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Sorting list of tuples (Python)

I am new in python and for practice reasons, I am trying to solve the following task:
Given a list of tuples
A = [(17, 8), (17, 12), (7, 2), (9, 15), (9, 17), (1, 4), (3, 9), (12, 14)]
My goal is: to sort in (n log n) time in descending order the list according to the first element of the set and if the first element of two sets are the same, sorting in descending order according to the second element. --> x <= x', y <= y'
So, I want to get the result:
A = [(17, 12), (17, 8), (12, 14), (9, 17), (9, 15), (7, 2), (3, 9), (1, 4)]
I have tried using the following code:
A.sort(reverse = True, key=lambda x: x[0] )
But it just sorts according to the first element and I do know if it is in n log n time.
Could you please help me with that?
Thank you!

To sort based on both values, remove the key function:
>>> A.sort(reverse = True)
>>> A
[(17, 12), (17, 8), (12, 14), (9, 17), (9, 15), (7, 2), (3, 9), (1, 4)]
And yes, Python sorts in O(n log n), for more, check out Timsort on Wikipedia, the algorithm Python uses for sorting.

Python's inbuilt sorted() method with a comparator function(Similar to C++) will work
sorted(tup, key = lambda x: x[0])

Tuples are naturally sorted by the first element, then the second:
A.sort(reverse = True)
Output as requested.

There you go:
A = [(17, 8), (17, 12), (7, 2), (9, 15), (9, 17), (1, 4), (3, 9), (12, 14)]
reverse_sorted = []
for item in reversed(sorted(A)):
reverse_sorted.append(item)
print(reverse_sorted)

As to add to the other's excellent answers. Search on Google with "python list sort time complexity" will return for you what you seek regarding python's sort time complexities (that others have already written above).
In addition, if you need to move away from using python's inbuilt sort function. There are many different sorting algorithms created over the years that you can recreate that have the desired time complexities. It would be a good "training" exercise if you're new to programming in general.

Group items in a list and calculate sums

I have a list with weekly figures and need to obtain the grouped totals by month.
The following code does the job, but there should be a more pythonic way of doing it with using the standard libraries.
The drawback of the code below is that the list needs to be in sorted order.
#Test data (not sorted)
sum_weekly=[('2020/01/05', 59), ('2020/01/19', 88), ('2020/01/26', 95), ('2020/02/02', 89),
('2020/02/09', 113), ('2020/02/16', 90), ('2020/02/23', 68), ('2020/03/01', 74), ('2020/03/08', 85),
('2020/04/19', 6), ('2020/04/26', 5), ('2020/05/03', 14),
('2020/05/10', 5), ('2020/05/17', 20), ('2020/05/24', 28),('2020/03/15', 56), ('2020/03/29', 5), ('2020/04/12', 2),]
month = sum_weekly[0][0].split('/')[1]
count=0
out=[]
for item in sum_weekly:
m_sel = item[0].split('/')[1]
if m_sel!=month:
out.append((month, count))
count=item[1]
else:
count+=item[1]
month = m_sel
out.append((month, count))
# monthly sums output as ('01', 242), ('02', 360), ('03', 220), ('04', 13), ('05', 67)
print (out)

You could use defaultdict to store the result instead of a list. The keys of the dictionary would be the months and you can simply add the values with the same month (key).
Possible implementation:
# Test Data
from collections import defaultdict
sum_weekly = [('2020/01/05', 59), ('2020/01/19', 88), ('2020/01/26', 95), ('2020/02/02', 89),
('2020/02/09', 113), ('2020/02/16', 90), ('2020/02/23', 68), ('2020/03/01', 74), ('2020/03/08', 85),
('2020/03/15', 56), ('2020/03/29', 5), ('2020/04/12', 2), ('2020/04/19', 6), ('2020/04/26', 5),
('2020/05/03', 14),
('2020/05/10', 5), ('2020/05/17', 20), ('2020/05/24', 28)]
results = defaultdict(int)
for date, count in sum_weekly: # used unpacking to make it clearer
month = date.split('/')[1]
# because we use a defaultdict if the key does not exist it
# the entry for the key will be created and initialize at zero
results[month] += count
print(results)

You can use itertools.groupby (it is part of standard library) - it does pretty much what you did under the hood (grouping together sequences of elements for which the key function gives same output). It can look like the following:
import itertools
def select_month(item):
return item[0].split('/')[1]
def get_value(item):
return item[1]
result = [(month, sum(map(get_value, group)))
for month, group in itertools.groupby(sorted(sum_weekly), select_month)]
print(result)

Terse, but maybe not that pythonic:
import calendar, functools, collections
{calendar.month_name[i]: val for i, val in functools.reduce(lambda a, b: a + b, [collections.Counter({datetime.datetime.strptime(time, '%Y/%m/%d').month: val}) for time, val in sum_weekly]).items()}

a method using pyspark
from pyspark import SparkContext
sc = SparkContext()
l = sc.parallelize(sum_weekly)
r = l.map(lambda x: (x[0].split("/")[1], x[1])).reduceByKey(lambda p, q: (p + q)).collect()
print(r) #[('04', 13), ('02', 360), ('01', 242), ('03', 220), ('05', 67)]

You can accomplish this with a Pandas dataframe. First, you isolate the month, and then use groupby.sum().
import pandas as pd
sum_weekly=[('2020/01/05', 59), ('2020/01/19', 88), ('2020/01/26', 95), ('2020/02/02', 89), ('2020/02/09', 113), ('2020/02/16', 90), ('2020/02/23', 68), ('2020/03/01', 74), ('2020/03/08', 85), ('2020/04/19', 6), ('2020/04/26', 5), ('2020/05/03', 14), ('2020/05/10', 5), ('2020/05/17', 20), ('2020/05/24', 28),('2020/03/15', 56), ('2020/03/29', 5), ('2020/04/12', 2)]
df= pd.DataFrame(sum_weekly)
df.columns=['Date','Sum']
df['Month'] = df['Date'].str.split('/').str[1]
print(df.groupby('Month').sum())

How to filter dictionary by value? [duplicate]

This question already has answers here:
How to filter a dictionary according to an arbitrary condition function?
(7 answers)
Closed 4 years ago.
I have dictionary in format "site_mame": (side_id, frequency):
d=[{'fpdownload2.macromedia.com': (1, 88),
'laposte.net': (2, 23),
'www.laposte.net': (3, 119),
'www.google.com': (4, 5441),
'match.rtbidder.net': (5, 84),
'x2.vindicosuite.com': (6, 37),
'rp.gwallet.com': (7, 88)}]
Is there a smart way to filter dictionary d by value so that I have only those positions, where frequency is less than 100? For example:
d=[{'fpdownload2.macromedia.com': (1, 88),
'laposte.net': (2, 23),
'match.rtbidder.net': (5, 84),
'x2.vindicosuite.com': (6, 37),
'rp.gwallet.com': (7, 88)}]
I don't want to use loops, just looking for smart and efficient solution...

You can use a dictionary comprehension with unpacking for a more Pythonic result:
d=[{'fpdownload2.macromedia.com': (1, 88),
'laposte.net': (2, 23),
'www.laposte.net': (3, 119),
'www.google.com': (4, 5441),
'match.rtbidder.net': (5, 84),
'x2.vindicosuite.com': (6, 37),
'rp.gwallet.com': (7, 88)}]
new_data = [{a:(b, c) for a, (b, c) in d[0].items() if c < 100}]
Output:
[{'laposte.net': (2, 23), 'fpdownload2.macromedia.com': (1, 88), 'match.rtbidder.net': (5, 84), 'x2.vindicosuite.com': (6, 37), 'rp.gwallet.com': (7, 88)}]

You can use a dictionary comprehension to do the filtering:
d = {
'fpdownload2.macromedia.com': (1, 88),
'laposte.net': (2, 23),
'www.laposte.net': (3, 119),
'www.google.com': (4, 5441),
'match.rtbidder.net': (5, 84),
'x2.vindicosuite.com': (6, 37),
'rp.gwallet.com': (7, 88),
}
d_filtered = {
k: v
for k, v in d.items()
if v[1] < 100
}

What you want is a dictionary comprehension. I'll show it with a different example:
d = {'spam': 120, 'eggs': 20, 'ham': 37, 'cheese': 101}
d = {key: value for key, value in d.items() if value >= 100}
If you don't already understand list comprehensions, this probably looks like magic that you won't be able to maintain and debug, so I'll show you how to break it out into an explicit loop statement that you should be able to understand easily:
new_d = {}
for key, value in d.items():
if value >= 100:
new_d[key] = value
If you can't figure out how to turn that back into the comprehension, just use the statement version until you learn a bit more; it's a bit more verbose, but better to have code you can think through in your head.
Your problem is slightly more complicated, because the values aren't just a number but a tuple of two numbers (so you want to filter on value[1], not value). And because you have a list of one dict rather than just a dict (so you may need to do this for each dict in the list). And of course my filter test isn't the same as yours. But hopefully you can figure it out from here.

Storing python list of tuples in a redis sorted set where second element of tuple is used as score

I have a Python list of tuples like so:
lst = [(22, -150.0), (23, -150.0), (18, -148.5), (15, -99.4), (5, -75.75), (4, -49.2), (13, -49.0), (9, -41.3), (20, -25.5), (17, -22.3), (10, -13.1), (16, -12.5), (14, -9.8), (3, -8.5), (1, -8.4), (12, -1.5), (7, -0.6), (2, -0.4), (6, 1.7), (21, 2.7)]
How do I pass that into a redis sorted set, such that the second value in each tuple is used as the score? I tried zadd(sorted_set, *lst), but I'm getting the error value is not a valid float.

The zadd method of redis.StrictRedis expects arguments in the form of
zadd(key, score1, name1, score2, name2, ...)
In other words, you need to reverse and unpack the inner tuples as well:
In [6]: r = redis.StrictRedis(...)
In [8]: r.zadd('sset', *(y for x in lst for y in x[::-1]))
Out[8]: 20
Redis.zadd and StrictRedis.zadd behave differently -- see README for more details. So if you use redis.Redis instead of redis.StrictRedis (you shouldn't), do not reverse tuples:
r.zadd('sset', *(y for x in lst for y in x))

Sort dict of tuples by either tuple value with customised comparator

I'm working on some python dicts of tuples. Each tuple containing 2 ints. The first digit in the tuple is reffered to as value and the second digit is referred to as work. I have 3 different comparators and I need to sort the dicts into descending order. This order should be determined by which comparator is called. i.e. the dict can be sorted 3 different ways. I've tried as many different ways as I could find to get this to work. I can do it without using the comparator by just breaking it up into a list and sorting by slicing the tuples but if anyone can shed some light on the syntax to sort using the comparators it would be greatly appreciated. Mine seems to be returning correctly for cmpWork but the other 2 aren't reversed.
Also it would be great if I could get the dict sorted by the tuple values.
I got a sort working with
sortedSubjects = sorted(tmpSubjects.iteritems(), key = operator.itemgetter(1), reverse = True)
but this doesn't let me slice the tuples.
First time posting noob so apologies for any mistakes.
def cmpValue(subInfo1, subInfo2):
return cmp(subInfo2[0] , subInfo1[0])
def cmpWork(subInfo1, subInfo2):
return cmp(subInfo1[1] , subInfo2[1])
def cmpRatio(subInfo1, subInfo2):
return cmp((float(subInfo2[0]) / subInfo2[1]) , (float(subInfo1[0]) / subInfo1[1]))
def greedyAdvisor(subjects, comparator):
tmpSubjects = subjects.copy()
sortedSubjects = sorted(tmpSubjects.values(), comparator, reverse = True)
print sortedSubjects
smallCatalog = {'6.00': (16, 8),'1.00': (7, 7),'6.01': (5, 3),'15.01': (9, 6)}
greedyAdvisor(smallCatalog, cmpRatio)
greedyAdvisor(smallCatalog, cmpValue)
greedyAdvisor(smallCatalog, cmpWork)
[(7, 7), (9, 6), (5, 3), (16, 8)]
[(5, 3), (7, 7), (9, 6), (16, 8)]
[(16, 8), (7, 7), (9, 6), (5, 3)]
ps
The line
sortedSubjects = sorted(tmpSubjects.iteritems(), key = operator.itemgetter(1), reverse = True)
returns
[('6.00', (16, 8)), ('15.01', (9, 6)), ('1.00', (7, 7)), ('6.01', (5, 3))]
which is almost exactly what I'm looking for except that I can't sort by the second value in the tuple and I can't sort by cmpRatio either.

but this doesn't let me slice the tuples
Starting with your example:
sortedSubjects = sorted(tmpSubjects.iteritems(),
key=operator.itemgetter(1),
cmp=comparator, # What about specifying the comparison?
reverse=True)

If you need to sort dictionary - use collections.OrderedDict
E.g., sort by 1st element of tuple
OrderedDict(sorted(smallCatalog.items(), key=lambda e:e[1][0]))
Out[109]: OrderedDict([('6.01', (5, 3)), ('1.00', (7, 7)), ('15.01', (9, 6)), ('6.00', (16, 8))])