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How to get a list of dictionaries from the following code?


How to get a list of dictionaries from the following code?

By : Vialeila
Date : October 25 2020, 07:10 PM
With these it helps This is because you are always updating the same dict instance.
A way to solve this, is to create a new local instance with the updated fields:
code :
for item in main_list:
    updated_dict = dict(dict2, **dict(zip(d, item)))
    list1.append(updated_dict)
list1 = [dict(dict2, **dict(zip(d, item))) for item in main_list]


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How can I iterate over a list of dictionaries and merge dictionaries to form new shorter list of dicts?

How can I iterate over a list of dictionaries and merge dictionaries to form new shorter list of dicts?


By : Roberto Albertoni
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further Assuming order of items in the merged list does not matter, simply go through each item in the list and copy it if you haven't seen it before or merge the fields if you have.
code :
merged = {}

for item in original:
    key = (item['price'], item['tickettypecode'], item['oneway'])
    if key in merged:
        for mergekey in ['inboundJourneys','outboundJourneys']:
            # assign extended copy rather than using list.extend()
            merged[key][mergekey] = merged[key][mergekey] + item[mergekey]
    else:
        merged[key] = item.copy()

mergedlist = merged.values()
I have a list of tuples containing dictionaries. How do I edit my code to find the values in a separate list, inside the

I have a list of tuples containing dictionaries. How do I edit my code to find the values in a separate list, inside the


By : Jean-Paul D
Date : March 29 2020, 07:55 AM
wish helps you You have the code checking if value contains all the items in input_list the wrong way around. all(v in input_list for v in value) checks that all the items in value can be found from input_list. If you change it the other way around it will work as you expected:
code :
all(v in value for v in input_list)
L = [(0, {0: [0], 1: [0, 1], 2: [0, 2], 3: [0, 3], 4: [0, 4]}), (1, {0: [1, 0], 1: [1], 2: [5, 1, 2,], 3: [1, 3], 4: [1, 4]}), (2, {0: [2, 0], 1: [2, 1], 2: [2], 3: [2, 3], 4: [2, 4]}), (3, {0: [3, 0], 1: [3, 1], 2: [3, 2], 3: [3], 4: [3, 4]}), (4, {0: [4, 0], 1: [4, 1], 2: [4, 2], 3: [4, 3], 4: [4]})]

input_list = set([0,1])

for tupl in L:
    dict_a = tupl[1]
    matching_key = [key for key, value in dict_a.items() if input_list <= set(value)]
    print('Node: ' + str(tupl[0]) + ' Match at key(s): ' + str(matching_key))
Node: 0 Match at key(s): [1]
Node: 1 Match at key(s): [0]
Node: 2 Match at key(s): []
Node: 3 Match at key(s): []
Node: 4 Match at key(s): []
Creating a unique list of dictionaries from a list of dictionaries which contains same keys but different values

Creating a unique list of dictionaries from a list of dictionaries which contains same keys but different values


By : Mauricio
Date : March 29 2020, 07:55 AM
seems to work fine suppose i have been given a list of dictionaries like this : , Here's a solution with plain python.
code :
dlst = [{"id":1, "symbol":'A', "num":4}, {"id":2, "symbol":'A', "num":3},
    {"id":1, "symbol":'A', "num":5}, {"id":2, "symbol":'B', "num":1}]


# Create a dict where keys are tuples of (id,symbol), values are num

combined_d = {}

for d in dlst:
    id_sym = (d["id"], d["symbol"])
    if id_sym in combined_d:
        combined_d[id_sym] += d["num"]
    else:
        combined_d[id_sym] = d["num"]

# create a list of dictionaries from the tuple-keyed dict

result = []

for k, v in combined_d.items():
    d = {"id": k[0], "symbol": k[1], "num": v}
    result.append(d)

print(result)
Convert list of dictionaries containing another list of dictionaries with multiple values to dataframe

Convert list of dictionaries containing another list of dictionaries with multiple values to dataframe


By : Richard
Date : March 29 2020, 07:55 AM
wish help you to fix your issue I think by changing my answer to your previous question, you can achieve what you want. Still start by filling nan with empty list:
code :
df['actions'][df['actions'].isnull()] = df['actions'][df['actions'].isnull()].apply(lambda x: [])
def find_action (list_action, action_type, what):
    for action in list_action:
        # for each action, see if the key action_type is the one wanted and what in the keys
        if action['action_type'] == action_type and what in action.keys():
            return action[what]
    # if not the right action type found, then empty
    return ''
df['a2cart_view'] = df['actions'].apply(find_action, args=(['add_to_cart','view']))
df['a2cart_click'] = df['actions'].apply(find_action, args=(['add_to_cart','click']))
df['pur_view'] = df['actions'].apply(find_action, args=(['purchase','view']))
df['pur_click'] = df['actions'].apply(find_action, args=(['purchase','click']))
df = df.drop('actions',axis=1)
Merging list of dictionaries into a smaller list of dictionaries dependent on recurring values

Merging list of dictionaries into a smaller list of dictionaries dependent on recurring values


By : user3687333
Date : March 29 2020, 07:55 AM
Hope this helps This could be done with a for-loop, iterating over list_of_dict and checking if the key from the dictionary is in the latest dictionary in the return list. If not, append it to the latest dictionary using the ** operator to merge the dictionaries, otherwise start a new dictionary in the return list.
Code:
code :
list_of_dict = [{'a' : '1'}, {'b' : 'something'}, {'c' : 'else'},
                {'a' : '2'}, {'b' : 'anything'}, {'c' : 'if'},
                {'a' : '3'}, {'b' : 'nothing'}, {'c' : 'matters'}]

rv = [{}]

for d in list_of_dict:
    if list(d)[0] not in rv[-1]:
        rv[-1] = {**rv[-1], **d}
    else:
        rv.append(d)
>>> rv
[{'a': '1', 'b': 'something', 'c': 'else'},
 {'a': '2', 'b': 'anything', 'c': 'if'},
 {'a': '3', 'b': 'nothing', 'c': 'matters'}]
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