So basically I am trying to replace this:
board = {
0:[0, 1, 2, 9, 10, 11, 18, 19, 20],
1:[3, 4, 5, 12, 13, 14, 21, 22, 23],
2:[6, 7, 8, 15, 16, 17, 24, 25, 26]}
with a for loop that will automatically create it. Sorry if this seems obvious, but I'm a bit of a noob and I'm having a lot of trouble with this.
It looks like you're generating the first 27 integers (starting at 0) and then grouping them. Let's write it like that.
def group_by_threes(n=27, group_count=3):
# The end result will be a dict of lists. The number of lists
# is determined by `group_count`.
result = {}
for x in range(group_count):
result[x] = []
# Since we're using subgroups of 3, we iterate by threes:
for x in range(n // 3):
base = 3 * x
result[x % 3] += [base, base + 1, base + 2]
# And give back the answer!
return result
This code could be made better by making the size of groups (three in this case) an argument, but I'll leave that as an exercise to the reader. ;)
The advantage of this method is that it's much more modular and adaptable than just writing a one-off method that generates the exact list you're looking for. After all, if you only wanted to ever generate that one list you showed, then it'd probably be better to hardcode it!
def create_list(x):
a = [x,x+1,x+2,x+9,x+9+1,x+9+2,x+18,x+18+1,x+18+2]
return a
output = {}
for i in range(3):
output[i*3] = create_list(i*3)
print output
please try this you get desired output
def create_list(x):
res = []
for i in xrange(0,3):
for j in xrange(0,3):
res.append(3*x+ 9*i + j)
return res
dictionary={}
for i in xrange(0,3):
dictionary[i]=create_list(i)
print dictionary
Result:
{0: [0, 1, 2, 9, 10, 11, 18, 19, 20], 1: [3, 4, 5, 12, 13, 14, 21, 22, 23], 2: [6, 7, 8, 15, 16, 17, 24, 25, 26]}
Related
I have two lists:
l1: 38510 entries
l2: 6384 entries
I want to extract only values, which are present in both lists.
So far that was my approach:
equals = []
for quote in l2:
for quote2 in l1:
if quote == quote2:
equals.append(quote)
len(equals)) = 4999
len(set(equals))) = 4452
First of all, I have the feeling this approach is pretty inefficient, because I am checking every value in l1 several times ..
Furthermore, it seems that I get still duplicates. Is this due to the inner-loop for l1?
Thank you!!
You can use list comprehension and the in operator.
a = [1, 2, 3, 4, 5, 6, 7, 8, 9]
b = [2, 4, 6, 8, 0]
[x for x in a if x in b]
#[2, 4, 6, 8]
You were on the right track by using sets. One of set's coolest features is that you can get the intersection between two sets. An intersection is another way to say the values that occur in both sets. You can read about it more in the docs
Here is my example:
l1_set = set(l1)
l2_set = set(l2)
equals = l1_set & l2_set
#If you really want it as a list
equals = list(equals)
print(equals)
The & operator tells python to return a new set that only has values in both sets. At the end, I went ahead and converted equals back to a list because that's what your original example wanted. You can omit that if you don't need it.
1. This is the simplest method where we haven’t used any built-in functions.
# Two lists in most simple way of showing the intersection
def intersection(list_one, list_two):
temp_list = [value for value in list_one if value in list_two]
return temp_list
# Illustrate the intersection
list_one = [4, 9, 1, 17, 11, 26, 28, 54, 69]
list_two = [9, 9, 74, 21, 45, 11, 63, 28, 26]
print(intersection(list_one, list_two))
# [123, 3, 23, 15]
2. You can use the python set() method.
# Two lists using set() method
def intersection(list_one, list_two):
return list(set(list_one) & set(list_two))
# Illustrate the intersection
list_one = [15, 13, 123, 23, 31, 10, 3, 311, 738, 25, 124, 19]
list_two = [12, 14, 1, 15, 36, 123, 23, 3, 315, 87]
print(intersection(list_one, list_two))
# [123, 3, 23, 15]
3. In this technique, we can use the built-in function called intersection() to compute the intersected list.
First, we need to use set() for a larger list then compute the intersection.
# Two lists using set() and intersection()
def intersection_list(list_one, list_two):
return list(set(list_one).intersection(list_two))
# Illustrate the intersection
list_one = [15, 13, 123, 23, 31, 10, 3, 311, 738, 25, 124, 19]
list_two = [12, 14, 1, 15, 36, 123, 23, 3, 315, 87, 978, 4, 13, 19, 20, 11]
if len(list_one) < len(list_two):
list_one, list_two = list_two, list_one
print(intersection_list(list_one, list_two))
# [3, 13, 15, 19, 23, 123]
Additional you can follow the bellow tutorials
Geeksforgeeks
docs.python.org
LearnCodingFast
Let's assume that all the entries in both of your lists are integers. If so, computing the intersection between the 2 lists would be more efficient than using list comprehension:
import timeit
l1 = [i for i in range(0, 38510)]
l2 = [i for i in range(0, 6384)]
st1 = timeit.default_timer()
# Using list comprehension
l3 = [i for i in l1 if i in l2]
ed1 = timeit.default_timer()
# Using set
st2 = timeit.default_timer()
l4 = list(set(l1) & set(l2))
ed2 = timeit.default_timer()
print(ed1-st1) # 5.7621682 secs
print(ed2-st2) # 0.004478600000000554 secs
As you have such long lists, you might want to use numpy which is specialized in providing efficient list processing for Python.
You can enjoy the fast processing with its numpy function. For your case, you can use numpy.intersect1d() to get the sorted, unique values that are in both of the input arrays, as follows:
import numpy as np
l1 = [1, 3, 5, 10, 11, 12]
l2 = [2, 3, 4, 10, 12, 14, 16, 18]
l_uniques = np.intersect1d(l1, l2)
print(l_uniques)
[ 3 10 12]
You can keep the resulting list as numpy array for further fast processing or further convert it back to Python list by:
l_uniques2 = l_uniques.tolist()
I working on a problem where I'm supposed to generate ten random but unique numbers that range from 1 to 15 inclusive. The thing is, I'm supposed to write everything in one line and to also get this output:
[2, 4, 6, 7, 8, 9, 11, 12, 13, 15]
Below I have some code I wrote but, it's not getting the output I want. What am I doing wrong and can I perhaps see a solution with a break so I know how to do this going down the road?
import random
print(sorted(random.sample(range(1,16),15)))
Output:
[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
The output I want is:
[2,4,6,7,8,9,11,12,13,15]
How do I get this in one line of code?
>>> help(random.sample)
sample(population, k): method of random.Random instance
Chooses k unique random elements from a population sequence or set.
I'm supposed to write everything in one line and to also get this output:
[2, 4, 6, 7, 8, 9, 11, 12, 13, 15]
>>> sorted(__import__('random').Random(4225).sample(range(1, 16), 10))
[2, 4, 6, 7, 8, 9, 11, 12, 13, 15]
If you want to generate ten numbers in range 1-15, change
print(sorted(random.sample(range(1,16),15)))
to
print(sorted(random.sample(range(1,16),10)))
# From the documentation :
# random.sample(population, k)
import random
population = range(16)
how_may_sample = 10
random.sample(population, how_many_sample)
# Now in one line
random.sample(range(16), 10)
I am quite new to programming and have a string with integrated list values. I am trying to isolate the numerical values in the string to be able to use them later.
I have tried to split the string, and change it back to a list and remove the EU variables with a loop. The initial definition produces the indexes of the duplicates and writes them in a list/string format that I am trying to change.
This is the csv file extract example:
Country,Population,Number,code,area
,,,,
Canada,8822267,83858,EU15,central
Denmark,11413058,305010,EU6,west
Southafrica,705034,110912,EU6,south
We are trying to add up repeating EU number populations.
def duplicates(listed, number):
return [i for i,x in enumerate(listed) if x == number]
a=list((x, duplicates(EUlist, x)) for x in set(EUlist) if EUlist.count(x) > 1)
str1 = ''.join(str(e) for e in a)
for x in range (6,27):
str2=str1.replace("EUx","")
#split=str1.split("EUx")
#Here is where I tried to split it as a list. Changing str1 back to a list. str1= [x for x in split]
This is what the code produces:
('EU6', [1, 9, 10, 14, 17, 19])('EU12', [21, 25])('EU25', [4, 5, 7, 12, 15, 16, 18, 20, 23, 24])('EU27', [2, 22])('EU9', [6, 13])('EU15', [0, 8, 26])
I am trying to isolate the numbers in the square brackets so it prints:
[1, 9, 10, 14, 17, 19]
[21, 25]
[4, 5, 7, 12, 15, 16, 18, 20, 23, 24]
[2, 22]
[6, 13]
[0, 8, 26]
This will allow me to isolate the indexes for further use.
I'm not sure without example data but I think this might do the trick:
def duplicates(listed, number):
return [i for i,x in enumerate(listed) if x == number]
a=list((x, duplicates(EUlist, x)) for x in set(EUlist) if EUlist.count(x) > 1)
for item in a:
print(item[1])
At least I think this should print what you asked for in the question.
As an alternative you can use pandas module and save some typing. Remove the four commas on second line and then:
import pandas as pd
csvfile = r'C:\Test\pops.csv'
df = pd.read_csv(csvfile)
df.groupby('membership')['Population'].sum()
Will output:
membership
Brexit 662307
EU10 10868
EU12 569219
EU15 8976639
EU25 17495803
EU27 900255
EU28 41053
EU6 13694963
EU9 105449
I am newbie in Python. I want to insert new node into the current route to check whether it makes the route shorter or not. However my code doesn't run well, please show me the mistake. The steps are following:
1. Create random subtour (example: 0-2-0)
2. Get randomly the node which is not visited and check this node in each pair of nodes in current route. If the node satisfy the shorter requirement, we insert it into current node (example: 0-4-2-0).
3. Continue until all nodes inserted into the route.
import random
distMatrix = [
[100, 14, 20, 10, 35, 18, 5],
[6, 100, 7, 35, 17, 9, 24],
[8, 35, 100, 36, 27, 3, 15],
[21, 7, 12, 100, 7, 4, 26],
[33, 25, 6, 18, 100, 19, 11],
[6, 2, 22, 30, 9, 100, 8],
[24, 3, 12, 5,17, 16, 100],
]
def get_total_distance(route,d):
total = 0
for i in range (len(route)-1):
pre = route[i]
succ = route[i+1]
total += d[pre][succ]
return total
def insertion(d):
numNodes = len(d)
notVisited = list(range(1, numNodes))
first_random_node = random.choice(notVisited)
route = [0]
route.append(first_random_node)
notVisited.remove(first_random_node)
route.append(0) #create first subtour
print("1st",route)
location = 0
while len(notVisited) != 0:
for j in notVisited:
for i in range (len(route)-1):
pre = route[i]
succ = route[i+1]
check_route = d[pre][j] + d[j][succ]
current_distance = d[pre][succ]
if check_route <= current_distance:
print(j)
route.insert(i + 1, j)
notVisited.remove(j)
print("2nd", route)
return route
solution = insertion(distMatrix)
print("The solution for the route is:",solution)
print("The total distance is:", get_total_distance(solution,distMatrix))
As it is, your code doesn't run: there's a syntax error because you named a local function the same as a package you're trying to use: random. Inside your function random, you can no longer access the package of that name, because you redefined the name.
Once that is resolved, your code hangs with this output:
1st [0, 5, 0]
3
2nd [0, 3, 5, 0]
6
2nd [0, 6, 3, 5, 0]
You have a logic problem with this combination:
while len(notVisited) != 0:
print("WHILE", len(notVisited))
for j in notVisited:
....
if check_route <= current_distance:
print(j)
route.insert(i + 1, j)
notVisited.remove(j)
print("2nd", route)
Your only exit from the while loop is to remove everything from notVisited. There's no logical guarantee that will happen -- and when it doesn't, you're stuck in an infinite loop trying to find a shorter route that doesn't exist.
I have seen this question been asked before but I'm looking for an answer with a twist:
Consider I have a range like 1-100 and i want to generate a list, with a specific stepsize like: numbers(1,100,5). This would return [1,25,50,75,100]. However, I would like it to return [1,100,50,25,75] or [1,100,50,75,25] Another example would be numbers(1,10,10) which would give me something similar to [1,10,5,2,7,3,8,4,9].
Is this even possible to do? The reason for this would be to be able to render image sequences without going from frame 1 to frame 2 to frame 3 and so forth. Instead I want to render the first frame, the last frame, the middle frame, the middle of the middle frame until all frames are accounted for.
Your question is kind of underspecified, but this should help you get started.
def numbers(first, last, count):
nums = [first, last]
width = last - first
while len(nums) < count:
width /= 2
stepper = first
while stepper < last:
rounded = round(stepper)
if rounded not in nums:
nums.append(rounded)
if len(nums) == count:
break
stepper += width
return nums
Then:
>>> numbers(0, 100, 5)
[0, 100, 50, 25, 75]
>>> numbers(0, 10, 10)
[0, 10, 5, 2, 8, 1, 4, 6, 9, 3, 7]
>>> numbers(0, 50, 50)
[0, 50, 25, 12, 38, 6, 19, 31, 44, 3, 9, 16, 22, 28, 34, 41, 47, 2, 5, 8, 11, 14, 17, 20, 23, 27, 30, 33, 36, 39, 42, 45, 48, 1, 4, 7, 10, 13, 15, 18, 21, 24, 26, 29, 32, 35, 37, 40, 43, 46]
The basic algorithm is as follows:
Start with a list of nums containing the two endpoints
Initialize width to the distance between the two endpoints
Then, loop:
Halve width
Step through first, first+width, first+2*width, ..., last-width, last, and add whichever among those are not already in nums to nums (so, for numbers(0, 100, 5), the first loop iteration will try 0, 50, and 100, and only add 50 since that wasn't already present; the second iteration will try 0, 25, 50, 75, and 100, and only add 25 and 75).
If we have enough numbers in nums, we're done
Return nums
Ok, so your desired frames are kind of wierd, especially as the elements in the first example aren't evenly ranged, eg. 100-75=25, 75-50=25, 50-25=25 , 25-1=24.
But, if we assume that you always want the start and end value in the frame and want the evenly spaced values pegged against the maximum value, we can do this:
def numbers(start,stop,step=1):
frame = [start]+range(y,stop,(start-stop)/(step-1))
return frame
The random module includes a shuffle() method that takes an array and shuffles it in-place.
Which means the function becomes:
from random import shuffle
def numbers(start,stop,step=1):
frame = [start]+range(y,stop,(start-stop)/(step-1))
shuffle(frame)
return frame
Giving us the following test runs:
>>> numbers(1,100,5)
[100, 50, 75, 1, 25]
>>> numbers(1,10,10)
[1, 3, 10, 9, 6, 5, 8, 4, 7, 2]
Original (but wrong) answer
The random module includes a shuffle() method that takes an array and shuffles it in-place.
For example:
from random import shuffle
def numbers(start,stop,step=1):
frame = range(start,stop,step)
shuffle(frame)
return frame
Then calling this function we get:
>>> numbers(1,100,25)
[1, 51, 26, 76]
>>> numbers(1,100,25)
[76, 26, 1, 51]
Note, that as per the range() function, the step value is repeatedly added to start such that the final array is of the form [start, start+1*step, start+2*step ... start + n*step], where start+n*step is less than stop while start+(n+1)*step is greater than stop.