Sort timestamp in python dictionary

Sort timestamp in python dictionary

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Sort timestamp in python dictionary
Tag : python , By : user98832
Date : November 29 2020, 09:01 AM

To fix the issue you can do ,
mydict.sort(key=lambda x:x['Timestamp'])

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Sort Multidimensional Dictionary by value and return newly sorted dictionary in Python

Tag : python , By : Adam Hill
Date : March 29 2020, 07:55 AM
it fixes the issue Dictionaries are unordered up to (and including) . Only for a few releases of dictionaries are ordered. As is written in the documentation:
from collections import OrderedDict

OrderedDict(sorted(my_dict.items(),key=lambda x:int(x[1]['marks']),reverse=True))
>>> OrderedDict(sorted(my_dict.items(),key=lambda x:int(x[1]['marks']),reverse=True))
OrderedDict([('005', {'name': 'EEE', 'date_accessed': '2017-07-12 17:43:42', 'class': '99', 'marks': '999'}), ('004', {'name': 'DDD', 'date_accessed': '2017-07-12 17:43:24', 'class': '50', 'marks': '100'}), ('003', {'name': 'CCC', 'date_accessed': '2017-07-12 17:43:24', 'class': '13', 'marks': '90'}), ('002', {'name': 'BBB', 'date_accessed': '2017-07-12 17:43:24', 'class': '10', 'marks': '90'}), ('001', {'name': 'AAA', 'date_accessed': '2017-07-12 17:43:24', 'class': '9', 'marks': '80'})])

Python Dictionary sort keys by a value with timestamp

Tag : python , By : vitorcoliveira
Date : March 29 2020, 07:55 AM
this will help By nature a dictionary is not sorted, so you can instead create a list of tuples with your key and values.
lst = sorted(d.items(), key = lambda kvp: kvp[1][1])

[('00:00:00:00:00:01', ['Aug 18, 2017 18:15:56.081727942 CEST', 'Aug 18, 2017 18:25:37.669160369 CEST', '-60', 18, 'SSID1']), ...]
lst = sorted(d.items(), key = lambda kvp: to_datetime(kvp[1][1]))

In Python, given a dictionary with lists in the values, how do I sort the dictionary based on the amount of items in tha

Tag : python , By : Tim Coffman
Date : March 29 2020, 07:55 AM
around this issue you need to use an OrderedDict see https://docs.python.org/3/library/collections.html#collections.OrderedDict
Based on their example
from collections import OrderedDict

d={'a': ['apple'], 'd': ['dog', 'dance', 'dragon'], 'r': ['robot'], 'c': ['cow', 'cotton']}
ordered = OrderedDict(sorted(d.items(),key=lambda t: len(t[1]),reverse=True))

In python, I want to sort a dictionary based on the dictionary of dictionary values

Tag : python , By : Chris Lomax
Date : March 29 2020, 07:55 AM
help you fix your problem In python I have a dictionary of dictionary such as , You can use a custom key for sorting:
print(dict(sorted(my_dict.items(), key=lambda x: x[1]['mark3'], reverse=True)))
{'C': {'mark1': '11', 'mark2': '19', 'mark3': '25'}, 'A': {'mark1': '15', 'mark2': '12', 'mark3': '20'}, 'B': {'mark1': '10', 'mark2': '90', 'mark3': '14'}}
from collections import OrderedDict
my_dict = OrderedDict([('A', {'mark1': '15', 'mark2': '12', 'mark3': '20'}), ('B', {'mark1': '10', 'mark2': '90', 'mark3': '14'}), ('C', {'mark1': '11', 'mark2': '19', 'mark3': '25'})])
print(OrderedDict(sorted(my_dict.items(), key=lambda x: x[1]['mark3'], reverse=True)))
print(dict(sorted(my_dict.items(), key=lambda x: (x[1]['mark3'], x[1]['mark2']), reverse=True)))

How to sort a dictionary inside a dictionary based on the key(alphabetically) of the inner dictionary in python

Tag : python , By : Bo.
Date : January 02 2021, 06:48 AM
To fix this issue It sounds like you need to call the sorted function on each of the "inner" dictionaries individually, as well as on the "outer" dictionary. You would also need to cast your inner dictionaries to have the OrderedDict type for them to be sortable. Here's what I could come up with, but there might be shorter/more idiomatic ways to do this:
# Sort the outer dictionary by key from high to low
OrderedDict(sorted(d.items(), key=lambda t: t[0], reverse=True))

# Sort the inner dictionaries by key
for item in d.items(): 
    d[item[0]] = OrderedDict(sorted(item[1].items(), key=lambda x: x[0])) 
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