['A2', 'A1'], 'X_P2': ['A3', 'A2'], 'X_P1_3': ['A2', 'A1'], 'VDD': ['A3', 'A3', 'A3', 'A3', 'A3', 'A3', 'A3', 'B', 'A3'], 'X': ['B', 'A1', 'A1', 'A1', 'A1', 'A1', 'A1', 'A1', 'B', 'A1']}
result = {key : { e: count for e, count in Counter(values).items() if count > 1} for key, values in data.items() if any(value > 1 for value in Counter(values).values())}
print(result)
Output
{'VDD': {'A3': 8}, 'X': {'B': 2, 'A1': 8}}
Or if you prefer the values as a list of tuples:
from collections import Counter
data = {'VSS': ['A2', 'A3', 'A1'], 'X_P1_1': ['A2', 'A1'], 'X_P2': ['A3', 'A2'], 'X_P1_3': ['A2', 'A1'],
'VDD': ['A3', 'A3', 'A3', 'A3', 'A3', 'A3', 'A3', 'B', 'A3'],
'X': ['B', 'A1', 'A1', 'A1', 'A1', 'A1', 'A1', 'A1', 'B', 'A1']}
result = {key: [(element, count) for element, count in counts.items() if count > 1] for key, counts in map(lambda x: (x[0], Counter(x[1])), data.items()) if
any(count > 1 for count in counts.values())}
print(result)
Output
{'VDD': [('A3', 8)], 'X': [('A1', 8), ('B', 2)]}
answered Nov 12 '18 at 0:42
Daniel Mesejo
13.4k11027
add a comment |
2
You could use Counter:
from collections import Counter
data = {'VSS': ['A2', 'A3', 'A1'], 'X_P1_1': ['A2', 'A1'], 'X_P2': ['A3', 'A2'], 'X_P1_3': ['A2', 'A1'], 'VDD': ['A3', 'A3', 'A3', 'A3', 'A3', 'A3', 'A3', 'B', 'A3'], 'X': ['B', 'A1', 'A1', 'A1', 'A1', 'A1', 'A1', 'A1', 'B', 'A1']}
result = {key : { e: count for e, count in Counter(values).items() if count > 1} for key, values in data.items() if any(value > 1 for value in Counter(values).values())}
print(result)
Output
{'VDD': {'A3': 8}, 'X': {'B': 2, 'A1': 8}}
Or if you prefer the values as a list of tuples:
from collections import Counter
data = {'VSS': ['A2', 'A3', 'A1'], 'X_P1_1': ['A2', 'A1'], 'X_P2': ['A3', 'A2'], 'X_P1_3': ['A2', 'A1'],
'VDD': ['A3', 'A3', 'A3', 'A3', 'A3', 'A3', 'A3', 'B', 'A3'],
'X': ['B', 'A1', 'A1', 'A1', 'A1', 'A1', 'A1', 'A1', 'B', 'A1']}
result = {key: [(element, count) for element, count in counts.items() if count > 1] for key, counts in map(lambda x: (x[0], Counter(x[1])), data.items()) if
any(count > 1 for count in counts.values())}
print(result)
Output
{'VDD': [('A3', 8)], 'X': [('A1', 8), ('B', 2)]}
answered Nov 12 '18 at 0:42
Daniel Mesejo
13.4k11027
add a comment |
2
2
2
You could use Counter:
from collections import Counter
data = {'VSS': ['A2', 'A3', 'A1'], 'X_P1_1': ['A2', 'A1'], 'X_P2': ['A3', 'A2'], 'X_P1_3': ['A2', 'A1'], 'VDD': ['A3', 'A3', 'A3', 'A3', 'A3', 'A3', 'A3', 'B', 'A3'], 'X': ['B', 'A1', 'A1', 'A1', 'A1', 'A1', 'A1', 'A1', 'B', 'A1']}
result = {key : { e: count for e, count in Counter(values).items() if count > 1} for key, values in data.items() if any(value > 1 for value in Counter(values).values())}
print(result)
Output
{'VDD': {'A3': 8}, 'X': {'B': 2, 'A1': 8}}
Or if you prefer the values as a list of tuples:
from collections import Counter
data = {'VSS': ['A2', 'A3', 'A1'], 'X_P1_1': ['A2', 'A1'], 'X_P2': ['A3', 'A2'], 'X_P1_3': ['A2', 'A1'],
'VDD': ['A3', 'A3', 'A3', 'A3', 'A3', 'A3', 'A3', 'B', 'A3'],
'X': ['B', 'A1', 'A1', 'A1', 'A1', 'A1', 'A1', 'A1', 'B', 'A1']}
result = {key: [(element, count) for element, count in counts.items() if count > 1] for key, counts in map(lambda x: (x[0], Counter(x[1])), data.items()) if
any(count > 1 for count in counts.values())}
print(result)
Output
{'VDD': [('A3', 8)], 'X': [('A1', 8), ('B', 2)]}
answered Nov 12 '18 at 0:42
Daniel Mesejo
13.4k11027
You could use Counter:
from collections import Counter
data = {'VSS': ['A2', 'A3', 'A1'], 'X_P1_1': ['A2', 'A1'], 'X_P2': ['A3', 'A2'], 'X_P1_3': ['A2', 'A1'], 'VDD': ['A3', 'A3', 'A3', 'A3', 'A3', 'A3', 'A3', 'B', 'A3'], 'X': ['B', 'A1', 'A1', 'A1', 'A1', 'A1', 'A1', 'A1', 'B', 'A1']}
result = {key : { e: count for e, count in Counter(values).items() if count > 1} for key, values in data.items() if any(value > 1 for value in Counter(values).values())}
print(result)
Output
{'VDD': {'A3': 8}, 'X': {'B': 2, 'A1': 8}}
Or if you prefer the values as a list of tuples:
from collections import Counter
data = {'VSS': ['A2', 'A3', 'A1'], 'X_P1_1': ['A2', 'A1'], 'X_P2': ['A3', 'A2'], 'X_P1_3': ['A2', 'A1'],
'VDD': ['A3', 'A3', 'A3', 'A3', 'A3', 'A3', 'A3', 'B', 'A3'],
'X': ['B', 'A1', 'A1', 'A1', 'A1', 'A1', 'A1', 'A1', 'B', 'A1']}
result = {key: [(element, count) for element, count in counts.items() if count > 1] for key, counts in map(lambda x: (x[0], Counter(x[1])), data.items()) if
any(count > 1 for count in counts.values())}
print(result)
Output
{'VDD': [('A3', 8)], 'X': [('A1', 8), ('B', 2)]}
answered Nov 12 '18 at 0:42
Daniel Mesejo
13.4k11027
edited Nov 12 '18 at 0:47
answered Nov 12 '18 at 0:42
Daniel Mesejo
13.4k11027
answered Nov 12 '18 at 0:42
Daniel Mesejo
13.4k11027
answered Nov 12 '18 at 0:42
Daniel Mesejo
13.4k11027
13.4k11027
add a comment |
add a comment |
0
The first step is to reverse the mapping.
valuekeydict = {}
for k, vs in orig_dict.items():
for v in vs:
valuekeydict.setdefault(v, ).append(v)
Then grab a set of all values where len > 1
result = {v for vs in valuekeydict.values() for v in vs if len(vs) > 1}
answered Nov 12 '18 at 0:44
Adam Smith
33.2k53174
add a comment |
0
The first step is to reverse the mapping.
valuekeydict = {}
for k, vs in orig_dict.items():
for v in vs:
valuekeydict.setdefault(v, ).append(v)
Then grab a set of all values where len > 1
result = {v for vs in valuekeydict.values() for v in vs if len(vs) > 1}
answered Nov 12 '18 at 0:44
Adam Smith
33.2k53174
add a comment |
0
0
0
The first step is to reverse the mapping.
valuekeydict = {}
for k, vs in orig_dict.items():
for v in vs:
valuekeydict.setdefault(v, ).append(v)
Then grab a set of all values where len > 1
result = {v for vs in valuekeydict.values() for v in vs if len(vs) > 1}
answered Nov 12 '18 at 0:44
Adam Smith
33.2k53174
The first step is to reverse the mapping.
valuekeydict = {}
for k, vs in orig_dict.items():
for v in vs:
valuekeydict.setdefault(v, ).append(v)
Then grab a set of all values where len > 1
result = {v for vs in valuekeydict.values() for v in vs if len(vs) > 1}
answered Nov 12 '18 at 0:44
Adam Smith
33.2k53174
answered Nov 12 '18 at 0:44
Adam Smith
33.2k53174
answered Nov 12 '18 at 0:44
Adam Smith
33.2k53174
answered Nov 12 '18 at 0:44
Adam Smith
33.2k53174
33.2k53174
add a comment |
add a comment |
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would you accept pandas?
– W-B
Nov 12 '18 at 0:38
@W-B what is pandas?
– M.salameh
Nov 12 '18 at 0:40