I´m trying to extract several keys/values out of a List.
My List:
a = [
{
"id": "1",
"system": "2",
},
{
"id": "3",
"system": "4",
}
]
Now i need to parse this into a function (next function) and it returns a[current] or a[0].
But now is a[current] or a[0] a dict.
Next step is just to extract the ID and the value of it. But this below only works if a is a list! So i need to convert the a[current] or a[0] into a list. The code below has to be the same because it´s a function and if i cannot change this for several reasons, so i need to convert the dict a into a list a.
c = list()
for data in a:
value = dict()
value["id"] = data.get("id")
c.append(value)
And here i stuck, i tried several methods like .keys(), .values(), but i can´t put them together to a list anymore. It needs to be scaleable/configurable because a changes from time to time (not a[0]["id"], ...). Currently a[0] looks like this: {'id': '1', 'system': '2'}, but it needs to be like this: [{'id': '1', 'system': '2'},], that i can parse it to my search function.
I need a new list like c:
c = [
{
"id": "1",
},
{
"id": "3",
}
]
Is this your your expected output:
a = [
{
"id": "1",
"system": "2",
},
{
"id": "3",
"system": "4",
}
]
c = list()
for data in a:
value={}
value["id"]=data.get("id")
c.append([value])
# c.extend([value])
print(c)
# [[{'id': '1'}], [{'id': '3'}]]
# print(c) # Extend's output
# [{'id': '1'},{'id': '3'}]
Or you can try one-line solution
print([[{"id":val.get("id")}] for val in a])
# [[{'id': '1'}], [{'id': '3'}]]
Or as your comment if you just want
[{'id': '1', 'system': '2'},]
Then :
print([a[0]])
code updated:
a = [
{
"id": "1",
"system": "2",
},
{
"id": "3",
"system": "4",
}
]
print([[value] for value in a ])
Result:
[[{'id': '1', 'system': '2'}], [{'id': '3', 'system': '4'}]]
Here is a function to filter your dicts list:
def filter_dicts(key, a):
return [{key: element[key]} for element in a]
Use it like this:
c = filter_dicts("id", a)
Note that this will cause an error if there is a dict without the specified key, which may or may not be what you want. To avoid this, replace element[key] with element.get(key, None).
Related
I have some json that I would like to transform from this:
[
{
"name":"field1",
"intValue":"1"
},
{
"name":"field2",
"intValue":"2"
},
...
{
"name":"fieldN",
"intValue":"N"
}
]
into this:
{ "field1" : "1",
"field2" : "2",
...
"fieldN" : "N",
}
For each pair, I need to change the value of the name field to a key, and the values of the intValue field to a value. This doesn't seem like flattening or denormalizing. Are there any tools that might do this out-of-the-box, or will this have to be brute-forced? What's the most pythonic way to accomplish this?
parameters = [ # assuming this is loaded already
{
"name":"field1",
"intValue":"1"
},
{
"name":"field2",
"intValue":"2"
},
{
"name":"fieldN",
"intValue":"N"
}
]
field_int_map = dict()
for p in parameters:
field_int_map[p['name']] = p['intValue']
yields {'field1': '1', 'field2': '2', 'fieldN': 'N'}
or as a dict comprehension
field_int_map = {p['name']:p['intValue'] for p in parameters}
This works to combine the name attribute with the intValue as key:value pairs, but the result is a dictionary instead of the original input type which was a list.
Use dictionary comprehension:
json_dct = {"parameters":
[
{
"name":"field1",
"intValue":"1"
},
{
"name":"field2",
"intValue":"2"
},
{
"name":"fieldN",
"intValue":"N"
}
]}
dct = {d["name"]: d["intValue"] for d in json_dct["parameters"]}
print(dct)
# {'field1': '1', 'field2': '2', 'fieldN': 'N'}
I have the below python dictionary stored as dictPython
{
"paging": {"count": 10, "start": 0, "links": []},
"elements": [
{
"organizationalTarget~": {
"vanityName": "vv",
"localizedName": "ViV",
"name": {
"localized": {"en_US": "ViV"},
"preferredLocale": {"country": "US", "language": "en"},
},
"primaryOrganizationType": "NONE",
"locations": [],
"id": 109,
},
"role": "ADMINISTRATOR",
},
],
}
I need to get the values of vanityName, localizedName and also the values from name->localized and name->preferredLocale.
I tried dictPython.keys() and it returned dict_keys(['paging', 'elements']).
Also I tried dictPython.values() and it returned me what is inside of the parenthesis({}).
I need to get [vv, ViV, ViV, US, en]
I am writing this in a form of answer, so I can get to explain it better without the comments characters limit
a dict in python is an efficient key/value structure or data type
for example dict_ = {'key1': 'val1', 'key2': 'val2'} to fetch key1 we can do it in 2 different ways
dict_.get(key1) this returns the value of the key in this case val1, this method has its advantage, that if the key1 is wrong or not found it returns None so no exceptions are raised. You can do dict_.get(key1, 'returning this string if the key is not found')
dict_['key1'] doing the same .get(...) but will raise a KeyError if the key is not found
So to answer your question after this introduction,
a dict can be thought of as nested dictionaries and/or objects inside of one another
to get your values you can do the following
# Fetch base dictionary to make code more readable
base_dict = dict_["elements"][0]["organizationalTarget~"]
# fetch name_dict following the same approach as above code
name_dict = base_dict["name"]
localized_dict = name_dict["localized"]
preferred_locale_dict = name_dict ["preferredLocale"]
so now we fetch all of the wanted data in their corresponding locations from your given dictionary, now to print the results, we can do the following
results_arr = []
for key1, key2 in zip(localized_dict, preferredLocale_dict):
results_arr.append(localized_dict.get(key1))
results_arr.append(preferred_locale_dict.get(key2))
print(results_arr)
What about:
dic = {
"paging": {"count": 10, "start": 0, "links": []},
"elements": [
{
"organizationalTarget~": {
"vanityName": "vv",
"localizedName": "ViV",
"name": {
"localized": {"en_US": "ViV"},
"preferredLocale": {"country": "US", "language": "en"},
},
"primaryOrganizationType": "NONE",
"locations": [],
"id": 109,
},
"role": "ADMINISTRATOR",
},
],
}
base = dic["elements"][0]["organizationalTarget~"]
c = base["name"]["localized"]
d = base["name"]["preferredLocale"]
output = [base["vanityName"], base["localizedName"]]
output.extend([c[key] for key in c])
output.extend([d[key] for key in d])
print(output)
outputs:
['vv', 'ViV', 'ViV', 'US', 'en']
So something like this?
[[x['organizationalTarget~']['vanityName'],
x['organizationalTarget~']['localizedName'],
x['organizationalTarget~']['name']['localized']['en_US'],
x['organizationalTarget~']['name']['preferredLocale']['country'],
x['organizationalTarget~']['name']['preferredLocale']['language'],
] for x in s['elements']]
I have a list of dicts:
input = [{'name':'A', 'Status':'Passed','id':'x1'},
{'name':'A', 'Status':'Passed','id':'x2'},
{'name':'A','Status':'Failed','id':'x3'},
{'name':'B', 'Status':'Passed','id':'x4'},
{'name':'B', 'Status':'Passed','id':'x5'}]
I want an output like :
output = [{'name':'A', 'Passed':'2', 'Failed':'1', 'Total':'3', '%Pass':'66%'},
{'name':'B', 'Passed':'2', 'Failed':'0', 'Total':'2', '%Pass':'100%'},
{'name':'Total', 'Passed':'4', 'Failed':'1', 'Total':'5', '%Pass':'80%'}]\
i started retrieving the different names by using a lookup :
lookup = {(d["name"]): d for d in input [::-1]}
names= [e for e in lookup.values()]
names= names[::-1]
and after using the list comprehension something like :\
for name in names :
name_passed = sum(["Passed" and "name" for d in input if 'Status' in d and name in d])
name_faled = sum(["Failed" and "name" for d in input if 'Status' in d and name in d])\
But i am not sure if there is a smartest way ? a simple loop and comparing dict values will be more simple!?
Assuming your input entries will always be grouped according to the "name" key-value pair:
entries = [
{"name": "A", "Status": "Passed", "id": "x1"},
{"name": "A", "Status": "Passed", "id": "x2"},
{"name": "A", "Status": "Failed", "id": "x3"},
{"name": "B", "Status": "Passed", "id": "x4"},
{"name": "B", "Status": "Passed", "id": "x5"}
]
def to_grouped(entries):
from itertools import groupby
from operator import itemgetter
for key, group_iter in groupby(entries, key=itemgetter("name")):
group = list(group_iter)
total = len(group)
passed = sum(1 for entry in group if entry["Status"] == "Passed")
failed = total - passed
perc_pass = (100 // total) * passed
yield {
"name": key,
"Passed": str(passed),
"Failed": str(failed),
"Total": str(total),
"%Pass": f"{perc_pass:.0f}%"
}
print(list(to_grouped(entries)))
Output:
[{'name': 'A', 'Passed': '2', 'Failed': '1', 'Total': '3', '%Pass': '66%'}, {'name': 'B', 'Passed': '2', 'Failed': '0', 'Total': '2', '%Pass': '100%'}]
This will not create the final entry you're looking for, which sums the statistics of all other entries. Though, that shouldn't be too hard to do.
I have two dictionaries, as below. Both dictionaries have a list of dictionaries as the value associated with their properties key; each dictionary within these lists has an id key. I wish to merge my two dictionaries into one such that the properties list in the resulting dictionary only has one dictionary for each id.
{
"name":"harry",
"properties":[
{
"id":"N3",
"status":"OPEN",
"type":"energetic"
},
{
"id":"N5",
"status":"OPEN",
"type":"hot"
}
]
}
and the other list:
{
"name":"harry",
"properties":[
{
"id":"N3",
"type":"energetic",
"language": "english"
},
{
"id":"N6",
"status":"OPEN",
"type":"cool"
}
]
}
The output I am trying to achieve is:
"name":"harry",
"properties":[
{
"id":"N3",
"status":"OPEN",
"type":"energetic",
"language": "english"
},
{
"id":"N5",
"status":"OPEN",
"type":"hot"
},
{
"id":"N6",
"status":"OPEN",
"type":"cool"
}
]
}
As id: N3 is common in both the lists, those 2 dicts should be merged with all the fields. So far I have tried using itertools and
ds = [d1, d2]
d = {}
for k in d1.keys():
d[k] = tuple(d[k] for d in ds)
Could someone please help in figuring this out?
Here is one of the approach:
a = {
"name":"harry",
"properties":[
{
"id":"N3",
"status":"OPEN",
"type":"energetic"
},
{
"id":"N5",
"status":"OPEN",
"type":"hot"
}
]
}
b = {
"name":"harry",
"properties":[
{
"id":"N3",
"type":"energetic",
"language": "english"
},
{
"id":"N6",
"status":"OPEN",
"type":"cool"
}
]
}
# Create dic maintaining the index of each id in resp dict
a_ids = {item['id']: index for index,item in enumerate(a['properties'])} #{'N3': 0, 'N5': 1}
b_ids = {item['id']: index for index,item in enumerate(b['properties'])} #{'N3': 0, 'N6': 1}
# Loop through one of the dict created
for id in a_ids.keys():
# If same ID exists in another dict, update it with the key value
if id in b_ids:
b['properties'][b_ids[id]].update(a['properties'][a_ids[id]])
# If it does not exist, then just append the new dict
else:
b['properties'].append(a['properties'][a_ids[id]])
print (b)
Output:
{'name': 'harry', 'properties': [{'id': 'N3', 'type': 'energetic', 'language': 'english', 'status': 'OPEN'}, {'id': 'N6', 'status': 'OPEN', 'type': 'cool'}, {'id': 'N5', 'status': 'OPEN', 'type': 'hot'}]}
It might help to treat the two objects as elements each in their own lists. Maybe you have other objects with different name values, such as might come out of a JSON-formatted REST request.
Then you could do a left outer join on both name and id keys:
#!/usr/bin/env python
a = [
{
"name": "harry",
"properties": [
{
"id":"N3",
"status":"OPEN",
"type":"energetic"
},
{
"id":"N5",
"status":"OPEN",
"type":"hot"
}
]
}
]
b = [
{
"name": "harry",
"properties": [
{
"id":"N3",
"type":"energetic",
"language": "english"
},
{
"id":"N6",
"status":"OPEN",
"type":"cool"
}
]
}
]
a_names = set()
a_prop_ids_by_name = {}
a_by_name = {}
for ao in a:
an = ao['name']
a_names.add(an)
if an not in a_prop_ids_by_name:
a_prop_ids_by_name[an] = set()
for ap in ao['properties']:
api = ap['id']
a_prop_ids_by_name[an].add(api)
a_by_name[an] = ao
res = []
for bo in b:
bn = bo['name']
if bn not in a_names:
res.append(bo)
else:
ao = a_by_name[bn]
bp = bo['properties']
for bpo in bp:
if bpo['id'] not in a_prop_ids_by_name[bn]:
ao['properties'].append(bpo)
res.append(ao)
print(res)
The idea above is to process list a for names and ids. The names and ids-by-name are instances of a Python set. So members are always unique.
Once you have these sets, you can do the left outer join on the contents of list b.
Either there's an object in b that doesn't exist in a (i.e. shares a common name), in which case you add that object to the result as-is. But if there is an object in b that does exist in a (which shares a common name), then you iterate over that object's id values and look for ids not already in the a ids-by-name set. You add missing properties to a, and then add that processed object to the result.
Output:
[{'name': 'harry', 'properties': [{'id': 'N3', 'status': 'OPEN', 'type': 'energetic'}, {'id': 'N5', 'status': 'OPEN', 'type': 'hot'}, {'id': 'N6', 'status': 'OPEN', 'type': 'cool'}]}]
This doesn't do any error checking on input. This relies on name values being unique per object. So if you have duplicate keys in objects in both lists, you may get garbage (incorrect or unexpected output).
I want to compare two lists of dictionaries in python, I have a list sent from the frontend and a query result stored in the same function, so all I want is to compare both of the lists with the key barcode and if they matches I want to append the name from the second dictionary to the first one
for example:
data_from_frontend = [
{ barcode: '1', name_en: 'milk' },
{ barcode: '2', name_en: 'water' },
{ barcode: '3', name_en: 'cheese' },
{ barcode: '10', name_en: 'pepsi' },
]
result_from_query = [
{ barcode: '1', name: 'PID012343' },
{ barcode: '2', name: 'PID123454' },
{ barcode: '10', name: 'PID123432' },
]
I want to compare both of the lists and by the barcode and if they match I want to merge the the pair of both variables to a new one + adding the one the doesn't match to another list, so the outcome would be two new variables with the [matched + name] and not_found, how can I achieve that ?
Here is what i've tried
equal = []
not_equal = []
no_barcode = []
x = [ { "age": "22" }, { "name": "John Doe" }, { "name": "Jane Doe" }, { "name": "Doctor" }, { "name": "Engineer" } ]
y = [ { "name": "Engineer" }, { "name": "Jane Doe" }, { "name": "Doctor" } ]
x_sort = sorted(x, key=lambda k: ("name" not in k, k.get("name", None)))
y_sort = sorted(y, key=lambda k: ("name" not in k, k.get("name", None)))
print(y_sort)
for x_val in x_sort:
if "name" not in x_val.keys():
no_barcode.append(x_val)
else:
for y_val in y_sort:
if x_val["name"] == y_val["name"]:
equal.append(x_val)
mapped = map(lambda k: k["name"], y_sort)
if x_val["name"] not in mapped:
not_equal.append(x_val)
print('equal')
print(equal)
print('not equal')
print(not_equal)
First of all you should fix your dict keys and enclose them into quotes.
Then you can use generator expression to find items, for example:
print('initial dict:')
pprint.pprint(data_from_frontend)
for item in result_from_query:
item_found = next((i for i in data_from_frontend if i['barcode'] == item['barcode']), False)
if item_found:
item_found['name'] = item['name']
print('dict after search:')
pprint.pprint(data_from_frontend)
will produce:
initial dict:
[{'barcode': '1', 'name_en': 'milk'},
{'barcode': '2', 'name_en': 'water'},
{'barcode': '3', 'name_en': 'cheese'},
{'barcode': '10', 'name_en': 'pepsi'}]
dict after search:
[{'barcode': '1', 'name': 'PID012343', 'name_en': 'milk'},
{'barcode': '2', 'name': 'PID123454', 'name_en': 'water'},
{'barcode': '3', 'name_en': 'cheese'},
{'barcode': '10', 'name': 'PID123432', 'name_en': 'pepsi'}]
Using False in generator will avoid error when searching by barcode value not existing in target dict.
P.S. dont forget to import pprint if you want to use it
P.P.S. and sure you can create new dict instead of modifying existing one, using same logic
Your barcode matching results can be get like this.
barcode = 'barcode'
name_en = 'name_en'
name = 'name'
matching_result = data_from_frontend[:] #get a copy of front end data to use as the output
for i in range(len(data_from_frontend)):
for j in range(len(result_from_query)):
if(data_from_frontend[i][barcode] == result_from_query[j][barcode]):
matching_result[i][name] = result_from_query[j][name]
break
print(matching_result)