I have a list:
[{'key':'key1', 'version':'1'}, {'key':'key2', 'version':'2'}, {'key':'key3', 'version':'3'}]
I want to remove all other keys from the dictionaries in the list and only to have 'key' and its value so the list would look like this:
[{'key': 'key1'}, {'key': 'key2'}, {'key': 'key3'}]
How can I extract it?
Here is an easy solution. This uses the .pop() method to remove the key that you do not want. -
lst = [{'key':'key1','version':'1'},{'key':'key2','version':'2'},{'key':'key3','version':'3'}]
for dicts in lst:
dicts.pop('version')
print(lst)
Result:
[{'key': 'key1'}, {'key': 'key2'}, {'key': 'key3'}]
This removes all version keys. Or you could use this method to remove everything except desired key:
new_l = []
for d in lst:
for key in d:
if key == 'key':
new_l.append({key: d[key]})
print(new_l)
Or even simpler, you can use a list and dictionary comprehension to solve this. I recommend you to do this because it is more readable and efficient.:
new_l = [{key: d[key] for key in d if key == 'key'} for d in lst]
print(new_l)
You can try this
in just one line!
yourList = [{'key':'key1','version':'1'},{'key':'key2','version':'2'},{'key':'key3','version':'3'}]
resultList = [{'key':dic['key']} for dic in yourList if 'key' in dic]
print(resultList)
Try this:
lst = [{'key':'key1','version':'1'},{'key':'key2','version':'2'},{'key':'key3','version':'3'}]
[item.pop('version') for item in lst]
Or, removing all possible keys that does not match 'key':
lst = [{'key':'key1','version':'1'},{'key':'key2','version':'2'},{'key':'key3','version':'3'}]
for item in lst:
for key in list(item.keys()):
item.pop(key) if key!='key' else None
You can try this:
l1 = [{'key':'key1','version':'1'},{'key':'key2','version':'2'},{'key':'key3','version':'3'}]
li2 = [{'key': j[x]} for j in l1 for x in j if x=='key']
print(li2)
Related
I have the following:
my_list = ["7777777", "888888", "99999"]
my_dict = {21058199: '500', 7777777: '500', 21058199: '500'}
I am trying to create a new dictionary which will include the dictionary value (from the original dictionary) that matches the list entry to the dictionary key (in the original dictionary)
for k in my_dict.keys():
if k in my_list:
new_dict.append(k)
print(new_dict)
should return
7777777: '500'
But I'm returning an empty set. I'm not sure what I'm doing wrong here
A dictionary comprehension would provide you what you need.
You need to make sure the types agree (int vs. str)
Unless the list is significantly longer than the dict, it will be much more efficient to iterate over the list and check that key is in the dict than the other way around.
E.g.:
In []:
new_dict = {k: my_dict[k] for k in map(int, my_list) if k in my_dict}
print(new_dict)
Out[]:
{7777777: '500'}
Try like this :
my_list = ["7777777", "888888", "99999"]
my_dict = {21058199: '500', 7777777: '500', 21058199: '500'}
new_dict = {k:my_dict[k] for k in my_dict.keys() if str(k) in my_list}
print(new_dict)
# {7777777: '500'}
Update:
You can also do this with project function from funcy library.
from funcy import project
new_dict = project(my_dict, map(int, my_list))
print(new_dict)
You have to int() the iterator.
my_list = ["7777777", "888888", "99999"]
my_dict = {21058199: '500', 7777777: '500', 21058199: '500'}
for l_i in my_list:
if l_i in my_dict:
print(my_dict[int(l_i)])
Your dictionary keys are of the type 'Int', while your list items are strings. You need to convert one into the other.
for example:
new_dict = {}
for k in my_dict.keys():
if str(k) in my_list:
new_dict[k] = my_dict[k]
Note that you cannot use .append() to add key-value pairs to a dictionary.
How can I get a dictionary value in a list of dictionaries, based on the dictionary satisfying some condition? For instance, if one of the dictionaries in the list has the id=5, I want to print the value corresponding to the name key of that dictionary:
list = [{'name': 'Mike', 'id': 1}, {'name': 'Ellen', 'id': 5}]
id = 5
if any(m['id'] == id for m in list):
print m['name']
This won't work because m is not defined outside the if statement.
You have a list of dictionaries, so you can use a list comprehension:
[d for d in lst if d['id'] == 5]
# [{'id': 5, 'name': 'Ellen'}]
new_list = [m['name'] for m in list if m['id']==5]
print '\n'.join(new_list)
This will be easy to accomplish with a single for-loop:
for d in list:
if 'id' in d and d['in'] == 5:
print(d['name'])
There are two key concepts to learn here. The first is that we used a for loop to "go through each element of the list". The second, is that we used the in word to check if a dictionary had a certain key.
How about the following?
for entry in list:
if entry['id']==5:
print entry['name']
It doesn't exist in Python2, but a simple solution in Python3 would be to use a ChainMap instead of a list.
import collections
d = collections.ChainMap(*[{'name':'Mike', 'id': 1}, {'name':'Ellen', 'id': 5}])
if 'id' in d:
print(d['id'])
You can do it by using the filter function:
lis = [ {'name': 'Mike', 'id': 1}, {'name':'Ellen', 'id': 5}]
result = filter(lambda dic:dic['id']==5,lis)[0]['name']
print(result)
I have the following Python list:
list1 = ['EW:G:B<<LADHFSSFAFFF', 'CB:E:OWTOWTW', 'PP:E:A,A<F<AF', 'GR:A:OUO-1-XXX-EGD:forthyFive:1:HMJeCXX:7', 'SX:F:-111', 'DS:f:115.5', 'MW:AA:0', 'MA:A:0XT:i:0', 'EY:EE:KJERWEWERKJWE']
I would like to take the entries of this list and create a dictionary of key-values pairs that looks like
dictionary_list1 = {'EW':'G:B<<LADHFSSFAFFF', 'CB':'E:OWTOWTW', 'PP':'E:A,A<F<AF', 'GR':'A:OUO-1-XXX-EGD:forthyFive:1:HMJeCXX:7', 'SX':'F:-111', 'DS':'f:115.5', 'MW':'AA:0', 'MA':'A:0XT:i:0', 'EW':'EE:KJERWEWERKJWE'}
How does one parse/split the list above list1 to do this? My first instinct was to try try1 = list1.split(":"), but then I think it is impossible to retrieve the "key" for this list, as there are multiple colons :
What is the most pythonic way to do this?
You can specify a maximum number of times to split with the second argument to split.
list1 = ['EW:G:B<<LADHFSSFAFFF', 'CB:E:OWTOWTW', 'PP:E:A,A<F<AF', 'GR:A:OUO-1-XXX-EGD:forthyFive:1:HMJeCXX:7', 'SX:F:-111', 'DS:f:115.5', 'MW:AA:0', 'MA:A:0XT:i:0', 'EW:EE:KJERWEWERKJWE']
d = dict(item.split(':', 1) for item in list1)
Result:
>>> import pprint
>>> pprint.pprint(d)
{'CB': 'E:OWTOWTW',
'DS': 'f:115.5',
'EW': 'EE:KJERWEWERKJWE',
'GR': 'A:OUO-1-XXX-EGD:forthyFive:1:HMJeCXX:7',
'MA': 'A:0XT:i:0',
'MW': 'AA:0',
'PP': 'E:A,A<F<AF',
'SX': 'F:-111'}
If you'd like to keep track of values for non-unique keys, like 'EW:G:B<<LADHFSSFAFFF' and 'EW:EE:KJERWEWERKJWE', you could add keys to a collections.defaultdict:
import collections
d = collections.defaultdict(list)
for item in list1:
k,v = item.split(':', 1)
d[k].append(v)
Result:
>>> pprint.pprint(d)
{'CB': ['E:OWTOWTW'],
'DS': ['f:115.5'],
'EW': ['G:B<<LADHFSSFAFFF', 'EE:KJERWEWERKJWE'],
'GR': ['A:OUO-1-XXX-EGD:forthyFive:1:HMJeCXX:7'],
'MA': ['A:0XT:i:0'],
'MW': ['AA:0'],
'PP': ['E:A,A<F<AF'],
'SX': ['F:-111']}
You can also use str.partition
list1 = ['EW:G:B<<LADHFSSFAFFF', 'CB:E:OWTOWTW', 'PP:E:A,A<F<AF', 'GR:A:OUO-1-XXX-EGD:forthyFive:1:HMJeCXX:7', 'SX:F:-111', 'DS:f:115.5', 'MW:AA:0', 'MA:A:0XT:i:0', 'EW:EE:KJERWEWERKJWE']
d = dict([t for t in x.partition(':') if t!=':'] for x in list1)
# or more simply as TigerhawkT3 mentioned in the comment
d = dict(x.partition(':')[::2] for x in list1)
for k, v in d.items():
print('{}: {}'.format(k, v))
Output:
MW: AA:0
CB: E:OWTOWTW
GR: A:OUO-1-XXX-EGD:forthyFive:1:HMJeCXX:7
PP: E:A,A<F<AF
EW: EE:KJERWEWERKJWE
SX: F:-111
DS: f:115.5
MA: A:0XT:i:0
I have two lists:
list1 = [IM12345, IM12346, IM12347, IM12348]
list2 = [ID300, ID404, ID300, ID601]
list2 associates with the corresponding list1 values. list1 has unique values where as list2 has duplicates.
I want to make list2 unique corresponding associated value will add in the that list2 value.
Dict= {ID300: {IM12345, IM12347}, ID404: IM12346, ID601: IM12348}
Above pattern can be in list, set or dictionary.
Which algorithm in python should I use to get the above result?
You could try collections.defaultdict:
from collections import defaultdict
d = defaultdict(set)
list1 = ['IM12345', 'IM12346', 'IM12347', 'IM12348']
list2 = ['ID300', 'ID404', 'ID300', 'ID601']
for key, value in zip(list2, list1):
d[key].add(value)
Demo:
>>> d
defaultdict(<class 'set'>, {'ID300': {'IM12345', 'IM12347'}, 'ID404': {'IM12346'}, 'ID601': {'IM12348'}})
>>>
>>>
>>> for i, j in d.items():
... print(i, j)
...
...
ID601 {'IM12348'}
ID300 {'IM12345', 'IM12347'}
ID404 {'IM12346'}
You can create a dict to save the dataset
list1 = ["IM12345", "IM12346", "IM12347", "IM12348"]
list2 = ["ID300", "ID404", "ID300", "ID601"]
dictResult=dict()
i=0
for item in list2:
print item
if dictResult.has_key(item):
dictResult[item].append(list1[i])
else:
dictResult[item]=[list1[i]]
i=i+1
print dictResult
Result:
{'ID404': ['IM12346'], 'ID300': ['IM12345', 'IM12347'], 'ID601': ['IM12348']}
Might not completely be a pythonic way but - here it goes:
Map the input:
map = dict(zip(list1, list2))
Now you can do an inverse mapping:
inv_map = {}
for k, v in map.iteritems():
inv_map[v] = inv_map.get(v, [])
inv_map[v].append(k)
Result for the example above:
>>> inv_map
{'ID404': ['IM12346'], 'ID300': ['IM12345', 'IM12347'], 'ID601': ['IM12348']}
Another way of doing it could be with list operations.
yourList = ["1","2","3","4","1","2"]
newList = []
for f in yourList:
if f not in newList:
newList.append(f)
Simple solution
from collections import defaultdict
list1 = ['IM12345', 'IM12346', 'IM12347', 'IM12348']
list2 = ['ID300', 'ID404', 'ID300', 'ID601']
d = defaultdict(list)
for n in range(len(list2)):
d[list2[n]].append(list1[n])
print d.items()
Result:
[('ID404', ['IM12346']), ('ID300', ['IM12345', 'IM12347']), ('ID601', ['IM12348'])]
Python2.7 Documentation----defaultdict
dict1 = {"1":"a" "2":"b" "3":"c"}
for dict2 in all_dict:
if compare_dicts(dict1, dict2):
...
...
I need the index of the dict inside all_dict which is exactly the same like dict1.
Does the for loop go over all_dict sequentially so I can count the iterations inside the for-loop?
You can write a function yielding all indices of matching objects in a list using enumerate():
def findall(lst, value):
for i, x in enumerate(lst):
if x == value:
yield i
You can apply this to your use case like this:
matching_indices = list(findall(all_dicts, dict1))
If you are just looking for a single match, the list.index() method is all you need:
matching_index = all_dicts.index(dict1)
Use filter:
filter(lambda x: x == dict1, all_dict)
This returns a list of all dictionaries you're looking for. Example:
>>> all_dict = [{'a':1}, {'b':2}, {'a':1}]
>>> dict1 = {'a':1}
>>> filter(lambda x: x == dict1, all_dict)
[{'a': 1}, {'a': 1}]