How to add names to teams? - python-3.x

I am trying to put them into three groups of two people each. I'm not sure where I need to go from here. I don't know how I can add two people to a group efficiently.
so it should look like this
group_one = {'2': dan, '8': tom}
group_two = {'10': james, '12': emily}
group_three = {'7': kim , '13': jones}

You could slice your list within dict comprehension to create each variable:
group_one = {i[1]:i[0] for i in my_list[0:2]}
group_two = {i[1]:i[0] for i in my_list[2:4]}
group_three = {i[1]:i[0] for i in my_list[4:6]}
>>> group_one
{2: 'dan', 8: 'tom'}
>>> group_two
{10: 'james', 12: 'emily'}
>>> group_three
{7: 'kim', 13: 'jones'}

#Sacul code is fine. But if you want an even smaller solution do:
>>> my_list = [['dan',2],['tom',8],['james',10],['emily',12],['kim',7],['jones',13]]
>>> teams = [{i[1] : i[0] for i in my_list[n:n+2]} for n in range(0, len(my_list), 2)]
[{2: 'dan', 8: 'tom'}, {10: 'james', 12: 'emily'}, {7: 'kim', 13: 'jones'}]
Now your teams are split and stored on a list of dictionaries.
To have them on different variables like group_one, group_two and group_three use unpacking.
>>> group_one, group_two, group_three = teams # Or simply use the code of teams

Related

Python3, Pandas, Extract associated Data from Dataframe to dictionaries

I've been working with a CSV file. I created a dataframe based on the file:
and added index names. and printed the dataframe (with added index names):
#Here is the code before failed computations.
import pandas as pd
import csv
colName = ['carIndex', 'carMake', 'Floatnum']
data2 = pd.read_csv('cars.csv', names=colName)
print(data2)
I have been trying to extract data to reach my goals although I have had some difficulty. My goals are as follows:
extract data and write alphabetical Dictionary with "carIndex" as key and a value of 0 - 2 (associated with Car A - C)
extract data and write alphabetical Dictionary with "carMake" as key and a value of 0 - 1 (Associated With Make X & Y)
Create (three) key-value pairs for the make "X" & "Y"'s values (associated with carIndex A-C) If a value doesn't exist the index should be None. append all three to a list of lists.
Finally take all three fields (First Dictionary, Second Dictionary, List-of-lists) and add them to a tuple for exportation
Anyone have suggestions for how I can extract the data as I want? Thanks in advance.
In Response to:
Will you please add two things to the question: 1. a text version of your dataframe (preferrably from print(df.to_dict())), and 2. a sample dataframe containing your expected output?
print(data2.to_dict()) (outputs) --> {'carIndex': {0: 'Car C', 1: 'Car A', 2: 'Car B', 3: 'Car B', 4: 'Car A'}, 'carMake': {0: ' Make X', 1: ' Make X', 2: ' Make X', 3: ' Make Y', 4: ' Make Y'}, 'Floatnum': {0: 2.0, 1: 2.5, 2: 1.5, 3: 4.0, 4: 3.5}}
Output Tuple I want: print(my_tup) (outputs) -->
({'Car A': 0, 'Car B': 1, 'Car C': 2}, {'Make X': 0, ' Make Y': 1}, [[2.5, 3.5], [1.5, 4.0], [1.0, None]])
Extract data and write alphabetical Dictionary with "carIndex" as key and a value of 0 - 2 (associated with Car A - C)
sorted = data2.sort_values('carIndex').drop_duplicates(subset='carIndex').reset_index()
carIndexDict = sorted['carIndex'].to_dict()
This will output
{0: 'Car A', 1: 'Car B', 2: 'Car C'}
Extract data and write alphabetical Dictionary with "carMake" as key and a value of 0 - 1 (Associated With Make X & Y)
Use the same strategy:
sorted = data2.sort_values('carMake').drop_duplicates(subset='carMake').reset_index()
carMakeDict = sorted['carMake'].to_dict()
Output:
{0: 'Make X', 1: 'Make Y'}
To make the list:
carIndexes = carIndexDict.values()
carMakes = carMakeDict.values()
full_list = []
for idx in carIndexes:
idx_search = data2.loc[df['carIndex'] == idx]
car_list = []
for make in carMakes:
make_search = idx_search.loc[idx_search['carMake'] == make]
if not make_search.empty:
car_list.append(make_search['Floatnum'].iloc[0])
else:
car_list.append(None)
full_list.append(car_list)
Outputs:
[[2.5, 3.5], [1.5, 4.0], [2.0, None]]
And finally the tuple:
myTuple = (carIndexDict, carMakeDict, full_list)
Outputs:
({0: 'Car A', 1: 'Car B', 2: 'Car C'}, {0: 'Make X', 1: 'Make Y'}, [[2.5, 3.5], [1.5, 4.0], [2.0, None]])

How can I replace a key-value pair in a nested dictionary with the value from the same key-value pair?

I want to replace the key-value pair in a directory with the value from the same key-value pair. In simple words, I want to replace
{k1:{key1:value1},k2:{Key2:value2}} with {k1:value1,k2:value2}
in simple words: {key1:value1} gets replaced with value1
My input is {'times': {0: 3284566678.449864},
'C0PLUSA': {0: 2.4529042386185244e-06},
'C1PLUSA': {0: 2.215571719760855e-06},
'I0PLUSA': {0: 1200.000000000029},
'M1BA': {0: 1.1971391933999447e-05},
'M1DA': {0: 2.5217060699404884e-06},
'M1GA': {0: 1.0922075194286918e-05},
'M1SA': {0: 0.0006776339108859284},
'M2BA': {0: 1.7152638293476106e-05}}
Code:
for i, key,value in a.items:
a[key][value] = value.values()
But I get an error saying this object is not iterable.
for node,value in a.items():
a = a.replace(node.values(), value.values())
but it still gives me an error.
I am expecting to get an output as:
{'times':3284566678.449864,
'C0PLUSA': 2.4529042386185244e-06,
'C1PLUSA': 2.215571719760855e-06,
'I0PLUSA': 1200.000000000029,
'M1BA': 1.1971391933999447e-05,
'M1DA': 2.5217060699404884e-06,
'M1GA': 1.0922075194286918e-05,
'M1SA': 0.0006776339108859284,
'M2BA': 1.7152638293476106e-05}
If all the keys in the inner dictionaries are 0 you can just do:
{k:v[0] for k,v in data.items()}
In case you want to ignore the keys of the inner dictionaries without knowing the value, you can see how this example works:
>>> data={'k1': {'key1': 'value1'}, 'k2': {'key2': 'value2'}}
>>> {k:list(v.values())[0] for k,v in data.items()}
{'k1': 'value1', 'k2': 'value2'}

Create dictionary using dictionary comprehension and assign multiple values to each key

Need to turn a list of words into a dictionary using dictionary comprehension. The keys would be the length of the words, and the values are the set of words from the original list whose length are that of its keys.
I am able to create a regular function for this purpose, but unable to do so in one dictionary comprehension.
As an example, I created a list of names of different lengths.
word_list = ["Amber", "Steven", "Carol", "Tuan", "Michael", "sam", "Wayne", "Anna", "Kay", "Jim", "D", "Belinda", "CharlieYu"]
def word_lengths_notit(input_list):
wl_dict = {}
for word in input_list:
if len(word) not in wl_dict.keys():
wl_dict[len(word)]=[] #create key that is the length of the word
wl_dict[len(word)].append(word.lower())
else:
if word.lower() not in wl_dict[len(word)]:
wl_dict[len(word)].append(word.lower())
print(wl_dict)
word_lengths_notit(word_list)
My output:
{5: ['amber', 'carol', 'wayne'], 6: ['steven'], 4: ['tuan', 'anna'], 7: ['michael', 'belinda'], 3: ['sam', 'kay', 'jim'], 1: ['d'], 9: ['charlieyu']}
This might not be the cleanest/most efficient code (I just started learning two weeks ago), but the output is correct.
Below, I tried dictionary comprehension, but it keeps overwriting my previous values instead of appending to it. I'm thinking I might have to use a list comprehension to collect all the words of the same length, but I'm not sure how (or if I can) create multiple lists of words of different lengths in one list comprehension.
def word_lengths(input_list):
wl_dict = {len(word):word for word in input_list]
print(wl_dict)
word_lengths(word_list)
Output: {5: 'Wayne', 6: 'Steven', 4: 'Anna', 7: 'Belinda', 3: 'Jim', 1: 'D', 9: 'CharlieYu'}
So you're looking to make a dict where each key is an integer and each value is a list, and you're looking to do it via dict comprehension. My advice for doing it in vanilla python is to simply nest a list comprehension (to filter words by name) inside of the dict comprehension:
word_list = ["Amber", "Steven", "Carol", "Tuan", "Michael", "sam", "Wayne", "Anna", "Kay", "Jim", "D", "Belinda", "CharlieYu"]
word_lengths = {n: [word for word in word_list if len(word) == n]
for n in range(10)}
If you want to avoid cases like 0: [], you could throw a ternary if clause on the end there to filter them out (e.g. if len([word for word in word_list if len(word) == n])). Alternatively, you could simply make a set of all the unique lengths that are present and iterate over that:
word_list = ["Amber", "Steven", "Carol", "Tuan", "Michael", "sam", "Wayne", "Anna", "Kay", "Jim", "D", "Belinda", "CharlieYu"]
possible_lengths = set([len(word) for word in word_list])
word_lengths = {n: [word for word in word_list if len(word) == n]
for n in possible_lengths}
The above code outputs the following on my machine:
>>> print(word_lengths)
{1: ['D'], 3: ['sam', 'Kay', 'Jim'], 4: ['Tuan', 'Anna'], 5: ['Amber', 'Carol', 'Wayne'], 6: ['Steven'], 7: ['Michael', 'Belinda'], 9: ['CharlieYu']}
Note that this solution is O(n^2) complexity. Look into the collections library, which almost certainly has some clever things in it you can do to get a faster solution.

Sorted a list of tuple and return first element of tuple in python [duplicate]

This question's answers are a community effort. Edit existing answers to improve this post. It is not currently accepting new answers or interactions.
I have a dictionary of values read from two fields in a database: a string field and a numeric field. The string field is unique, so that is the key of the dictionary.
I can sort on the keys, but how can I sort based on the values?
Note: I have read Stack Overflow question here How do I sort a list of dictionaries by a value of the dictionary? and probably could change my code to have a list of dictionaries, but since I do not really need a list of dictionaries I wanted to know if there is a simpler solution to sort either in ascending or descending order.
Python 3.7+ or CPython 3.6
Dicts preserve insertion order in Python 3.7+. Same in CPython 3.6, but it's an implementation detail.
>>> x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
>>> {k: v for k, v in sorted(x.items(), key=lambda item: item[1])}
{0: 0, 2: 1, 1: 2, 4: 3, 3: 4}
or
>>> dict(sorted(x.items(), key=lambda item: item[1]))
{0: 0, 2: 1, 1: 2, 4: 3, 3: 4}
Older Python
It is not possible to sort a dictionary, only to get a representation of a dictionary that is sorted. Dictionaries are inherently orderless, but other types, such as lists and tuples, are not. So you need an ordered data type to represent sorted values, which will be a list—probably a list of tuples.
For instance,
import operator
x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sorted_x = sorted(x.items(), key=operator.itemgetter(1))
sorted_x will be a list of tuples sorted by the second element in each tuple. dict(sorted_x) == x.
And for those wishing to sort on keys instead of values:
import operator
x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sorted_x = sorted(x.items(), key=operator.itemgetter(0))
In Python3 since unpacking is not allowed we can use
x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sorted_x = sorted(x.items(), key=lambda kv: kv[1])
If you want the output as a dict, you can use collections.OrderedDict:
import collections
sorted_dict = collections.OrderedDict(sorted_x)
As simple as: sorted(dict1, key=dict1.get)
Well, it is actually possible to do a "sort by dictionary values". Recently I had to do that in a Code Golf (Stack Overflow question Code golf: Word frequency chart). Abridged, the problem was of the kind: given a text, count how often each word is encountered and display a list of the top words, sorted by decreasing frequency.
If you construct a dictionary with the words as keys and the number of occurrences of each word as value, simplified here as:
from collections import defaultdict
d = defaultdict(int)
for w in text.split():
d[w] += 1
then you can get a list of the words, ordered by frequency of use with sorted(d, key=d.get) - the sort iterates over the dictionary keys, using the number of word occurrences as a sort key .
for w in sorted(d, key=d.get, reverse=True):
print(w, d[w])
I am writing this detailed explanation to illustrate what people often mean by "I can easily sort a dictionary by key, but how do I sort by value" - and I think the original post was trying to address such an issue. And the solution is to do sort of list of the keys, based on the values, as shown above.
You could use:
sorted(d.items(), key=lambda x: x[1])
This will sort the dictionary by the values of each entry within the dictionary from smallest to largest.
To sort it in descending order just add reverse=True:
sorted(d.items(), key=lambda x: x[1], reverse=True)
Input:
d = {'one':1,'three':3,'five':5,'two':2,'four':4}
a = sorted(d.items(), key=lambda x: x[1])
print(a)
Output:
[('one', 1), ('two', 2), ('three', 3), ('four', 4), ('five', 5)]
Dicts can't be sorted, but you can build a sorted list from them.
A sorted list of dict values:
sorted(d.values())
A list of (key, value) pairs, sorted by value:
from operator import itemgetter
sorted(d.items(), key=itemgetter(1))
In recent Python 2.7, we have the new OrderedDict type, which remembers the order in which the items were added.
>>> d = {"third": 3, "first": 1, "fourth": 4, "second": 2}
>>> for k, v in d.items():
... print "%s: %s" % (k, v)
...
second: 2
fourth: 4
third: 3
first: 1
>>> d
{'second': 2, 'fourth': 4, 'third': 3, 'first': 1}
To make a new ordered dictionary from the original, sorting by the values:
>>> from collections import OrderedDict
>>> d_sorted_by_value = OrderedDict(sorted(d.items(), key=lambda x: x[1]))
The OrderedDict behaves like a normal dict:
>>> for k, v in d_sorted_by_value.items():
... print "%s: %s" % (k, v)
...
first: 1
second: 2
third: 3
fourth: 4
>>> d_sorted_by_value
OrderedDict([('first': 1), ('second': 2), ('third': 3), ('fourth': 4)])
Using Python 3.5
Whilst I found the accepted answer useful, I was also surprised that it hasn't been updated to reference OrderedDict from the standard library collections module as a viable, modern alternative - designed to solve exactly this type of problem.
from operator import itemgetter
from collections import OrderedDict
x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sorted_x = OrderedDict(sorted(x.items(), key=itemgetter(1)))
# OrderedDict([(0, 0), (2, 1), (1, 2), (4, 3), (3, 4)])
The official OrderedDict documentation offers a very similar example too, but using a lambda for the sort function:
# regular unsorted dictionary
d = {'banana': 3, 'apple':4, 'pear': 1, 'orange': 2}
# dictionary sorted by value
OrderedDict(sorted(d.items(), key=lambda t: t[1]))
# OrderedDict([('pear', 1), ('orange', 2), ('banana', 3), ('apple', 4)])
Pretty much the same as Hank Gay's answer:
sorted([(value,key) for (key,value) in mydict.items()])
Or optimized slightly as suggested by John Fouhy:
sorted((value,key) for (key,value) in mydict.items())
As of Python 3.6 the built-in dict will be ordered
Good news, so the OP's original use case of mapping pairs retrieved from a database with unique string ids as keys and numeric values as values into a built-in Python v3.6+ dict, should now respect the insert order.
If say the resulting two column table expressions from a database query like:
SELECT a_key, a_value FROM a_table ORDER BY a_value;
would be stored in two Python tuples, k_seq and v_seq (aligned by numerical index and with the same length of course), then:
k_seq = ('foo', 'bar', 'baz')
v_seq = (0, 1, 42)
ordered_map = dict(zip(k_seq, v_seq))
Allow to output later as:
for k, v in ordered_map.items():
print(k, v)
yielding in this case (for the new Python 3.6+ built-in dict!):
foo 0
bar 1
baz 42
in the same ordering per value of v.
Where in the Python 3.5 install on my machine it currently yields:
bar 1
foo 0
baz 42
Details:
As proposed in 2012 by Raymond Hettinger (cf. mail on python-dev with subject "More compact dictionaries with faster iteration") and now (in 2016) announced in a mail by Victor Stinner to python-dev with subject "Python 3.6 dict becomes compact and gets a private version; and keywords become ordered" due to the fix/implementation of issue 27350 "Compact and ordered dict" in Python 3.6 we will now be able, to use a built-in dict to maintain insert order!!
Hopefully this will lead to a thin layer OrderedDict implementation as a first step. As #JimFasarakis-Hilliard indicated, some see use cases for the OrderedDict type also in the future. I think the Python community at large will carefully inspect, if this will stand the test of time, and what the next steps will be.
Time to rethink our coding habits to not miss the possibilities opened by stable ordering of:
Keyword arguments and
(intermediate) dict storage
The first because it eases dispatch in the implementation of functions and methods in some cases.
The second as it encourages to more easily use dicts as intermediate storage in processing pipelines.
Raymond Hettinger kindly provided documentation explaining "The Tech Behind Python 3.6 Dictionaries" - from his San Francisco Python Meetup Group presentation 2016-DEC-08.
And maybe quite some Stack Overflow high decorated question and answer pages will receive variants of this information and many high quality answers will require a per version update too.
Caveat Emptor (but also see below update 2017-12-15):
As #ajcr rightfully notes: "The order-preserving aspect of this new implementation is considered an implementation detail and should not be relied upon." (from the whatsnew36) not nit picking, but the citation was cut a bit pessimistic ;-). It continues as " (this may change in the future, but it is desired to have this new dict implementation in the language for a few releases before changing the language spec to mandate order-preserving semantics for all current and future Python implementations; this also helps preserve backwards-compatibility with older versions of the language where random iteration order is still in effect, e.g. Python 3.5)."
So as in some human languages (e.g. German), usage shapes the language, and the will now has been declared ... in whatsnew36.
Update 2017-12-15:
In a mail to the python-dev list, Guido van Rossum declared:
Make it so. "Dict keeps insertion order" is the ruling. Thanks!
So, the version 3.6 CPython side-effect of dict insertion ordering is now becoming part of the language spec (and not anymore only an implementation detail). That mail thread also surfaced some distinguishing design goals for collections.OrderedDict as reminded by Raymond Hettinger during discussion.
It can often be very handy to use namedtuple. For example, you have a dictionary of 'name' as keys and 'score' as values and you want to sort on 'score':
import collections
Player = collections.namedtuple('Player', 'score name')
d = {'John':5, 'Alex':10, 'Richard': 7}
sorting with lowest score first:
worst = sorted(Player(v,k) for (k,v) in d.items())
sorting with highest score first:
best = sorted([Player(v,k) for (k,v) in d.items()], reverse=True)
Now you can get the name and score of, let's say the second-best player (index=1) very Pythonically like this:
player = best[1]
player.name
'Richard'
player.score
7
I had the same problem, and I solved it like this:
WantedOutput = sorted(MyDict, key=lambda x : MyDict[x])
(People who answer "It is not possible to sort a dict" did not read the question! In fact, "I can sort on the keys, but how can I sort based on the values?" clearly means that he wants a list of the keys sorted according to the value of their values.)
Please notice that the order is not well defined (keys with the same value will be in an arbitrary order in the output list).
If values are numeric you may also use Counter from collections.
from collections import Counter
x = {'hello': 1, 'python': 5, 'world': 3}
c = Counter(x)
print(c.most_common())
>> [('python', 5), ('world', 3), ('hello', 1)]
Starting from Python 3.6, dict objects are now ordered by insertion order. It's officially in the specifications of Python 3.7.
>>> words = {"python": 2, "blah": 4, "alice": 3}
>>> dict(sorted(words.items(), key=lambda x: x[1]))
{'python': 2, 'alice': 3, 'blah': 4}
Before that, you had to use OrderedDict.
Python 3.7 documentation says:
Changed in version 3.7: Dictionary order is guaranteed to be insertion
order. This behavior was implementation detail of CPython from 3.6.
In Python 2.7, simply do:
from collections import OrderedDict
# regular unsorted dictionary
d = {'banana': 3, 'apple':4, 'pear': 1, 'orange': 2}
# dictionary sorted by key
OrderedDict(sorted(d.items(), key=lambda t: t[0]))
OrderedDict([('apple', 4), ('banana', 3), ('orange', 2), ('pear', 1)])
# dictionary sorted by value
OrderedDict(sorted(d.items(), key=lambda t: t[1]))
OrderedDict([('pear', 1), ('orange', 2), ('banana', 3), ('apple', 4)])
copy-paste from : http://docs.python.org/dev/library/collections.html#ordereddict-examples-and-recipes
Enjoy ;-)
This is the code:
import operator
origin_list = [
{"name": "foo", "rank": 0, "rofl": 20000},
{"name": "Silly", "rank": 15, "rofl": 1000},
{"name": "Baa", "rank": 300, "rofl": 20},
{"name": "Zoo", "rank": 10, "rofl": 200},
{"name": "Penguin", "rank": -1, "rofl": 10000}
]
print ">> Original >>"
for foo in origin_list:
print foo
print "\n>> Rofl sort >>"
for foo in sorted(origin_list, key=operator.itemgetter("rofl")):
print foo
print "\n>> Rank sort >>"
for foo in sorted(origin_list, key=operator.itemgetter("rank")):
print foo
Here are the results:
Original
{'name': 'foo', 'rank': 0, 'rofl': 20000}
{'name': 'Silly', 'rank': 15, 'rofl': 1000}
{'name': 'Baa', 'rank': 300, 'rofl': 20}
{'name': 'Zoo', 'rank': 10, 'rofl': 200}
{'name': 'Penguin', 'rank': -1, 'rofl': 10000}
Rofl
{'name': 'Baa', 'rank': 300, 'rofl': 20}
{'name': 'Zoo', 'rank': 10, 'rofl': 200}
{'name': 'Silly', 'rank': 15, 'rofl': 1000}
{'name': 'Penguin', 'rank': -1, 'rofl': 10000}
{'name': 'foo', 'rank': 0, 'rofl': 20000}
Rank
{'name': 'Penguin', 'rank': -1, 'rofl': 10000}
{'name': 'foo', 'rank': 0, 'rofl': 20000}
{'name': 'Zoo', 'rank': 10, 'rofl': 200}
{'name': 'Silly', 'rank': 15, 'rofl': 1000}
{'name': 'Baa', 'rank': 300, 'rofl': 20}
Try the following approach. Let us define a dictionary called mydict with the following data:
mydict = {'carl':40,
'alan':2,
'bob':1,
'danny':3}
If one wanted to sort the dictionary by keys, one could do something like:
for key in sorted(mydict.iterkeys()):
print "%s: %s" % (key, mydict[key])
This should return the following output:
alan: 2
bob: 1
carl: 40
danny: 3
On the other hand, if one wanted to sort a dictionary by value (as is asked in the question), one could do the following:
for key, value in sorted(mydict.iteritems(), key=lambda (k,v): (v,k)):
print "%s: %s" % (key, value)
The result of this command (sorting the dictionary by value) should return the following:
bob: 1
alan: 2
danny: 3
carl: 40
You can create an "inverted index", also
from collections import defaultdict
inverse= defaultdict( list )
for k, v in originalDict.items():
inverse[v].append( k )
Now your inverse has the values; each value has a list of applicable keys.
for k in sorted(inverse):
print k, inverse[k]
You can use the collections.Counter. Note, this will work for both numeric and non-numeric values.
>>> x = {1: 2, 3: 4, 4:3, 2:1, 0:0}
>>> from collections import Counter
>>> #To sort in reverse order
>>> Counter(x).most_common()
[(3, 4), (4, 3), (1, 2), (2, 1), (0, 0)]
>>> #To sort in ascending order
>>> Counter(x).most_common()[::-1]
[(0, 0), (2, 1), (1, 2), (4, 3), (3, 4)]
>>> #To get a dictionary sorted by values
>>> from collections import OrderedDict
>>> OrderedDict(Counter(x).most_common()[::-1])
OrderedDict([(0, 0), (2, 1), (1, 2), (4, 3), (3, 4)])
The collections solution mentioned in another answer is absolutely superb, because you retain a connection between the key and value which in the case of dictionaries is extremely important.
I don't agree with the number one choice presented in another answer, because it throws away the keys.
I used the solution mentioned above (code shown below) and retained access to both keys and values and in my case the ordering was on the values, but the importance was the ordering of the keys after ordering the values.
from collections import Counter
x = {'hello':1, 'python':5, 'world':3}
c=Counter(x)
print( c.most_common() )
>> [('python', 5), ('world', 3), ('hello', 1)]
You can also use a custom function that can be passed to parameter key.
def dict_val(x):
return x[1]
x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sorted_x = sorted(x.items(), key=dict_val)
You can use a skip dict which is a dictionary that's permanently sorted by value.
>>> data = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
>>> SkipDict(data)
{0: 0.0, 2: 1.0, 1: 2.0, 4: 3.0, 3: 4.0}
If you use keys(), values() or items() then you'll iterate in sorted order by value.
It's implemented using the skip list datastructure.
Of course, remember, you need to use OrderedDict because regular Python dictionaries don't keep the original order.
from collections import OrderedDict
a = OrderedDict(sorted(originalDict.items(), key=lambda x: x[1]))
If you do not have Python 2.7 or higher, the best you can do is iterate over the values in a generator function. (There is an OrderedDict for 2.4 and 2.6 here, but
a) I don't know about how well it works
and
b) You have to download and install it of course. If you do not have administrative access, then I'm afraid the option's out.)
def gen(originalDict):
for x, y in sorted(zip(originalDict.keys(), originalDict.values()), key=lambda z: z[1]):
yield (x, y)
#Yields as a tuple with (key, value). You can iterate with conditional clauses to get what you want.
for bleh, meh in gen(myDict):
if bleh == "foo":
print(myDict[bleh])
You can also print out every value
for bleh, meh in gen(myDict):
print(bleh, meh)
Please remember to remove the parentheses after print if not using Python 3.0 or above
from django.utils.datastructures import SortedDict
def sortedDictByKey(self,data):
"""Sorted dictionary order by key"""
sortedDict = SortedDict()
if data:
if isinstance(data, dict):
sortedKey = sorted(data.keys())
for k in sortedKey:
sortedDict[k] = data[k]
return sortedDict
Here is a solution using zip on d.values() and d.keys(). A few lines down this link (on Dictionary view objects) is:
This allows the creation of (value, key) pairs using zip(): pairs = zip(d.values(), d.keys()).
So we can do the following:
d = {'key1': 874.7, 'key2': 5, 'key3': 8.1}
d_sorted = sorted(zip(d.values(), d.keys()))
print d_sorted
# prints: [(5, 'key2'), (8.1, 'key3'), (874.7, 'key1')]
As pointed out by Dilettant, Python 3.6 will now keep the order! I thought I'd share a function I wrote that eases the sorting of an iterable (tuple, list, dict). In the latter case, you can sort either on keys or values, and it can take numeric comparison into account. Only for >= 3.6!
When you try using sorted on an iterable that holds e.g. strings as well as ints, sorted() will fail. Of course you can force string comparison with str(). However, in some cases you want to do actual numeric comparison where 12 is smaller than 20 (which is not the case in string comparison). So I came up with the following. When you want explicit numeric comparison you can use the flag num_as_num which will try to do explicit numeric sorting by trying to convert all values to floats. If that succeeds, it will do numeric sorting, otherwise it'll resort to string comparison.
Comments for improvement welcome.
def sort_iterable(iterable, sort_on=None, reverse=False, num_as_num=False):
def _sort(i):
# sort by 0 = keys, 1 values, None for lists and tuples
try:
if num_as_num:
if i is None:
_sorted = sorted(iterable, key=lambda v: float(v), reverse=reverse)
else:
_sorted = dict(sorted(iterable.items(), key=lambda v: float(v[i]), reverse=reverse))
else:
raise TypeError
except (TypeError, ValueError):
if i is None:
_sorted = sorted(iterable, key=lambda v: str(v), reverse=reverse)
else:
_sorted = dict(sorted(iterable.items(), key=lambda v: str(v[i]), reverse=reverse))
return _sorted
if isinstance(iterable, list):
sorted_list = _sort(None)
return sorted_list
elif isinstance(iterable, tuple):
sorted_list = tuple(_sort(None))
return sorted_list
elif isinstance(iterable, dict):
if sort_on == 'keys':
sorted_dict = _sort(0)
return sorted_dict
elif sort_on == 'values':
sorted_dict = _sort(1)
return sorted_dict
elif sort_on is not None:
raise ValueError(f"Unexpected value {sort_on} for sort_on. When sorting a dict, use key or values")
else:
raise TypeError(f"Unexpected type {type(iterable)} for iterable. Expected a list, tuple, or dict")
I just learned a relevant skill from Python for Everybody.
You may use a temporary list to help you to sort the dictionary:
# Assume dictionary to be:
d = {'apple': 500.1, 'banana': 1500.2, 'orange': 1.0, 'pineapple': 789.0}
# Create a temporary list
tmp = []
# Iterate through the dictionary and append each tuple into the temporary list
for key, value in d.items():
tmptuple = (value, key)
tmp.append(tmptuple)
# Sort the list in ascending order
tmp = sorted(tmp)
print (tmp)
If you want to sort the list in descending order, simply change the original sorting line to:
tmp = sorted(tmp, reverse=True)
Using list comprehension, the one-liner would be:
# Assuming the dictionary looks like
d = {'apple': 500.1, 'banana': 1500.2, 'orange': 1.0, 'pineapple': 789.0}
# One-liner for sorting in ascending order
print (sorted([(v, k) for k, v in d.items()]))
# One-liner for sorting in descending order
print (sorted([(v, k) for k, v in d.items()], reverse=True))
Sample Output:
# Ascending order
[(1.0, 'orange'), (500.1, 'apple'), (789.0, 'pineapple'), (1500.2, 'banana')]
# Descending order
[(1500.2, 'banana'), (789.0, 'pineapple'), (500.1, 'apple'), (1.0, 'orange')]
Use ValueSortedDict from dicts:
from dicts.sorteddict import ValueSortedDict
d = {1: 2, 3: 4, 4:3, 2:1, 0:0}
sorted_dict = ValueSortedDict(d)
print sorted_dict.items()
[(0, 0), (2, 1), (1, 2), (4, 3), (3, 4)]
Iterate through a dict and sort it by its values in descending order:
$ python --version
Python 3.2.2
$ cat sort_dict_by_val_desc.py
dictionary = dict(siis = 1, sana = 2, joka = 3, tuli = 4, aina = 5)
for word in sorted(dictionary, key=dictionary.get, reverse=True):
print(word, dictionary[word])
$ python sort_dict_by_val_desc.py
aina 5
tuli 4
joka 3
sana 2
siis 1
If your values are integers, and you use Python 2.7 or newer, you can use collections.Counter instead of dict. The most_common method will give you all items, sorted by the value.
This works in 3.1.x:
import operator
slovar_sorted=sorted(slovar.items(), key=operator.itemgetter(1), reverse=True)
print(slovar_sorted)
For the sake of completeness, I am posting a solution using heapq. Note, this method will work for both numeric and non-numeric values
>>> x = {1: 2, 3: 4, 4:3, 2:1, 0:0}
>>> x_items = x.items()
>>> heapq.heapify(x_items)
>>> #To sort in reverse order
>>> heapq.nlargest(len(x_items),x_items, operator.itemgetter(1))
[(3, 4), (4, 3), (1, 2), (2, 1), (0, 0)]
>>> #To sort in ascending order
>>> heapq.nsmallest(len(x_items),x_items, operator.itemgetter(1))
[(0, 0), (2, 1), (1, 2), (4, 3), (3, 4)]

Find maximum value in a nested dictionary

I am having trouble understanding how to pull the maximum value of a nested dictionary when it is structured as so:
dict = {'City': {1: {'avg_dur': 10.58568297387339,
'n_trips': 1901,
'tot_dur': 20123.383333333313},
2: {'avg_dur': 12.25947507658035,
'n_trips': 2394,
'tot_dur': 29349.183333333356},
3: {'avg_dur': 12.95495652953303,
'n_trips': 3719,
'tot_dur': 48179.48333333334}}}
I am trying to extract the key for the maximum 'avg_trips' function. In the snippet above, I would expect the answer to return 3. I think I need to use lambda here, but I'm not sure how that works with nested dictionaries to this level.
Use max with key
Ex:
dict = {'City': {1: {'avg_dur': 10.58568297387339,
'n_trips': 1901,
'tot_dur': 20123.383333333313},
2: {'avg_dur': 12.25947507658035,
'n_trips': 2394,
'tot_dur': 29349.183333333356},
3: {'avg_dur': 12.95495652953303,
'n_trips': 3719,
'tot_dur': 48179.48333333334}}}
print(max(dict["City"].items(), key=lambda x: x[1]['n_trips'])[0])
Output:
3
You could also sort the keys by n_trips and take the last one:
>>> sorted(mydict['City'].keys(), key=lambda x: mydict['City'][x]['n_trips'])[-1]
3

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