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I have the following dictionary:
StudentGrades = {
'Ivan': [4.32, 3, 2],
'Martin': [3.45, 5, 6],
'Stoyan': [2, 5.67, 4],
'Vladimir': [5.63, 4.67, 6]
}
I want to make a function that prints the average of the grades of the students, i.e. the average of the values
This answer was intended for Python2, which is now dead
Okay, so let's iterate over all dictionary keys and average the items:
avgDict = {}
for k,v in StudentGrades.iteritems():
# v is the list of grades for student k
avgDict[k] = sum(v)/ float(len(v))
In Python3, the iteritems() method is no longer necessary, can use items() directly.
now you can just see :
avgDict
Out[5]:
{'Ivan': 3.106666666666667,
'Martin': 4.816666666666666,
'Stoyan': 3.89,
'Vladimir': 5.433333333333334}
From your question I think you're queasy about iteration over dicts, so here is the same with output as a list :
avgList = []
for k,v in StudentGrades.iteritems():
# v is the list of grades for student k
avgDict.append(sum(v)/ float(len(v)))
Be careful though : the order of items in a dictionary is NOT guaranteed; this is, the order of key/values when printing or iterating on the dictionary is not guaranteed (as dicts are "unsorted").
Looping over the same identical dictionary object(with no additions/removals) twice is guaranteed to behave identically though.
If you don't want to do the simple calculation use statistics.mean:
from statistics import mean
StudentGrades = {
'Ivan': [4.32, 3, 2],
'Martin': [3.45, 5, 6],
'Stoyan': [2, 5.67, 4],
'Vladimir': [5.63, 4.67, 6]
}
for st,vals in StudentGrades.items():
print("Average for {} is {}".format(st,mean(vals)))
from scipy import mean
map(lambda x: mean(StudentGrades[x]), StudentGrades)
Generates this output:
[3.1066666666666669,
3.8900000000000001,
5.4333333333333336,
4.8166666666666664]
If you prefer a non-scipy solution one could use sum and len like supposed by Jiby:
map(lambda x: sum(StudentGrades[x])/len(StudentGrades[x]), StudentGrades)
EDIT: I am terribly sorry, I forgot you want a Python 3.4 solution, therefore (because you would get a map object returned) you need, for example, an additional list command:
from scipy import mean
list(map(lambda x: mean(StudentGrades[x]), StudentGrades))
This will return the desired output.
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For a homework assignment I am given this task:
Read a list of numbers and generate a new list which squares the members of the original list.
generate a third list of ints that ranks the second list of ints.
What does it mean to "rank the list"? What is this asking me to do?
I am NOT asking for a solution to the programming assignment, please do not provide one.
To create a list of integers between 0 and N, a solution is to use the range(N) function.
Then you can use the append() method which adds an item to the end of the list while getting the square of each element.
Finally use the sort() method which sorts the list ascending by default. You can also make a function to decide the sorting criteria(s)
def printSortedList():
l = list()
for i in range(1,21):
l.append(i**2)
l.sort()
print(l)
printSortedList()
Ranking means to number the items in a list from lowest to highest, or vice-versa.
So, for example, “ranking” the list [639, 810, 150, 162, 461, 853, 648] produces the list [4, 6, 1, 2, 3, 7, 5], where 1 corresponds to the lowest value (150), 2 to the next-lowest value (162).
Or maybe you're supposed to rank them the other way, giving [4, 2, 7, 6, 5, 1, 3].
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How can I add a constant (ie 10) to a list without creating a new list?
For example,
list = [2, 4, 6, 8]
How can I overwrite the elements in this list without creating another list (append method)? I understand that there is a map lambda method. Any way to go about by using a for loop?
When x = 10, expected final output should be
list = [12, 14, 16, 18]
x = 10
for i in range(0, len(list)):
list[i] = list[i] + x
I kept the variable name as list, but I recommend you don't use built-in names as names for your variables.
You can for sure use the map function in Python as follows. I changed your list name to list1 to not confuse it with the datatype.
list1 = [2,4,6,8]
print(list(map(lambda x : x+10, list1)))
#Jax Teller 's solution is the standard.
Here is a generator, not changing the original, but instead yielding mapped values
Also, using the word list isn't good, because it is a Python construct. Use something else instead like my_list
my_list = [2,4,6,8]
val = 10
gen = (val + e for e in my_list)
This is actually the same as map.
Just notice you can't use list(gen) because it would create a new list, not just iterate.
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I have a list like this:
my_lst = [[1,2,3,4]]
I was wondering how can I remove [] from the list to get the following list:
my_lst = [1,2,3,4]
The double [[]] is because you have a list containing a list. To get out the inner list, just index it:
my_lst = my_lst[0]
or unpack it:
[my_lst] = my_lst
Unpacking has the mild advantage that it will give you an error if the outer list doesn't not contain exactly one value (while my_lst[0] will silently discard any extra values).
There are many other options.
my_lst = my_lst[0]
You have a list with one element, which is itself a list. You want the variable to just be that first element, the list 1,2,3,4.
you have a list of lists.
my_lst = my_lst[0] #get the first list of the list
You can use itertools to achieve that without writing your explicit loop if the length of the outer list could be more than one.
In [47]: import itertools
In [49]: list(itertools.chain.from_iterable([[1, 2, 3], [1, 2]]))
Out[49]: [1, 2, 3, 1, 2]
You can reduce 1 dimension using sum.
my_lst=sum(my_lst,[])
# [1,2,3,4]
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Hi I need a single list of nested lists in python without any inbuilt function. I did a lot of research but not able to find without flatten or ravel function().
for ex:
input = [1,2,[2,3,[4,5]],[6,7],[8,[9]],10]
output = [1,2,2,3,4,5,6,7,8,9,10]
You can use recursion to implement your own flatten function. Here's a nice tutorial explaining the concepts of recursion.
The idea here is that you loop through each element in your list and if you encounter a sublist you recursively call the same method with the current list. If it's a valid element you add it to to your output list.
Example:
def flatten(input_list, output_list=[]):
for i in input_list:
if isinstance(i, list):
flatten(i, output_list)
else:
output_list.append(i)
return output_list
input_list = [1,2,[2,3,[4,5]],[6,7],[8,[9]],10]
print(flatten(input_list))
Outputs:
[1, 2, 2, 3, 4, 5, 6, 7, 8, 9, 10]
I found a solution on my own. Below is the solution:
c = []
def test(a):
for val in a:
if type(val) == list:
test(val)
else:
c.append(val)
return c
a=[1,2,[2,3,[4,5]],[6,7],[8,[9]],10]
print test(a)
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Hi guys i have a dictionary that looks like this:
[[['Test', 0,1],['Test2', 0,4],['Test3', 0,5],['Test4', 0,2],['Test5', 0,6],...]]
How can I get the top two (for example)?
Expected result:
[['Test3', 0,5],['Test5',0,6]]
Sort by the third value:
>>> sorted(l[0], key=lambda x: x[2])[-2:]
[['Test3', 0, 5], ['Test5', 0, 6]]
test = [[['Test', 0,1],['Test2', 0,4],['Test3', 0,5],['Test4', 0,2],['Test5', 0,6]]]
result = sorted(test[0], key=lambda x: x[2])[-2:]
print(result)
Output:
[['Test3', 0, 5], ['Test5', 0, 6]]
Something like this?
(I am assuming that you actually mean ['Test', 0, 1] to be a list with 3 elements which is to be sorted by the third element rather than a two element list with the second element being a float, in which case it should be ['Test', 0.1] and you can look at #jonrsharpe's answer. I also assumed that the output order (lowest to highest) is relevant.)
That is not a dictionary, it is a list ([]) containing another list which contains lists. Each sub-sub-list contains three elements:
['Test2', 0, 4]
#^0 ^1 ^2
Note that Python doesn't allow the user of commas as decimal points; I think you want 0.4.
Assuming that, sorting the sub-list and extracting the two highest sub-sub-lists is easy:
from operator import itemgetter
output = sorted(lst[0], key=itemgetter(1), reverse=True)[:2]
If I am wrong about the comma, make the argument to itemgetter 2.