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I was trying to create this [80000, 104000, 135000...] list in Python. Its the value, starting at 80,000 multiplied by 1.3 each time I want
What i've tried:
a = [num*1.5 for num in ??? if num>=80000] #???--> i've tried range(10)
I should be able to do this but I can't find any solutions rn..
I must use list-comprehensions, if possible.
Some help would be nice, thank you!
There is a very basic mathematical operation that represents multiplying by the same value many time: power.
a = [80000 * (1.3**n) for n in range(100)]
You could write your own generator then use that in conjunction with a list comprehension.
def numgen(start, factor, limit):
for _ in range(limit):
yield int(start)
start *= factor
mylist = [value for value in numgen(80_000, 1.3, 10)]
print(mylist)
Output:
[80000, 104000, 135200, 175760, 228488, 297034, 386144, 501988, 652584, 848359]
import numpy as np
print(80000 * 1.3**np.arange(3))
# [ 80000. 104000. 135200.]
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A friend asked me to make her coding more "pythonic" but I'm pretty new at it myself. This is what I came up with, and I'm a little concerned that it won't hit all of the numbers (6, 7, 8, 9, 10 and 11). I also KNOW that there is a better way, but I just don't know what it is. Can you help?
prob = 0
for r in range(6,11):
prob += binom.pmf(k=r, n=11, p=0.2)
print(‘The probability is {}’,format(prob))
I think that your code is perfectly pythonic, for what that's worth
You could make it a little more compact and more performant with a list comprehension, the sum() function, and an f-string. Also, I took into account that you wanted 11 inclusive, so your range should be from 6 to 12.
probs = [binom.pmf(k=r, n=11, p=0.2) for r in range(6, 12)]
print(f'The probability is {sum(probs)}')
Use list comprenhension:
def binom(n,r,p):
# dummy function
return n+r+p
prob = sum([binom(r,12, 0.2) for r in range(6,11)])
print(f"The probability is {prob}")
Edit:
To make 11 inclusive, increase your range to 12
Edit 2:
As suggested by OneCricketeer, if only need to compute the sum, you can remove the square brackets, like so:
prob = sum(binom(r,12, 0.2) for r in range(6,11))
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I'm trying to find all possible groupings for a list so that the resulting list has a specified length, here's an example:
group([2,3,5,6,8], length=3)
would give
[[2,[3,5],[6,8]], [[2,3],5,[6,8]], [2,3,[5,6,8]], ...
what would be the best way to approach this?
Try this recursive solution, warning: if your list grows larger, this will grow exponentially! Also, this will have some duplicate in that the orders are different.
from itertools import permutations
master = []
def group(l, num, res=[]):
if num == 1:
res.append(l)
master.append(res)
return
for i in range(len(l)-num + 1):
firstgroup = list({e[:i+1] for e in permutations(l)})
for each in firstgroup:
myres = res[:]
myres.append(list(each))
lcp = l[:]
for j in each:
lcp.remove(j)
group(lcp, num-1, myres)
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I can't figure out how to do this without using complex functions, please help. this is the docstring of the code:
'''
finds all numbers in the list below a certain threshold
:param numList: a list of numbers
:threshold: the cutoff (only numbers below this will be included)
:returns: a new list of all numbers from numList below the threshold
'''
One approach
def filterList(numList, threshold):
return list(filter(lambda x: x < threshold, numList))
Another approach:
filteredList = [x for x in numList if x < threshold]
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I have a list a=['abc','cdv','fasdf'] and also have a constant n which says the amount of time print each elements two times.
For example, n=2 should return a=['abc','abc','cdv','cdv']; or n=4 will return a=['abc','abc','cdv','cdv','fasdf','fasdf','abc','abc'].
Here is one way using itertools.chain and a generator comprehension:
from itertools import chain
a = ['abc','cdv','fasdf']
n = 4
res = list(chain.from_iterable([a[i % len(a)]]*2 for i in range(n)))
# ['abc', 'abc', 'cdv', 'cdv', 'fasdf', 'fasdf', 'abc', 'abc']
it looks like you'll need to recycle elements if n is larger than the length of the list. An easy way to deal with this is to duplicated the array as many times as needed.
import math
n_over = math.ceil(len(a)/n)
n_reps = 1 + n_over
a_long = a * n_reps
we can iterate over the new array to build the new one
a_rep = []
for e in a_long[0:n]:
a_new += [e]*n
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Please help I have no idea on how to write this function. I tried a ceaser cypher function and it didn't work. Any ideas?
Write a function cycle( S, n ) that takes in a string S of '0's and '1's and an integer n and returns the a string in which S has shifted its last character to the initial position n times. For example, cycle('1110110000', 2) would return '0011101100'.
The function you are looking for is:
def cycle(s, n):
return s[-n:] + s[:-n]
You could use Python's deque data type as follows:
import collections
def cycle(s, n):
d = collections.deque(s)
d.rotate(n)
return "".join(d)
print cycle('1110110000', 2)
This would display:
0011101100