How do I isolate dicts from a list if the id is not found in a second list of dicts (in python)?
up vote
0
down vote
favorite
I have two lists of dicts
list1 =
[
{"name": "Maria",
"id": "16a",
},
{"name": "Tania",
"id": "13b",
},
{"name": "Steve",
"id": "5a",
}
]
list2 =
[
{"name": "Eric",
"id": "16a",
},
{"name": "Mike",
"id": "7b",
},
{"name": "Steve",
id: "57a",
}
]
I want to be able to return a list of dicts from list2, if the same id is not found in list1
For example, it should return
[
{"name": "Mike",
"id": "7b",
},
{"name": "Steve",
"id": "57a",
}
]
I tried a few suggestions here on stack overflow but haven't been able to get it right.
python
asked Nov 8 at 16:43
rose
2391310
add a comment |
up vote
0
down vote
favorite
I have two lists of dicts
list1 =
[
{"name": "Maria",
"id": "16a",
},
{"name": "Tania",
"id": "13b",
},
{"name": "Steve",
"id": "5a",
}
]
list2 =
[
{"name": "Eric",
"id": "16a",
},
{"name": "Mike",
"id": "7b",
},
{"name": "Steve",
id: "57a",
}
]
I want to be able to return a list of dicts from list2, if the same id is not found in list1
For example, it should return
[
{"name": "Mike",
"id": "7b",
},
{"name": "Steve",
"id": "57a",
}
]
I tried a few suggestions here on stack overflow but haven't been able to get it right.
python
asked Nov 8 at 16:43
rose
2391310
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have two lists of dicts
list1 =
[
{"name": "Maria",
"id": "16a",
},
{"name": "Tania",
"id": "13b",
},
{"name": "Steve",
"id": "5a",
}
]
list2 =
[
{"name": "Eric",
"id": "16a",
},
{"name": "Mike",
"id": "7b",
},
{"name": "Steve",
id: "57a",
}
]
I want to be able to return a list of dicts from list2, if the same id is not found in list1
For example, it should return
[
{"name": "Mike",
"id": "7b",
},
{"name": "Steve",
"id": "57a",
}
]
I tried a few suggestions here on stack overflow but haven't been able to get it right.
python
asked Nov 8 at 16:43
rose
2391310
I have two lists of dicts
list1 =
[
{"name": "Maria",
"id": "16a",
},
{"name": "Tania",
"id": "13b",
},
{"name": "Steve",
"id": "5a",
}
]
list2 =
[
{"name": "Eric",
"id": "16a",
},
{"name": "Mike",
"id": "7b",
},
{"name": "Steve",
id: "57a",
}
]
I want to be able to return a list of dicts from list2, if the same id is not found in list1
For example, it should return
[
{"name": "Mike",
"id": "7b",
},
{"name": "Steve",
"id": "57a",
}
]
I tried a few suggestions here on stack overflow but haven't been able to get it right.
python
python
asked Nov 8 at 16:43
rose
2391310
asked Nov 8 at 16:43
rose
2391310
asked Nov 8 at 16:43
rose
2391310
asked Nov 8 at 16:43
rose
2391310
asked Nov 8 at 16:43
rose
2391310
2391310
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
Use a list-comprehension that iterates through list2 checking the id with ids in list1:
list1 = [
{'name': "Maria",
'id': "16a",
},
{'name': "Tania",
'id': "13b",
},
{'name': "Steve",
'id': "5a",
}
]
list2 = [
{'name': "Eric",
'id': "16a",
},
{'name': "Mike",
'id': "7b",
},
{'name': "Steve",
'id': "57a",
}
]
list1_ids = [y['id'] for y in list1]
result = [x for x in list2 if x['id'] not in list1_ids]
# [{'name': 'Mike', 'id': '7b'}, {'name': 'Steve', 'id': '57a'}]
answered Nov 8 at 16:48
Austin
8,3563828
1
It's probably worthwhile to mention that Vasilis G's answer is preferable in terms of performance (O(1) vs O(N)).
– Demian Brecht
Nov 8 at 18:12
add a comment |
up vote
1
down vote
This should do:
[d2 for d2 in list2 if d2['id'] not in [d1['id'] for d1 in list1]]
Output:
[{'id': '7b', 'name': 'Mike'}, {'id': '57a', 'name': 'Steve'}]
answered Nov 8 at 16:53
Alexandre Nixon
47611
Copy of answer 1
– Ian Kirkpatrick
Nov 8 at 16:54
@IanKirkpatrick didn't even check the answers
– Alexandre Nixon
Nov 8 at 16:55
add a comment |
up vote
1
down vote
You can also do it using filter function:
list1 = [
{"name": "Maria",
"id": "16a",
},
{"name": "Tania",
"id": "13b",
},
{"name": "Steve",
"id": "5a",
}
]
list2 = [
{"name": "Eric",
"id": "16a",
},
{"name": "Mike",
"id": "7b",
},
{"name": "Steve",
"id": "57a",
}
]
IDs = set(value["id"] for value in list1)
output = list(filter(lambda elem: elem["id"] not in IDs, list2))
print(output)
Output:
[{'name': 'Mike', 'id': '7b'}, {'name': 'Steve', 'id': '57a'}]
answered Nov 8 at 16:52
Vasilis G.
2,6492621
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Use a list-comprehension that iterates through list2 checking the id with ids in list1:
list1 = [
{'name': "Maria",
'id': "16a",
},
{'name': "Tania",
'id': "13b",
},
{'name': "Steve",
'id': "5a",
}
]
list2 = [
{'name': "Eric",
'id': "16a",
},
{'name': "Mike",
'id': "7b",
},
{'name': "Steve",
'id': "57a",
}
]
list1_ids = [y['id'] for y in list1]
result = [x for x in list2 if x['id'] not in list1_ids]
# [{'name': 'Mike', 'id': '7b'}, {'name': 'Steve', 'id': '57a'}]
answered Nov 8 at 16:48
Austin
8,3563828
1
It's probably worthwhile to mention that Vasilis G's answer is preferable in terms of performance (O(1) vs O(N)).
– Demian Brecht
Nov 8 at 18:12
add a comment |
up vote
3
down vote
accepted
Use a list-comprehension that iterates through list2 checking the id with ids in list1:
list1 = [
{'name': "Maria",
'id': "16a",
},
{'name': "Tania",
'id': "13b",
},
{'name': "Steve",
'id': "5a",
}
]
list2 = [
{'name': "Eric",
'id': "16a",
},
{'name': "Mike",
'id': "7b",
},
{'name': "Steve",
'id': "57a",
}
]
list1_ids = [y['id'] for y in list1]
result = [x for x in list2 if x['id'] not in list1_ids]
# [{'name': 'Mike', 'id': '7b'}, {'name': 'Steve', 'id': '57a'}]
answered Nov 8 at 16:48
Austin
8,3563828
1
It's probably worthwhile to mention that Vasilis G's answer is preferable in terms of performance (O(1) vs O(N)).
– Demian Brecht
Nov 8 at 18:12
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Use a list-comprehension that iterates through list2 checking the id with ids in list1:
list1 = [
{'name': "Maria",
'id': "16a",
},
{'name': "Tania",
'id': "13b",
},
{'name': "Steve",
'id': "5a",
}
]
list2 = [
{'name': "Eric",
'id': "16a",
},
{'name': "Mike",
'id': "7b",
},
{'name': "Steve",
'id': "57a",
}
]
list1_ids = [y['id'] for y in list1]
result = [x for x in list2 if x['id'] not in list1_ids]
# [{'name': 'Mike', 'id': '7b'}, {'name': 'Steve', 'id': '57a'}]
answered Nov 8 at 16:48
Austin
8,3563828
Use a list-comprehension that iterates through list2 checking the id with ids in list1:
list1 = [
{'name': "Maria",
'id': "16a",
},
{'name': "Tania",
'id': "13b",
},
{'name': "Steve",
'id': "5a",
}
]
list2 = [
{'name': "Eric",
'id': "16a",
},
{'name': "Mike",
'id': "7b",
},
{'name': "Steve",
'id': "57a",
}
]
list1_ids = [y['id'] for y in list1]
result = [x for x in list2 if x['id'] not in list1_ids]
# [{'name': 'Mike', 'id': '7b'}, {'name': 'Steve', 'id': '57a'}]
answered Nov 8 at 16:48
Austin
8,3563828
edited Nov 8 at 17:08
answered Nov 8 at 16:48
Austin
8,3563828
answered Nov 8 at 16:48
Austin
8,3563828
answered Nov 8 at 16:48
Austin
8,3563828
8,3563828
1
It's probably worthwhile to mention that Vasilis G's answer is preferable in terms of performance (O(1) vs O(N)).
– Demian Brecht
Nov 8 at 18:12
add a comment |
1
It's probably worthwhile to mention that Vasilis G's answer is preferable in terms of performance (O(1) vs O(N)).
– Demian Brecht
Nov 8 at 18:12
1
1
It's probably worthwhile to mention that Vasilis G's answer is preferable in terms of performance (O(1) vs O(N)).
– Demian Brecht
Nov 8 at 18:12
It's probably worthwhile to mention that Vasilis G's answer is preferable in terms of performance (O(1) vs O(N)).
– Demian Brecht
Nov 8 at 18:12
add a comment |
up vote
1
down vote
This should do:
[d2 for d2 in list2 if d2['id'] not in [d1['id'] for d1 in list1]]
Output:
[{'id': '7b', 'name': 'Mike'}, {'id': '57a', 'name': 'Steve'}]
answered Nov 8 at 16:53
Alexandre Nixon
47611
Copy of answer 1
– Ian Kirkpatrick
Nov 8 at 16:54
@IanKirkpatrick didn't even check the answers
– Alexandre Nixon
Nov 8 at 16:55
add a comment |
up vote
1
down vote
This should do:
[d2 for d2 in list2 if d2['id'] not in [d1['id'] for d1 in list1]]
Output:
[{'id': '7b', 'name': 'Mike'}, {'id': '57a', 'name': 'Steve'}]
answered Nov 8 at 16:53
Alexandre Nixon
47611
Copy of answer 1
– Ian Kirkpatrick
Nov 8 at 16:54
@IanKirkpatrick didn't even check the answers
– Alexandre Nixon
Nov 8 at 16:55
add a comment |
up vote
1
down vote
up vote
1
down vote
This should do:
[d2 for d2 in list2 if d2['id'] not in [d1['id'] for d1 in list1]]
Output:
[{'id': '7b', 'name': 'Mike'}, {'id': '57a', 'name': 'Steve'}]
answered Nov 8 at 16:53
Alexandre Nixon
47611
This should do:
[d2 for d2 in list2 if d2['id'] not in [d1['id'] for d1 in list1]]
Output:
[{'id': '7b', 'name': 'Mike'}, {'id': '57a', 'name': 'Steve'}]
answered Nov 8 at 16:53
Alexandre Nixon
47611
edited Nov 8 at 16:55
answered Nov 8 at 16:53
Alexandre Nixon
47611
answered Nov 8 at 16:53
Alexandre Nixon
47611
answered Nov 8 at 16:53
Alexandre Nixon
47611
47611
Copy of answer 1
– Ian Kirkpatrick
Nov 8 at 16:54
@IanKirkpatrick didn't even check the answers
– Alexandre Nixon
Nov 8 at 16:55
add a comment |
Copy of answer 1
– Ian Kirkpatrick
Nov 8 at 16:54
@IanKirkpatrick didn't even check the answers
– Alexandre Nixon
Nov 8 at 16:55
Copy of answer 1
– Ian Kirkpatrick
Nov 8 at 16:54
Copy of answer 1
– Ian Kirkpatrick
Nov 8 at 16:54
@IanKirkpatrick didn't even check the answers
– Alexandre Nixon
Nov 8 at 16:55
@IanKirkpatrick didn't even check the answers
– Alexandre Nixon
Nov 8 at 16:55
add a comment |
up vote
1
down vote
You can also do it using filter function:
list1 = [
{"name": "Maria",
"id": "16a",
},
{"name": "Tania",
"id": "13b",
},
{"name": "Steve",
"id": "5a",
}
]
list2 = [
{"name": "Eric",
"id": "16a",
},
{"name": "Mike",
"id": "7b",
},
{"name": "Steve",
"id": "57a",
}
]
IDs = set(value["id"] for value in list1)
output = list(filter(lambda elem: elem["id"] not in IDs, list2))
print(output)
Output:
[{'name': 'Mike', 'id': '7b'}, {'name': 'Steve', 'id': '57a'}]
answered Nov 8 at 16:52
Vasilis G.
2,6492621
add a comment |
up vote
1
down vote
You can also do it using filter function:
list1 = [
{"name": "Maria",
"id": "16a",
},
{"name": "Tania",
"id": "13b",
},
{"name": "Steve",
"id": "5a",
}
]
list2 = [
{"name": "Eric",
"id": "16a",
},
{"name": "Mike",
"id": "7b",
},
{"name": "Steve",
"id": "57a",
}
]
IDs = set(value["id"] for value in list1)
output = list(filter(lambda elem: elem["id"] not in IDs, list2))
print(output)
Output:
[{'name': 'Mike', 'id': '7b'}, {'name': 'Steve', 'id': '57a'}]
answered Nov 8 at 16:52
Vasilis G.
2,6492621
add a comment |
up vote
1
down vote
up vote
1
down vote
You can also do it using filter function:
list1 = [
{"name": "Maria",
"id": "16a",
},
{"name": "Tania",
"id": "13b",
},
{"name": "Steve",
"id": "5a",
}
]
list2 = [
{"name": "Eric",
"id": "16a",
},
{"name": "Mike",
"id": "7b",
},
{"name": "Steve",
"id": "57a",
}
]
IDs = set(value["id"] for value in list1)
output = list(filter(lambda elem: elem["id"] not in IDs, list2))
print(output)
Output:
[{'name': 'Mike', 'id': '7b'}, {'name': 'Steve', 'id': '57a'}]
answered Nov 8 at 16:52
Vasilis G.
2,6492621
You can also do it using filter function:
list1 = [
{"name": "Maria",
"id": "16a",
},
{"name": "Tania",
"id": "13b",
},
{"name": "Steve",
"id": "5a",
}
]
list2 = [
{"name": "Eric",
"id": "16a",
},
{"name": "Mike",
"id": "7b",
},
{"name": "Steve",
"id": "57a",
}
]
IDs = set(value["id"] for value in list1)
output = list(filter(lambda elem: elem["id"] not in IDs, list2))
print(output)
Output:
[{'name': 'Mike', 'id': '7b'}, {'name': 'Steve', 'id': '57a'}]
answered Nov 8 at 16:52
Vasilis G.
2,6492621
edited Nov 8 at 16:58
answered Nov 8 at 16:52
Vasilis G.
2,6492621
answered Nov 8 at 16:52
Vasilis G.
2,6492621
answered Nov 8 at 16:52
Vasilis G.
2,6492621
2,6492621
add a comment |
add a comment |
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