Getting values of each number to be not more 90 - python

I have a code which generates random number and put them in a list. The total of the values of these number must follow a defined value (in this case 6066). The numbers in the list also have to be a certain amount, meaning that i want 95 numbers to be generated randomly into a list, and the total of the values of these 95 numbers in the list is equals to 6066.
The code :
import random
def num(n, total):
dividers = sorted(random.sample(range(1, total), n - 1))
j= [a - b for a, b in zip(dividers + [total], [0] + dividers)]
return j
i=num(95,6066)
print (i)
The problem im facing is that i do not want any of the values of the 95 numbers in the list to exceed 85. How do i do this?
I have tried:
import random
def num(n, total):
dividers = sorted(random.sample(range(1, total), n - 1))
j= [a - b for a, b in zip(dividers + [total], [0] + dividers)]
for k in j:
if k>85:
j.remove(k)
return j
i=num(95,6066)
print (i)
But this only removes the number which are more than 85 from the list, i need to have 95 numbers in the list and total up to 6066

One solution will be to consider the problem like you are trying to distribute 6066 items between 95 buckets and each one has a capacity of 85, so you just loop over the items and each time choose a bucket that is not already full.
Here is a simple implementation. It won't be particularly fast, but it avoids the need to backtrack because there is no possibility of violating the rules (total sum incorrect or individual value exceeds the maximum).
Note that other solutions that are equally valid within your rules may have a different probability distribution, but you have not said anything about what probability distribution you require.
import random
def num(n, total, maxv):
if total > n * maxv:
raise ValueError("incompatible requirements")
vals = [0 for _ in range(n)]
not_full = list(range(n))
for _ in range(total):
index = random.choice(not_full)
vals[index] += 1
if vals[index] == maxv:
not_full.remove(index)
return vals
answer = num(95, 6066, 85)
print(answer)
print(max(answer))
print(sum(answer))
Gives:
[59, 59, 73, 63, 77, 58, 54, 71, 73, 67, 69, 67, 58, 79, 63, 59, 80, 58, 77, 64, 62, 64, 54, 50, 64, 72, 62, 69, 81, 61, 63, 50, 65, 56, 60, 51, 59, 61, 63, 56, 67, 69, 69, 64, 85, 66, 74, 66, 63, 63, 63, 68, 84, 66, 53, 82, 59, 66, 63, 58, 67, 58, 59, 58, 69, 56, 63, 61, 73, 58, 65, 60, 61, 53, 68, 51, 58, 57, 67, 60, 65, 73, 63, 59, 62, 49, 66, 59, 64, 56, 69, 58, 61, 67, 74]
85
6066

Related

given an random 100 number with duplicate. I want to count how many number is inside an interval of number in python

For Example
first_interval = [40, 50, 60, 70, 80, 90]
second_interval = [49, 59, 69, 79, 89, 99]
Data = [40, 42, 47, 49, 50, 52, 55, 56, 57, 59, 60, 61, 63, 65, 65, 65, 66, 68, 68, 69, 72, 74, 78, 79, 81, 85, 87, 88, 90, 98]
x = first_interval[0] <= data <= second_interval[0]
y = first_interval[1] <= data <= second_intercal[1] # and so on
I want to know how many numbers from data is between 40-49, 50-59, 60-69 and so on
frequency = [4, 6] # 4 is x and 6 is y
Iterate on the bounds using zip, then with a list comprehension you can filter the correct values
first_interval = [40, 50, 60, 70, 80, 90]
second_interval = [49, 59, 69, 79, 89, 99]
data = [40, 42, 47, 49, 50, 52, 55, 56, 57, 59, 60, 61, 63, 65, 65,
65, 66, 68, 68, 69, 72, 74, 78, 79, 81, 85, 87, 88, 90, 98]
result = {}
for start, end in zip(first_interval, second_interval):
result[(start, end)] = len([v for v in data if start <= v <= end])
print(result)
# {(40, 49): 4, (50, 59): 6, (60, 69): 10, (70, 79): 4, (80, 89): 4, (90, 99): 2}
print(result[(40, 49)])
# 4
The version with a list and len is easier to understand
result[(start, end)] = len([v for v in data if start <= v <= end])
But the following version would be more performant for bigger size, as it's a generator, it won't have to build the whole list to just forget it after
result[(start, end)] = sum((1 for v in data if start <= v <= end))
Another version, that doesn't use the predefined bounds, and so is much performant as it's complexity is O(n) and not O(n*m) as the first one : you iterate once on values, not on values for each bounds
result = defaultdict(int) # from collections import defaultdict
for value in data:
start = 10 * (value // 10)
result[(start, start + 9)] += 1
This may help you :
first_interval = [40, 50, 60, 70, 80, 90]
second_interval = [49, 59, 69, 79, 89, 99]
Data = [40, 42, 47, 49, 50, 52, 55, 56, 57, 59, 60, 61, 63, 65, 65, 65, 66, 68, 68, 69, 72, 74, 78, 79, 81, 85, 87, 88, 90, 98]
def find_occurence(start,end,data):
counter = 0
for i in data :
if start<=i<=end :
counter += 1
return counter
print(find_occurence(first_interval[0],second_interval[0],Data)) #this gives you the anser for x and the same thing for y
Note : start :means from where you want to start.
end : mean where you want to stop.
We can use numpy.histogram with bins defined by:
first_interval bins, but open on the right
max(second_interval) to determine the close of rightmost bin
Code
# Generate counts and bins (right most edge given by max(second_interval))
frequency, bins = np.histogram(data, bins = first_interval + [max(second_interval)])
# Show Results
for i in range(len(frequency)):
if i < len(frequency) - 1:
print(f'{bins[i]}-{bins[i+1]-1} : {frequency[i]}') # frequency doesn't include right edge
else:
print(f'{bins[i]}-{bins[i+1]} : {frequency[i]}') # frequency includes right edge in last bin
Output
40-49 : 4
50-59 : 6
60-69 : 10
70-79 : 4
80-89 : 4
90-99 : 2

Using a python for loop to print numbers from a list which are not in order of their index

I have a little query I'm currently trying to use a for loop to print the numbers in my list, I would like to print several numbers which are next to each other, but then I would also like to print several numbers which are further down the list. My code is below what I would like to do is print the numbers 51 and 52 as well as printing the numbers 56 and 57. Is it possible to do this within one loop or will I need to create another on?
Test = [50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62]
for i in range (1,3), (6, 8):
print (Test[i])
Test = [50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62]
indices = (1, 2, 6, 7)
for i in indices:
print(Test[i])
>>>51 52 56, 57 (on separate lines, of course)
If you always want to print in pairs, you can only specify the first element's index and print twice in the loop:
Test = [50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62]
indices = (1, 6)
for i in indices:
print(Test[i])
print(Test[i+1])
>>>51 52 56, 57 (on separate lines, of course)
You can do this output
Test = [50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62]
for i in range (1,3):
print (Test[i])
for i in range (6, 8):
print (Test[i])

How to generate a list of numbers in python

guys. I am now working on a python algorithm and I am new to python. I'd like to generate a list of numbers like 4, 7, 8, 11, 12, 13, 16, 17, 18, 19, 22, 23, 24, 25... with 2 for loops.
I've done some work to find some numbers and I am close to the result I want, which is generate a list contains this numbers
My code is here:
for x in range(0, 6, 1):
start_ind = int(((x+3) * (x+2)) / 2 + 1)
print("start index is ", [start_ind], x)
start_node = node[start_ind]
for y in range(0, x):
ind = start_ind + y + 1
ind_list = node[ind]
index = [ind_list]
print(index)
Node is a list:
node = ['n%d' % i for i in range(0, 36, 1)]
What I received from this code is:
start index is [7] 1
['n8']
start index is [11] 2
['n12']
['n13']
start index is [16] 3
['n17']
['n18']
['n19']
start index is [22] 4
['n23']
['n24']
['n25']
['n26']
start index is [29] 5
['n30']
['n31']
['n32']
['n33']
['n34']
This seems to give the same list: and I think it's much clearer what's happening!
val=4
result=[]
for i in range(1,7):
for j in range(val,val+i):
val = val+1
result.append(j)
val = j+3
print(result)
Do not think you need a loop for this, let alone two:
import numpy as np
dif = np.ones(100, dtype = np.int32)
dif[np.cumsum(np.arange(14))] = 3
(1+np.cumsum(dif)).tolist()
output
[4, 7, 8, 11, 12, 13, 16, 17, 18, 19, 22, 23, 24, 25, 26, 29, 30, 31, 32, 33, 34, 37, 38, 39, 40, 41, 42, 43, 46, 47, 48, 49, 50, 51, 52, 53, 56, 57, 58, 59, 60, 61, 62, 63, 64, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 121, 122, 123, 124, 125, 126, 127, 128, 129]
ind_list = []
start_ind = 4
for x in range(0, 6):
ind_list.append(start_ind)
for y in range(1, x+1):
ind_list.append(start_ind + y)
start_ind = ind_list[len(ind_list)-1]+3
print(ind_list)
You could probably use this. the print function works fine, the list I assume works fairly well for the numbers provided. It appends the new number at the beginning of the loop, with a cotinually longer loop each time for x. I'm assuming the number sequence is 4, 4+3, 4+3+1, 4+3+1+3, 4+3+1+3+1, 4+3+1+3+1+1, 4+3+1+3+1+1+3, ....

Bin sequential values in list

I have the list:
new_maks = [75, 76, 77, 78, 79, 80, 81, 85, 86, 87, 88, 89, 91]
I want to bin the elements into areas where the next element are sequentially increased by 1. My initial idea is basically to initialize two lists bin_start and bin_end and iterate through new_maks to check for sequential values.
bin_start = []
bin_end = []
counter = 0
for i in range(len(new_maks)):
if new_maks[i] == new_maks[0]:
bin_start.append(new_maks[i])
elif (new_maks[i] - new_maks[i-1]) ==1:
try:
bin_end[counter] = new_maks[i]
except:
bin_end.append(new_maks[i])
elif (new_maks[i] - new_maks[i-1]) >1:
if new_maks[i] != new_maks[-1]:
bin_start.append(new_maks[i])
counter +=1
Which produces the desired result of:
bin_start= [75, 85]
bin_end = [81, 89]
Is there a simpler/vectorized way to achieve this result?
Here's for performance efficiency with NumPy tools -
def start_stop_seq1(a):
m = np.r_[False,np.diff(a)==1,False]
return a[m[:-1]!=m[1:]].reshape(-1,2).T
Sample run -
In [34]: a # input array
Out[34]:
array([ 75, 76, 77, 78, 79, 80, 81, 85, 86, 87, 88, 89, 91,
92, 93, 100, 101, 110])
In [35]: start_stop_seq1(a)
Out[35]:
array([[ 75, 85, 91, 100],
[ 81, 89, 93, 101]])
Alternative #1 : One liner with one more np.diff
We can go one step further to achieve compactness -
In [43]: a[np.diff(np.r_[False,np.diff(a)==1,False])].reshape(-1,2).T
Out[43]:
array([[ 75, 85, 91, 100],
[ 81, 89, 93, 101]])
A simpler way could be to use groupby and count:
from itertools import groupby, count
counter = count(1)
new_mask = [75, 76, 77, 78, 79, 80, 81, 85, 86, 87, 88, 89]
generator = ((first, last) for key, (first, *_, last) in groupby(new_mask, key=lambda val: val - next(counter)))
bin_start, bin_end = zip(*generator)
print(bin_start)
print(bin_end)
Output
(75, 85)
(81, 89)
This is based in an old itertools recipe. If you fancy pandas you could do something like this:
import pandas as pd
new_mask = [75, 76, 77, 78, 79, 80, 81, 85, 86, 87, 88, 89]
s = pd.Series(data=new_mask)
result = s.groupby(s.values - s.index).agg(['first', 'last'])
bin_start, bin_end = zip(*result.itertuples(index=False))
print(bin_start)
print(bin_end)
Again this is based on the principle that consecutive increasing (by 1) values will have the same difference against an running sequence. As mentioned in the linked documentation:
The key to the solution is differencing with a range so that
consecutive numbers all appear in same group.

How to get all values possible from a list of number without itertools

I want to create an algorithm that find all values that can be created with the 4 basic operations + - * / from a list of number n, where 2 <= len(l) <= 6 and n >= 1
All numbers must be integers.
I have seen a lot of similar topics but I don't want to use the itertool method, I want to understand why my recursive program doesn't work
I tried to make a costly recursive program that makes an exhaustive search of all the possible combinations, like a tree with n=len(l) start and each tree depth is n.
L list of the starting number
C the current value
M the list of all possible values
My code:
def result(L,C,M):
if len(L)>0:
for i in range(len(L)) :
a=L[i]
if C>=a:
l=deepcopy(L)
l.remove(a)
m=[] # new current values
#+
m.append(C+a)
# * 1 is useless
if C !=1 or a !=1:
m.append(C*a)
# must be integer
if C%a==0 and a<=C: # a can't be ==0
m.append(C//a)
#0 is useless
if C!=a:
m.append(C-a)
for r in m: #update all values possible
if r not in M:
M.append(r)
for r in m: # call the fucntion again with new current values,and updated list of remaining number
result(l,r,M)
def values_possible(L) :
m=[]
for i in L:
l=deepcopy(L)
l.remove(i)
result(l,i,m)
m.sort()
return m
For small lists without duplicate numbers, my algorithm seems to work but with lists like [1,1,2,2,4,5] it misses some values.
It returns:
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21,
22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41,
42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61,
62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81,
82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 94, 95, 96, 97, 98, 99, 100, 101,
102, 104, 105, 110, 112, 115, 116, 118, 119, 120, 121, 122, 124, 125, 128, 130,
140, 160]
but it misses 93,108,114,117,123,126,132,135,150,180.
Let's take an even simpler example: [1, 1, 2, 2].
One of the numbers your algorithm can't find is 9 = (1 + 2) * (1 + 2).
Your algorithm simply cannot come up with this computation because it always deals with a "current" value C. You can start with C = 1 + 2, but you cannot find the next 1 + 2 because it has to be constructed separately.
So your recursion will have to do at least some kind of partitioning into two groups, finding all the answers for those and then doing combining them.
Something like this could work:
def partitions(L):
if not L:
yield ([], [])
else:
for l, r in partitions(L[1:]):
yield [L[0]] + l, r
yield l, [L[0]] + r
def values_possible(L):
if len(L) == 1:
return L
results = set()
for a, b in partitions(L):
if not a or not b:
continue
for va in values_possible(a):
for vb in values_possible(b):
results.add(va + vb)
results.add(va * vb)
if va > vb:
results.add(va - vb)
if va % vb == 0:
results.add(va // vb)
return results
Not too efficient though.

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