increase value of a dictionary using dictionary comprehension - python

I would like to write a simple code to store histogram data in a dictionary, I want to use dictionary comprehension build the final histogram dict. in python. I can achieve that in multiple ways as shown in code below but I have a problem with method two, here is how it works:
Enumerate the list of random letters to use that index to access a slice of the list starting at the current iteration to the end
check if letters exists in the slice
if yes, add the letter as index to the dictionary and increase its value by 1
if no, add the letter as index to the dictionary and set its value to 1
my problem is where is this temporary dictionary to add to before the dict. Comprehension is exhausted. I tried the final dict “histogram_dict” but it’s not the same.
Again, The main goal is to rewrite the first code in terms of dictionary comprehension as in the second code.
Thanks
import random
rnd_letters_list=[chr(random.randrange(97,122)) for i in range(21)]
histogram_dict={}
#=============first method ==============
# for letter in rnd_letters_list:
# if letter in histogram_dict:
# histogram_dict[letter]+=1
# else:
# histogram_dict[letter]=1
# print(histogram_dict)
#=============second method ==============
histogram_dict={letter: XXX_dict[letter]+=1
if letter in rnd_letters_list[indx::]
else XXX_dict[letter]=1
for indx,letter in enumerate(rnd_letters_list)}
print(histogram_dict)
#=============third method ==============
# histogram_dict={letter:rnd_letters_list.count(letter)
# for letter in rnd_letters_list}
# print(histogram_dict)
#=============fourth method ==============
# from collections import Counter
# print(dict(
# Counter(rnd_letters_list).most_common()))

I had the same problem a few weeks ago. I extendet the counter class.
Counter itself extends the dictionary.
Counter Docs
from typing import Counter, Dict
class Histogram(Counter):
"""
A that saves discrete values for a histogram in an efficient manor.
"""
def population(self) -> int:
return sum(self.values())
def mean(self) -> float:
population = 0
average_value = 0
for key, value in self.items():
average_value += key * value
population += value
return average_value / population
def y_mean(self) -> float:
entry_counts: int = len(self.items())
sum_of_values: int = sum(self.values())
return sum_of_values / (entry_counts * sum_of_values)
def median(self, alpha: float = 0.5) -> float:
position: int = round(alpha * self.population())
keys = sorted(self.keys())
for key in keys:
position -= self[key]
if position < 0:
return key
return float("NaN")
def variance(self) -> float:
pop: int = self.population()
if pop < 2:
return 0
var: float = 0
mean: float = self.mean()
for key, value in self.items():
var += float(value) * ((float(key) - mean) ** 2)
return var / (pop - 1)
def normed(self) -> Dict[int, float]:
dictionary: Dict[int, float] = {}
population = self.population()
for key, value in self.items():
dictionary[key] = value / population
return dictionary

Related

Finding and averaging elements of multiple dataframes that are within 10% of each other

I have objects that store values are dataframes. I have been able to compare if values from two dataframes are within 10% of each other. However, I am having difficulty extending this to multiple dataframes. Moreover, I am wondering how I should apporach this problem if dataframes are not the same size?
def add_well_peak(self, *other):
if len(self.Bell) == len(other.Bell): #if dataframes ARE the same size
for k in range(len(self.Bell)):
for j in range(len(other.Bell)):
if int(self.Size[k]) - int(self.Size[k])*(1/10) <= int(other.Size[j]) <= int(self.Size[k]) + int(self.Size[k])*(1/10):
#average all
For example, in the image below, there are objects that contain dataframes (i.e., self, other1, other2). The colors represent matches (i.e, values that are within 10% of each other). If a match exist, then average the values. If a match does not exist still include the unmatch number. I want to be able to generalize this for any number of objects greater or equal than 2 (other 1, other 2, other 3, other ....). Any help would be appreciated. Please let me know if anything is unclear. This is my first time posting. Thanks again.
matching data
Results:
Using my solution on the dataframes of your image, I get the following:
Threshold outlier = 0.2:
0
0 1.000000
1 1493.500000
2 5191.333333
3 35785.333333
4 43586.500000
5 78486.000000
6 100000.000000
Threshold outlier = 0.5:
0 1
0 1.000000 NaN
1 1493.500000 NaN
2 5191.333333 NaN
3 43586.500000 35785.333333
4 78486.000000 100000.000000
Explanations:
The lines are averaged peaks, the columns representing the different values obtained for these peaks. I assumed the average emanating from the biggest number of elements was the legitimate one, and the rest within the THRESHOLD_OUTLIER were the outliers (should be sorted, the more probable you are as a legitimate peak, the more you are on the left (the 0th column is the most probable)). For instance, on line 3 of the 0.5 outlier threshold results, 43586.500000 is an average coming from 3 dataframes, while 35785.333333 comes from only 2, thus the most probable is the first one.
Issues:
The solution is quite complicated. I assume a big part of it could be removed, but I can't see how for the moment, and as it works, I'll certainly leave the optimization to you.
Still, I tried commenting my best, and if you have any question, do not hesitate!
Files:
CombinationLib.py
from __future__ import annotations
from typing import Dict, List
from Errors import *
class Combination():
"""
Support class, to make things easier.
Contains a string `self.combination` which is a binary number stored as a string.
This allows to test every combination of value (i.e. "101" on the list `[1, 2, 3]`
would signify grouping `1` and `3` together).
There are some methods:
- `__add__` overrides the `+` operator
- `compute_degree` gives how many `1`s are in the combination
- `overlaps` allows to verify if combination overlaps (use the same value twice)
(i.e. `100` and `011` don't overlap, while `101` and `001` do)
"""
def __init__(self, combination:str) -> Combination:
self.combination:str = combination
self.degree:int = self.compute_degree()
def __add__(self, other: Combination) -> Combination:
if self.combination == None:
return other.copy()
if other.combination == None:
return self.copy()
if self.overlaps(other):
raise CombinationsOverlapError()
result = ""
for c1, c2 in zip(self.combination, other.combination):
result += "1" if (c1 == "1" or c2 == "1") else "0"
return Combination(result)
def __str__(self) -> str:
return self.combination
def compute_degree(self) -> int:
if self.combination == None:
return 0
degree = 0
for bit in self.combination:
if bit == "1":
degree += 1
return degree
def copy(self) -> Combination:
return Combination(self.combination)
def overlaps(self, other:Combination) -> bool:
for c1, c2 in zip(self.combination, other.combination):
if c1 == "1" and c1 == c2:
return True
return False
class CombinationNode():
"""
The main class.
The main idea was to build a tree of possible "combinations of combinations":
100-011 => 111
|---010-001 => 111
|---001-010 => 111
At each node, the combination applied to the current list of values was to be acceptable
(all within THREASHOLD_AVERAGING).
Also, the shorter a path, the better the solution as it means it found a way to average
a lot of the values, with the minimum amount of outliers possible, maybe by grouping
the outliers together in a way that makes sense, ...
- `populate` fills the tree automatically, with every solution possible
- `path` is used mainly on leaves, to obtain the path taken to arrive there.
"""
def __init__(self, combination:Combination) -> CombinationNode:
self.combination:Combination = combination
self.children:List[CombinationNode] = []
self.parent:CombinationNode = None
self.total_combination:Combination = combination
def __str__(self) -> str:
list_paths = self.recur_paths()
list_paths = [",".join([combi.combination.combination for combi in path]) for path in list_paths]
return "\n".join(list_paths)
def add_child(self, child:CombinationNode) -> None:
if child.combination.degree > self.combination.degree and not self.total_combination.overlaps(child.combination):
raise ChildDegreeExceedParentDegreeError(f"{child.combination} > {self.combination}")
self.children.append(child)
child.parent = self
child.total_combination += self.total_combination
def path(self) -> List[CombinationNode]:
path = []
current = self
while current.parent != None:
path.append(current)
current = current.parent
path.append(current)
return path[::-1]
def populate(self, combination_dict:Dict[int, List[Combination]]) -> None:
missing_degrees = len(self.combination.combination)-self.total_combination.degree
if missing_degrees == 0:
return
for i in range(min(self.combination.degree, missing_degrees), 0, -1):
for combination in combination_dict[i]:
if not self.total_combination.overlaps(combination):
self.add_child(CombinationNode(combination))
for child in self.children:
child.populate(combination_dict)
def recur_paths(self) -> List[List[CombinationNode]]:
if len(self.children) == 0:
return [self.path()]
paths = []
for child in self.children:
for path in child.recur_paths():
paths.append(path)
return paths
Errors.py
class ChildDegreeExceedParentDegreeError(Exception):
pass
class CombinationsOverlapError(Exception):
pass
class ToImplementError(Exception):
pass
class UncompletePathError(Exception):
pass
main.py
from typing import Dict, List, Set, Tuple, Union
import pandas as pd
from CombinationLib import *
best_depth:int = -1
best_path:List[CombinationNode] = []
THRESHOLD_OUTLIER = 0.2
THRESHOLD_AVERAGING = 0.1
def verif_averaging_pct(combination:Combination, values:List[float]) -> bool:
"""
For a given combination of values, we must have all the values within
THRESHOLD_AVERAGING of the average of the combination
"""
avg = 0
for c,v in zip(combination.combination, values):
if c == "1":
avg += v
avg /= combination.degree
for c,v in zip(combination.combination, values):
if c == "1"and (v > avg*(1+THRESHOLD_AVERAGING) or v < avg*(1-THRESHOLD_AVERAGING)):
return False
return True
def recursive_check(node:CombinationNode, depth:int, values:List[Union[float, int]]) -> None:
"""
Here is where we preferencially ask for a small number of bigger groups
"""
global best_depth
global best_path
# If there are more groups than the current best way to do, stop
if best_depth != -1 and depth > best_depth:
return
# If all the values of the combination are not within THRESHOLD_AVERAGING, stop
if not verif_averaging_pct(node.combination, values):
return
# If we finished the list of combinations, and this way is the best, keep it, stop
if len(node.children) == 0:
if best_depth == -1 or depth < best_depth:
best_depth = depth
best_path = node.path()
return
# If we are still not finished (not every value has been used), continue
for cnode in node.children:
recursive_check(cnode, depth+1, values)
def groups_from_list(values:List[Union[float, int]]) -> List[List[Union[float, int]]]:
"""
From a list of values, get the smallest list of groups of elements
within THRESHOLD_AVERAGING of each other.
It implies that we will try and recursively find the biggest group possible
within the unsused values (i.e. groups with combinations of size [3, 1] are prefered
over [2, 2])
"""
global best_depth
global best_path
groups:List[List[float]] = []
# Generate all the combinations (I used binary for this)
combination_dict:Dict[int, List[Combination]] = {}
for i in range(1, 2**len(values)):
combination = format(i, f"0{len(values)}b") # Here is the binary conversion
counter = 0
for c in combination:
if c == "1":
counter += 1
if counter not in combination_dict:
combination_dict[counter] = []
combination_dict[counter].append(Combination(combination))
# Generate of the combinations of combinations that use all values (without using one twice)
combination_trees:List[List[CombinationNode]] = []
for key in combination_dict:
for combination in combination_dict[key]:
cn = CombinationNode(combination)
cn.populate(combination_dict)
combination_trees.append(cn)
best_depth = -1
best_path = None
for root in combination_trees:
recursive_check(root, 0, values)
# print(",".join([combination.combination.combination for combination in best_path]))
for combination in best_path:
temp = []
for c,v in zip(combination.combination.combination, values):
if c == "1":
temp.append(v)
groups.append(temp)
return groups
def averages_from_groups(gs:List[List[Union[float, int]]]) -> List[float]:
"""Computing the averages of each group"""
avgs:List[float] = []
for group in gs:
avg = 0
for elt in group:
avg += elt
avg /= len(group)
avgs.append(avg)
return avgs
def end_check(ds:List[pd.DataFrame], ids:List[int]) -> bool:
"""Check if we finished consuming all the dataframes"""
for d,i in zip(ds, ids):
if i < len(d[0]):
return False
return True
def search(group:List[Union[float, int]], values_list:List[Union[float, int]]) -> List[int]:
"""Obtain all the indices corresponding to a set of values"""
# We will get all the indices in values_list of the values in group
# If a value is present in group, all the occurences of this value will be too,
# so we can use a set and search every occurence for each value.
indices:List[int] = []
group_set = set(group)
for value in group_set:
for i,v in enumerate(values_list):
if value == v:
indices.append(i)
return indices
def threshold_grouper(total_list:List[Union[float, int]]) -> pd.DataFrame:
"""Building a 2D pd.DataFrame with the averages (x) and the outliers (y)"""
result_list:List[List[Union[float, int]]] = [[total_list[0]]]
result_index = 0
total_index = 1
while total_index < len(total_list):
# Only checking if the bigger one is within THRESHOLD_OUTLIER of the little one.
# If it is the case, the opposite is true too.
# If yes, it is an outlier
if result_list[result_index][0]*(1+THRESHOLD_OUTLIER) >= total_list[total_index]:
result_list[result_index].append(total_list[total_index])
# Else it is a new peak
else:
result_list.append([total_list[total_index]])
result_index += 1
total_index += 1
result:pd.DataFrame = pd.DataFrame(result_list)
return result
def dataframes_merger(dataframes:List[pd.DataFrame]) -> pd.DataFrame:
"""Merging the dataframes, with THRESHOLDS"""
# Store the averages for the within 10% cells, in ascending order
result = []
# Keep tabs on where we are regarding each dataframe (needed for when we skip cells)
curr_indices:List[int] = [0 for _ in range(len(dataframes))]
# Repeat until all the cells in every dataframe has been seen once
while not end_check(dataframes, curr_indices):
# Get the values of the current indices in the dataframes
curr_values = [dataframe[0][i] for dataframe,i in zip(dataframes, curr_indices)]
# Get the largest 10% groups from the current list of values
groups = groups_from_list(curr_values)
# Compute the average of these groups
avgs = averages_from_groups(groups)
# Obtain the minimum average...
avg_min = min(avgs)
# ... and its index
avg_min_index = avgs.index(avg_min)
# Then get the group corresponding to the minimum average
avg_min_group = groups[avg_min_index]
# Get the indices of the values included in this group
indices_to_increment = search(avg_min_group, curr_values)
# Add the average to the result merged list
result.append(avg_min)
# For every element in the average we added, increment the corresponding index
for index in indices_to_increment:
curr_indices[index] += 1
# Re-assemble the dataframe, taking the threshold% around average into account
result = threshold_grouper(result)
print(result)
df1 = pd.DataFrame([1, 1487, 5144, 35293, 78486, 100000])
df2 = pd.DataFrame([1, 1500, 5144, 36278, 45968, 100000])
df3 = pd.DataFrame([1, 5286, 35785, 41205, 100000])
dataframes_merger([df3, df2, df1])

How to find the highest value element in a list with reference to a dictionary on python

How do I code a function in python which can:
iterate through a list of word strings which may contain duplicate words and referencing to a dictionary,
find the word with the highest absolute sum, and
output it along with the corresponding absolute value.
The function also has to ignore words which are not in the dictionary.
For example,
Assume the function is called H_abs_W().
Given the following list and dict:
list_1 = ['apples','oranges','pears','apples']
Dict_1 = {'apples':5.23,'pears':-7.62}
Then calling the function as:
H_abs_W(list_1,Dict_1)
Should give the output:
'apples',10.46
EDIT:
I managed to do it in the end with the code below. Looking over the answers, turns out I could have done it in a shorter fashion, lol.
def H_abs_W(list_1,Dict_1):
freqW = {}
for char in list_1:
if char in freqW:
freqW[char] += 1
else:
freqW[char] = 1
ASum_W = 0
i_word = ''
for a,b in freqW.items():
x = 0
d = Dict_1.get(a,0)
x = abs(float(b)*float(d))
if x > ASum_W:
ASum_W = x
i_word = a
return(i_word,ASum_W)
list_1 = ['apples','oranges','pears','apples']
Dict_1 = {'apples':5.23,'pears':-7.62}
d = {k:0 for k in list_1}
for x in list_1:
if x in Dict_1.keys():
d[x]+=Dict_1[x]
m = max(Dict_1, key=Dict_1.get)
print(m,Dict_1[m])
try this,
key, value = sorted(Dict_1.items(), key = lambda x : x[1], reverse=True)[0]
print(f"{key}, {list_1.count(key) * value}")
# apples, 10.46
you can use Counter to calculate the frequency(number of occurrences) of each item in the list.
max(counter.values()) will give us the count of maximum occurring element
max(counter, key=counter.get) will give the which item in the list is
associated with that highest count.
========================================================================
from collections import Counter
def H_abs_W(list_1, Dict_1):
counter = Counter(list_1)
count = max(counter.values())
item = max(counter, key=counter.get)
return item, abs(count * Dict_1.get(item))

Pythonic way to assign range of number to bucket

I'm developing an ABtest framework using django. I want to assign variant number based on bucket_id from cookies' request.
bucket_id is set by the front end with a range integer from 0-99.
So far, I have created the function name get_bucket_name:
def get_bucket_range(data):
range_bucket = []
first_val = 0
next_val = 0
for i, v in enumerate(data.split(",")):
v = int(v)
if i == 0:
first_val = v
range_bucket.append([0, first_val])
elif i == 1:
range_bucket.append([first_val, first_val + v])
next_val = first_val + v
else:
range_bucket.append([next_val, next_val + v])
next_val = next_val + v
return range_bucket
Data input for get_bucket_range is a comma delineated string which means we have 3 variants where each variant has its own weight e.g. data = "25,25,50" with first variant's weight being 25 etc.
I then created a function to assign the variant named,
def assign_variant(range_bucket, num):
for i in range(len(range_bucket)):
if num in range(range_bucket[i][0], range_bucket[i][1]):
return i
This function should have 2 parameters, range_bucket -> from get_bucket_range function, and num -> bucket_id from cookies.
With this function I can return which bucket_id belongs to the variant id.
For example, we have 25 as bucket_id, with data = "25,25,50". This means our bucket_id should belong to variant id 1. Or in the case that we have 25 as bucket_id, with data = "10,10,10,70". This should mean that our bucket_id will belong to variant id 2.
However, it feels like neither of my functions are pythonic or optimised. Does anyone here have any suggestions as to how I could improve my code?
Your functions could look like this for example:
def get_bucket_range(data):
last = 0
range_bucket = []
for v in map(int, data.split(',')):
range_bucket.append([last, last+v])
last += v
return range_bucket
def assign_variant(range_bucket, num):
for i, (low, high) in enumerate(range_bucket):
if low <= num < high:
return i
You can greatly reduce the lengths of your functions with the itertools.accumulate and bisect.bisect functions. The first function accumulates all the weights into sums (10,10,10,70 becomes 10,20,30,100), and the second function gives you the index of where that element would belong, which in your case is equivalent to the index of the group it belongs to.
from itertools import accumulate
from bisect import bisect
def get_bucket_range(data):
return list(accumulate(map(int, data.split(',')))
def assign_variant(range_bucket, num):
return bisect(range_bucket, num)

Python - List Binary Search without bisect

I have a list and I want to binary_search a key(number).
My code is below but I don't have a clue what to do where the bold text on code is:
(What to do with this? Is an other function? int imid = midpoint(imin, imax))
List = []
x = 1
#Import 20 numbers to list
for i in range (0,20):
List.append (i)
print (List)
key = input("\nGive me a number for key: ")
def midpoint(imin, imax):
return point((imin+imax)/2)
def binary_search(List,key,imin,imax,point):
while (imax >= imin):
int imid = midpoint(imin, imax)
if(List[imid] == key):
return imid;
elif (List[imid] < key):
imin = imid + 1;
else:
imax = imid - 1;
return KEY_NOT_FOUND;
print (binary_search(key))
midpoint(imin, imax)
binary_search(List,key,imin,imax,point)
It doesn't seem to be doing anything for you; remove the call to midpoint, and point, and just have
def binary_search(List,key,imin,imax,point):
while (imax >= imin):
imid = (imin + imax) / 2
(However, there are some things wrong with your code, and it won't work with just that change;
You create a list called List then try to append to an uninitialized variable called myList
You 'import 20 random' numbers, but range() is not random, it's a simple sequence 1, 2, 3, 4...
range already returns a list, no need to count through it and copy it, just use it
You call binary_search with an empty List, a key, and three uninitialized variables
binary_search assumes the list is sorted, which it is, but if the comment about 'random numbers' was correct, it wouldn't be.
)

Python: Concatenate similiar objects in List

I have a list containing strings as ['Country-Points'].
For example:
lst = ['Albania-10', 'Albania-5', 'Andorra-0', 'Andorra-4', 'Andorra-8', ...other countries...]
I want to calculate the average for each country without creating a new list. So the output would be (in the case above):
lst = ['Albania-7.5', 'Andorra-4.25', ...other countries...]
Would realy appreciate if anyone can help me with this.
EDIT:
this is what I've got so far. So, "data" is actually a dictionary, where the keys are countries and the values are list of other countries points' to this country (the one as Key). Again, I'm new at Python so I don't realy know all the built-in functions.
for key in self.data:
lst = []
index = 0
score = 0
cnt = 0
s = str(self.data[key][0]).split("-")[0]
for i in range(len(self.data[key])):
if s in self.data[key][i]:
a = str(self.data[key][i]).split("-")
score += int(float(a[1]))
cnt+=1
index+=1
if i+1 != len(self.data[key]) and not s in self.data[key][i+1]:
lst.append(s + "-" + str(float(score/cnt)))
s = str(self.data[key][index]).split("-")[0]
score = 0
self.data[key] = lst
itertools.groupby with a suitable key function can help:
import itertools
def get_country_name(item):
return item.split('-', 1)[0]
def get_country_value(item):
return float(item.split('-', 1)[1])
def country_avg_grouper(lst) :
for ctry, group in itertools.groupby(lst, key=get_country_name):
values = list(get_country_value(c) for c in group)
avg = sum(values)/len(values)
yield '{country}-{avg}'.format(country=ctry, avg=avg)
lst[:] = country_avg_grouper(lst)
The key here is that I wrote a function to do the change out of place and then I can easily make the substitution happen in place by using slice assignment.
I would probabkly do this with an intermediate dictionary.
def country(s):
return s.split('-')[0]
def value(s):
return float(s.split('-')[1])
def country_average(lst):
country_map = {}|
for point in lst:
c = country(pair)
v = value(pair)
old = country_map.get(c, (0, 0))
country_map[c] = (old[0]+v, old[1]+1)
return ['%s-%f' % (country, sum/count)
for (country, (sum, count)) in country_map.items()]
It tries hard to only traverse the original list only once, at the expense of quite a few tuple allocations.

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