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How to remove empty key-value from dictionary comprehension when applying filter

I am new to python and learning how to use a dictionary comprehension. I have a movie cast dictionary that I would like to filter on a specific value using the dictionary comprehension technique. I was able to get it work but for some reason I get empty dictionaries added as well if the condition is not met. Why does it do it? And how can I ensure these are not included?
movie_cast = [{'id': 90633,'name': 'Gal Gadot','cast_id': 0, 'order': 0},
{'id': 62064, 'name': 'Chris Pine','cast_id': 15, 'order': 1},
{'id': 41091, 'name': 'Kristen Wiig', 'cast_id': 12,'order': 2},
{'id': 41092, 'name': 'Pedro Pascal', 'cast_id': 13, 'order': 3},
{'id': 32, 'name': 'Robin Wright', 'cast_id': 78, 'order': 4}]
limit = 1
cast_limit = []
for dict in movie_cast:
d = {key:value for (key,value) in dict.items() if dict['order'] < limit}
cast_limit.append(d)
print(cast_limit)
current_result = [{'id': 90633,'name': 'Gal Gadot','cast_id': 0, 'order': 0},
{'id': 62064, 'name': 'Chris Pine','cast_id': 15, 'order': 1},{},{},{}]
desired_result = [{'id': 90633,'name': 'Gal Gadot','cast_id': 0, 'order': 0},
{'id': 62064, 'name': 'Chris Pine','cast_id': 15, 'order': 1}]

Try with this (you need a list comprehension, not a dict comprehension):
cast_limit = [dct for dct in movie_cast if dct['order'] < limit]
I.e., you need to filter out elements of the list, not elements of a dict.

remove duplicate dictionary python

I have a problem, I have a list like this:
[{'id': 34, 'questionid': 5, 'text': 'yes', 'score': 1}, {'id': 10, 'questionid': 5,
'text': 'test answer updated', 'score': 2}, {'id': 20, 'questionid': 5, 'text': 'no',
'score': 0}, {'id': 35, 'questionid': 5, 'text': 'yes', 'score': 1}]
and I want remove duplicate "questionid", "text" and "score", for example in this case I want output like this:
[{'id': 34, 'questionid': 5, 'text': 'yes', 'score': 1}, {'id': 10, 'questionid': 5,
'text': 'test answer updated', 'score': 2}, {'id': 20, 'questionid': 5, 'text': 'no',
'score': 0}]
How can I get this output in python?

We could create dictionary that has "questionid", "text" and "score" tuple as key and dicts as values and use this dictionary to check for duplicate values in data:
from operator import itemgetter
out = {}
for d in data:
key = itemgetter("questionid", "text", "score")(d)
if key not in out:
out[key] = d
out = list(out.values())
Output:
[{'id': 34, 'questionid': 5, 'text': 'yes', 'score': 1},
{'id': 10, 'questionid': 5, 'text': 'test answer updated', 'score': 2},
{'id': 20, 'questionid': 5, 'text': 'no', 'score': 0}]

Google Charts with data from MongoDB in Python

I'm trying to set data from database MongoDB for drawing Google Chart in python.
import pymongo
from pymongo import MongoClient
import pandas as pd
from string import Template
import pprint
cluster = MongoClient("mongodb+srv://saman:1234#cluster0-vhwfp.mongodb.net/test?retryWrites=true&w=majority")
db= cluster["Test"]
collection= db["Test"]
rm=list(collection.find({}))
chart_data_str= ''
results= collection.find({})
for x in results:
chart_data_str += '%s,\n' %x
print(chart_data_str)
I want to know, How can I change the folowing format
{'_id': 5, 'name': 'Mahdi', 'Score': 9},
{'_id': 1, 'name': 'Ali', 'Score': 3},
{'_id': 3, 'name': 'Christian', 'Score': 6},
{'_id': 4, 'name': 'Niklas', 'Score': 1},
{'_id': 2, 'name': 'Dominik', 'Score': 2},
{'_id': 0, 'name': 'Saman', 'Score': 5},
to thises format
['Mahdi', 9],
['Ali', 3],
['Christian', 6],
['Niklas', 1],
['Dominik', 2],
['Saman', 5]
thanks

If you just want to print that out as a string:
for result in collection.find({}, {'_id': 0}):
print(f"[{result.get('name')}, {result.get('Score')}],")
If this isn't what you want, please clarify your question.

Python update a value in a list of dictionaries from another list of dictionaries

If given two list of dictionaries (score_list and update_list) below, how do I update score_list from the list of dictionaries from update_list?
score_list = [{'id': 1, 'score': 123}, {'id': 2, 'score': 234}, {'id': 3, 'score': 345}]
update_list = [{'id': 1, 'score': 500}, {'id': 3, 'score': 300}]
# return this
score_list = [{'id': 1, 'score': 500}, {'id': 2, 'score': 234}, {'id': 3, 'score': 300}]

I highly recommend using a mapping when you have a unique key to match:
update_mapping = {d['id']: d for d in update_list}
score_list = [update_mapping.get(d['id'], d) for d in score_list]

Pythonic sort a list of dictionaries in a tricky order

I have a list of id's sorted in a proper oder:
ids = [1, 2, 4, 6, 5, 0, 3]
I also have a list of dictionaries, sorted in some random way:
rez = [{'val': 7, 'id': 1}, {'val': 8, 'id': 2}, {'val': 2, 'id': 3}, {'val': 0, 'id': 4}, {'val': -1, 'id': 5}, {'val': -4, 'id': 6}, {'val': 9, 'id': 0}]
My intention is to sort rez list in a way that corresponds to ids:
rez = [{'val': 7, 'id': 1}, {'val': 8, 'id': 2}, {'val': 0, 'id': 4}, {'val': -4, 'id': 6}, {'val': -1, 'id': 5}, {'val': 9, 'id': 0}, {'val': 2, 'id': 3}]
I tried:
rez.sort(key = lambda x: ids.index(x['id']))
However that way is too slow for me, as len(ids) > 150K, and each dict actually had a lot of keys (some values there are strings). Any suggestion how to do it in the most pythonic, but still fastest way?

You don't need to sort because ids specifies the entire ordering of the result. You just need to pick the correct elements by their ids:
rez_dict = {d['id']:d for d in rez}
rez_ordered = [rez_dict[id] for id in ids]
Which gives:
>>> rez_ordered
[{'id': 1, 'val': 7}, {'id': 2, 'val': 8}, {'id': 4, 'val': 0}, {'id': 6, 'val': -4}, {'id': 5, 'val': -1}, {'id': 0, 'val': 9}, {'id': 3, 'val': 2}]
This should be faster than sorting because it can be done in linear time on average, while sort is O(nlogn).
Note that this assumes that there will be one entry per id, as in your example.

I think you are on the right track. If you need to speed it up, because your list is too long and you are having quadratic complexity, you can turn the list into a dictionary first, mapping the ids to their respective indices.
indices = {id_: pos for pos, id_ in enumerate(ids)}
rez.sort(key = lambda x: indices[x['id']])
This way, indices is {0: 5, 1: 0, 2: 1, 3: 6, 4: 2, 5: 4, 6: 3}, and rez is
[{'id': 1, 'val': 7},
{'id': 2, 'val': 8},
{'id': 4, 'val': 0},
{'id': 6, 'val': -4},
{'id': 5, 'val': -1},
{'id': 0, 'val': 9},
{'id': 3, 'val': 2}]