How to replicate Pandas syntax? (To filter data frames) - python-3.x

How do I implement the syntax for filtering dataframes in Pandas? (df[df.column1 > someValue])
I am trying to make a class that have the same syntax of Pandas when filtering dataframes.
How do I replicate the syntax for a Dataframe df = DataFrame(someData) like this one:
df[df.column1 > someValue]
I implemented the methods __getattr__ and __getitem__ for the syntaxes of
df.column1
df['column1']
But I don't know how to link both together. Also, I could not find the function to copy from Pandas code.
Either an implementation to this problem or the reference to the function in Pandas would be of great help.
Edit:(Solution)
Following the hint on the answers I implemented the __getitem__ function as follows:
from tier tools import compress
def __getitem__(self, name):
"""Get items with [ and ]
"""
#If there is no expression, return a column
if isinstance(name, str):
return self.data[name]
#if there was an expression return the dataframe filtered
elif isinstance(name, list):
ind = list(compress(range(len(name)), name))
temp = DataFrame([[self.data[c].values[i]
for i in ind]
for c in self.columns],
columns=self.columns)
return temp
Note that I also had to implement the comparison methods for my column class (Series).
The full code can be seen here.

You need to implement __getitem__ to take a list of booleans and only return items when True. You will also need to implement the conditional operators (>, ==, etc.) to return that list of booleans, e.g. (proof of concept code):
class A(object):
def __init__(self, data):
self.data = data
def __getitem__(self, key):
return A([d for k, d in zip(key, self.data) if k])
def __gt__(self, value):
return [d > value for d in self.data]
def __repr__(self):
return str(self.__class__) + ' [' + ', '.join(str(d) for d in self.data) + ']'
>>> a = A(list(range(20)))
>>> a
<class '__main__.A'> [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
>>> a[a > 5]
<class '__main__.A'> [6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]

I think you basically want something that just wraps a recarray or structured array.
import numpy as np
myarray = np.array([("Hello",2.5,3),
("World",3.6,2),
('Foobar',2,7)]).T
df = np.core.records.fromarrays(myarray,
names='column1, column2, column3',
formats = 'S8, f8, i8')
print(df)
print(df[df.column3<=3])
While I don't use Pandas myself, the DataFrame seems like it is very similar to a recarray. If you wanted to roll your own, be sure to read about subclassing ndarray. numpy arrays can also be indexed with boolean mask variables such as
myarray = np.array([(1,2.5,3.),
(2,3.6,2.),
(3,2,7.)])
print(myarray[myarray[:,2]<=3.])

Related

Python Way - Processing multiple files & add the sum of all the results in Parallel

I have a list of 500 json files. Contents of the files are as follows
{'minute': '2022-11-16T02:29:00.000+00:00', 'mycount': [[0, 0], [1, 32], [2, 3456], [3, 446], [4, 534534], [5, 474], [6, 448], [7, 529], [8, 507], [9, 515], [10, 477], [11, 486], [12, 491], [13, 474], [14, 528], [15, 23]]}
I want to achieve the following using parallel processing ( may be processing 100 files in parallel)
For each file find the sum of second element of each element of mycount ( 0 + 32 +3456 +446+534534..]. Lets call it sum1
calculate sum1 for all the files and return total sum = sum1 + sum2+ sum3...
How can I achieve this using mutithreading in python?
If you don't mind of using multiprocessing instead of multithreading, you can adopt multiprocessing library and json decoder to parse the content of your files:
import multiprocessing as mp
import json
# Other libraries
import os
import warnings
def compute_file_sum(f):
"""Compute the sum for a file"""
try:
# Read the whole content
with open(f, 'r') as ff:
file_content = ff.readlines()
# Load as a JSON (mind the change of ' into ")
file_content = json.loads('\n'.join(file_content).replace("'", '"'))
# Compute the sum of second items of each element in 'mycount'
return sum(
c[1] for c in file_content['mycount']
)
except Exception as e:
# Handle exceptions
warnings.warn(f"Issues with file {f}, {str(e)}")
return 0
def get_filepaths(root_dir):
"""Get an iterator with the paths of the files of interest"""
return map(
lambda y: os.path.join(root_dir, y),
filter(
# Filter only files whose names match some conditions
lambda x: os.path.splitext(x)[0].startswith('bbb') and os.path.splitext(x)[1] == '.txt',
next(os.walk(root_dir))[2]
)
)
if __name__ == '__main__':
# Get the path of the folder with the files of interest
# Here is the folder with the python script
root_dir = os.path.dirname(__file__)
# Compute in parallel the sum for each file
with mp.Pool(processes=mp.cpu_count() - 1) as a:
file_sums = a.imap_unordered(compute_file_sum, get_filepaths(root_dir))
# Get the total sum
total_sum = sum(file_sums)

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

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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)]

Python defining iterators for dictionary comprehension for a class

I want to check with someone if I defined the iteration function properly. To explain, please consider the following example:
x=[{'n':'foo', 'a': [1,2,3], 'b':[2,3,5]}, {'n':'baz','a':[4,5,6], 'b':[7,8,9]},
{'n':'foo', 'a': [4,3,4], 'b':[1,5,6]}, {'n':'bar','a':[1,2,2], 'b':[2,5,6]}]
quick_dict = {key['n']: [sample['a'] for sample in x if sample['n']==key['n']] for key in x}
This works as expected and outputs:
{'foo': [[1, 2, 3], [4, 3, 4]], 'baz': [[4, 5, 6]], 'bar': [[1, 2, 2]]}
I am trying to do something similar for a class I defined using the __next__ and __iter__ methods. The class instance has many functions and attributes but for the purpose of this question, only the attribute samples is important because it is a list of dictionaries exactly as in the above example. I defined the methods as follows:
def __next__(self):
if self.itercounter < len(self.samples)-1:
self.itercounter +=1
return self.samples[self.itercounter]
else:
raise StopIteration
def __iter__(self):
self.itercounter = -1
return self
This seems to work for list comprehensions, but it fails for dictionary comprehensions.
If I do:
quick_dict = {key['Name']: [sample['CM'] for sample in data if sample['Name'] == key['Name']]
for key in data.samples}
then it works because it is directly accessing the list of dictionaries and it knows what to do. On the other hand if I do
quick_dict = {key['Name']: [sample['CM'] for sample in data if sample['Name'] == key['Name']]
for key in data}
then it is going through my functions, and it doesn't work. It just returns a dictionary with a single key. Here 'CM' is just a key like 'a' in the example.
What am I doing wrong in my definition of __iter__ and __next__?
Your second definition of quick_dict iterates over data with for sample in data while already iterating over it with for key in data. However, your __iter__ and __next__ implementation uses a single instance attribute to control iteration, meaning that nested iteration over data won't work because the second (nested) call to __iter__ resets the counter. To support nested iteration, eliminate __next__ and have __iter__ return a generator instead:
def __iter__(self):
i = -1
while i < len(self.samples)-1:
i += 1
yield self.samples[i]

Retrieve one value from default dict on python 3

I have a dict which I populate with data using setdefault method as follows:
if date not in call_dict:
call_dict.setdefault(date, [0, 0]).append(0)
else:
call_dict[date][0] += 1
if x[12] != 'ANSWERED':
call_dict[date][1] += 1
call_dict[date][2] = 100*(call_dict[date][1]/call_dict[date][0])
At the end of this process I have a dict which is structured like this:
{'key' : value0, value1, value2}
Then I have to plot only key and value2 (as value2 is a function of the key) but I can't find the way to access this value.
I tried myDict.values() but it did not work (as expected). I would really like to avoid creating another list/dict/anything because of script performance and also to achieve
that I would still have to reach for the value2 which would solve my problem.
Any ideas how to solve this problem?
Sample values of dict:
{'08:23': [45, 17, 37.77777777777778],
'08:24': [44, 15, 34.090909090909086],
'08:25': [46, 24, 52.17391304347826],
'08:48': [49, 19, 38.775510204081634],
You can get them from the dictionary with a list comprehension:
data = { "2018-07-01": [45, 17, 37.77777777777778],
"2018-07-02": [44, 15, 34.090909090909086],
"2018-07-03": [46, 24, 52.17391304347826],
"2018-07-04": [49, 19, 38.775510204081634]}
xy = [(x,data[x][2]) for x in data.keys()] # extract tuples of (date-key, 3rd value)
print(xy)
Output:
[('2018-07-01', 37.77777777777778), ('2018-07-02', 34.090909090909086),
('2018-07-03', 52.17391304347826), ('2018-07-04', 38.775510204081634)]
If you need them for plotting you might want to do:
x,y = zip(*xy)
print(x)
print(y)
Output:
('2018-07-01', '2018-07-02', '2018-07-03', '2018-07-04') # x
(37.77777777777778, 34.090909090909086, 52.17391304347826, 38.775510204081634) # y
and supply those to your plotting library as x and y data.
Doku: zip(*iterables)

Fast bit string generation given indices

Here a is a list, for example [34, 52, 57].
The function takes in this list and creates a bit string of length 64, where every index is a 0 except at the given indices.
So it would look like [0,0,....1,...1,..1,..0,0,0] where only at indices [34, 52, 57] we have ones.
def bit_string_gen(a):
bit_string = []
for key, value in enumerate(range(64)):
if key in a:
bit_string.append(1)
else:
bit_string.append(0)
return bit_string
Is there a better way to do this, maybe using lambda or map or itertools instead of enumerate.
The problem with your approach is that you:
use an if statement for every bit; and
use the in test which can be rather expensive.
A beter method can be:
def bit_string_gen(a):
bit_string = [0]*64
for value in a:
bit_string[value] = 1
return bit_string
So here you iterate only over the values of a, and you set these bits to 1.
Nevertheless it is a bit weird to encode this with a list of ints. A more compact way to do this is to encode this binary on an integer. For instance by using:
def bit_string_gen(a):
bit_string = 0
for value in a:
bit_string |= 1 << value
return bit_string
So in the last case if you set the bits like in your sample input, you obtain:
>>> bin(bit_string_gen([34, 52, 57]))
'0b1000010000000000000000010000000000000000000000000000000000'
>>> hex(bit_string_gen([34, 52, 57]))
'0x210000400000000'
If you're looking for a solution using map/lambda, here's a one-liner:
map(lambda x: 1 if x in [34, 52, 57] else 0, range(0, 64))

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