if I have a dictionary
x ={0: 0, 1: 4, 2: 0, 3: 2, 4: 2, 5: 4}
how do i get the keys of all the largest values
In this case they would be 1 and 5 .
Not a duplicate question. looking to find all the keys and not just the one.
x ={0: 0, 1: 4, 2: 0, 3: 2, 4: 2, 5: 4}
maximum = max(x.values())
keys = [key for key, value in x.items() if value == maximum]
print(keys) # => [1, 5]
There is a class in collections called Counter that does exactly what you want. It provides the exact functionality you require via it's most_common method:
from collections import counter
maxes = Counter(x).most_common(2)
print([x[0] for x in maxes])
[1, 5]
Now this is probably not exactly what you want because I hard coded in the number 2. You can get this by using another Counter on the values of your dictionary!
x = Counter(x) # This preserves x: just makes it a subclass of dict
max_count = Counter(x.values())[x.most_common(1)[0][1]]
maxes = x.most_common(max_count)
maxes = [x[0] for x in maxes]
Here, I compute the number of times that the most common value occurs by counting all the different values, and then checking the maximum one using x.most_common(1)[0][1].
Please do not select this answer. #BrightOne has the right answer. This is just a thing I did to see if I could avoid using anything but Counters. It is not actually a good idea.
Related
a = [0,0,0,0,0,0,0,0,0,1,2,3,4,0,0]
for i in a:
if i < 3:
del a[0]
print(a)
[0, 0, 0, 1, 2, 3, 4]
should be:
[3,4,0,0]
You are using i in a confusing way, since you're not using it as an index (which is kinda the norm) but as an iterator. Just for readability I would suggest changing the iterator name to something more descriptive.
If I understand you're question correctly, you want to delete the first element, until it is bigger or equal 3. If that is your question, you could do it like this:
a = [0,0,0,0,0,0,0,0,0,1,2,3,4,0,0]
while a[0]<3:
del a[0]
print(a) # [3,4,0,0]
a = [0,0,0,0,0,0,0,0,0,1,2,3,4]
n= []
for i in range(len(a)):
if(a[i]<3):
continue
else:
n.append(a[i])
print(n) # [3,4]
I have a dictionary with lots of data from a CSV file where the key is the row number and the value is a list that contains the column data.
What I want to do is to check from a data of each line (key) and from column 2, I take its data from column 4 and then look for this data in column 2 of another line (key) and take its data in column 4 and continue until it finds the last value of column 4 in column 2.
My code is this:
dict_FH_coord = self.dict_FH()
site_SI='some_site'
list_test = [site_SI]
while len(list_test) > 0:
for ele in list_test:
for cle, val in dict_FH_coord.items():
list_test = []
if val[2] == ele:
list_test.append(val[4])
list_def.append(val)
print(val)
But this code does not work because it stops at the first iteration and it finds that the elements linked to the starting site_SI only
Is there a way to do successive iterations with the list list_test which becomes dynamic to solve my problem?
If you want to modify list 'on air' you sohuld do something like
a = [1, 2, 3]
a[:] = [1, 2]
In your case the only way you may use this inside the loop (avoiding infinite list size increasement):
if val[2] == ele:
list_test[:] = list_test[1:]+[val[4]]
list_def.append(val)
else:
list_test[:] = list_test[1:]
But it wont work as intended because previous iteration ends at index 1 (for ele in list_test:), and list_test would never change in size.
Both this cases can not be merged with each other.
I suggest you to use Queue, but be careful to avoid infinite links looping inside your data:
from queue import Queue
dict_FH_coord = {
1: [0, 1, 'some_site', 3, 'some_site_2'],
2: [0, 1, 'some_site_2', 3, 'some_site_3'],
3: [0, 1, 'some_site_2', 3, 'some_site_4'],
4: [0, 1, 'some_site_3', 3, 'some_site_5'],
}
site_SI = 'some_site'
unvisited = Queue()
unvisited.put(site_SI)
list_def = list()
while not unvisited.empty():
ele = unvisited.get()
for cle, val in dict_FH_coord.items():
if val[2] == ele:
unvisited.put(val[4])
list_def.append(val)
print(val)
I am having list which as follows
input_list= [2, 3, 5, 2, 5, 1, 5]
I want to get all the indexes of maximum value. Need efficient solution. The output will be as follows.
output = [2,4,6] (The above list 5 is maximum value in a list)
I have tried by using below code
m = max(input_list)
output = [i for i, j in enumerate(a) if j == m]
I need to find any other optimum solution.
from collections import defaultdict
dic=defaultdict(list)
input_list=[]
for i in range(len(input_list)):
dic[input_list[i]]+=[i]
max_value = max(input_list)
Sol = dic[max_value]
You can use numpy (numpy arrays are very fast):
import numpy as np
input_list= np.array([2, 3, 5, 2, 5, 1, 5])
i, = np.where(input_list == np.max(input_list))
print(i)
Output:
[2 4 6]
Here's the approach which is described in comments. Even if you use some library, fundamentally you need to traverse at least once to solve this problem (considering input list is unsorted). So even lower bound for the algorithm would be Omega(size_of_list). If list is sorted we can leverage binary_search to solve the problem.
def max_indexes(l):
try:
assert l != []
max_element = l[0]
indexes = [0]
for index, element in enumerate(l[1:]):
if element > max_element:
max_element = element
indexes = [index + 1]
elif element == max_element:
indexes.append(index + 1)
return indexes
except AssertionError:
print ('input_list in empty')
Use a for loop for O(n) and iterating just once over the list resolution:
from itertools import islice
input_list= [2, 3, 5, 2, 5, 1, 5]
def max_indexes(l):
max_item = input_list[0]
indexes = [0]
for i, item in enumerate(islice(l, 1, None), 1):
if item < max_item:
continue
elif item > max_item:
max_item = item
indexes = [i]
elif item == max_item:
indexes.append(i)
return indexes
Here you have the live example
Think of it in this way, unless you iterate through the whole list once, which is O(n), n being the length of the list, you won't be able to compare the maximum with all values in the list, so the best you can do is O(n), which you already seems to be doing in your example.
So I am not sure you can do it faster than O(n) with the list approach.
Hi so im currently taking discrete structures and algorithm course and have to work with python for the first time so im having a little trouble getting my function find the max value in the list can you take a look at my code because im trying to also convert to pseudocode:
def max_search(numbers):
numbers = [1, 5, 9, 3, 4, 6]
max = numbers = [0]
for i in range(1, len(numbers)):
if numbers[i] > max:
max = numbers[i]
max_search(numbers)
print(max)
Use the max method provided for list
max(numbers)
When you write the code for maximum number in a list, start by thinking of base cases, which will be.
Maximum can be pre-defined constant, say -1 if the list is empty
Maximum is the first element in the list, if the list only has one element.
After that, if the list is longer, you assign the first element of the list as maximum, and then you iterate through the list, updating the maximum if you find a number which is greater than the maximum.
def max_search(numbers):
#Maximum of an empty list is undefined, I defined it as -1
if len(numbers) == 0:
return -1
#Maximum of a list with one element is the element itself
if len(numbers) == 1:
return numbers[0]
max = numbers[0]
#Iterate through the list and update maximum on the fly
for num in numbers:
if num >= max:
max = num
return max
In your case, you are overwriting the numbers argument with another list inside the function [1, 5, 9, 3, 4, 6], and you are recursively calling the same functions with same arguments, which will lead to Stack Overflow
I have made some changes
def max_search(numbers):
max = -1 # if numbers contains all positive number
for i in range(len(numbers)):
if numbers[i] > max:
max = numbers[i]
max = max_search([1, 5, 9, 3, 4, 6])
print(max)
I am creating a function which is supposed to return a dictionary with keys and values from different lists. But I amhavin problems in getting the mean of a list o numbers as values of the dictionary. However, I think I am getting the keys properly.
This is what I get so far:
def exp (magnitudes,measures):
"""return for each magnitude the associated mean of numbers from a list"""
dict_expe = {}
for mag in magnitudes:
dict_expe[mag] = 0
for mea in measures:
summ = 0
for n in mea:
summ += n
dict_expe[mag] = summ/len(mea)
return dict_expe
print(exp(['mag1', 'mag2', 'mag3'], [[1,2,3],[3,4],[5]]))
The output should be:
{mag1 : 2, mag2: 3.5, mag3: 5}
But what I am getting is always 5 as values of all keys. I thought about the zip() method but im trying to avoid it as because the it requieres the same length in both lists.
An average of a sequence is sum(sequence) / len(sequence), so you need to iterate through both magnitudes and measures, calculate these means (arithmetical averages) and store it in a dictionary.
There are much more pythonic ways you can achieve this. All of these examples produce {'mag1': 2.0, 'mag2': 3.5, 'mag3': 5.0} as result.
Using for i in range() loop:
def exp(magnitudes, measures):
means = {}
for i in range(len(magnitudes)):
means[magnitudes[i]] = sum(measures[i]) / len(measures[i])
return means
print(exp(['mag1', 'mag2', 'mag3'], [[1, 2, 3], [3, 4], [5]]))
But if you need both indices and values of a list you can use for i, val in enumerate(sequence) approach which is much more suitable in this case:
def exp(magnitudes, measures):
means = {}
for i, mag in enumerate(magnitudes):
means[mag] = sum(measures[i]) / len(measures[i])
return means
print(exp(['mag1', 'mag2', 'mag3'], [[1, 2, 3], [3, 4], [5]]))
Another problem hides here: i index belongs to magnitudes but we are also getting values from measures using it, this is not a big deal in your case if you have magnitudes and measures the same length but if magnitudes will be larger you will get an IndexError. So it seems to me like using zip function is what would be the best choice here (actually as of python3.6 it doesn't require two lists to be the same length, it will just use the length of shortest one as the length of result):
def exp(magnitudes, measures):
means = {}
for mag, mes in zip(magnitudes, measures):
means[mag] = sum(mes) / len(mes)
return means
print(exp(['mag1', 'mag2', 'mag3'], [[1, 2, 3], [3, 4], [5]]))
So feel free to use the example which suits your requirements of which one you like and don't forget to add docstring.
More likely you don't need such pythonic way but it can be even shorter when dictionary comprehension comes into play:
def exp(magnitudes, measures):
return {mag: sum(mes) / len(mes) for mag, mes in zip(magnitudes, measures)}
print(exp(['mag1', 'mag2', 'mag3'], [[1, 2, 3], [3, 4], [5]]))