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I am new to python programming. I was trying to write a program to find how many 9s are present in the list.enter image description here
seq = [1, 2, 3, 4, 9, 6, 8, 13, 9, 12, 19]
n = 0
t = []
y = len(seq)
for x in range (0, y-1):
if seq[x] == 9:
n += 1
t.append(x+1)
else: continue
print (n, "numbers of 9s are present at", t, "position" )'
output: 2 numbers of 9s are present at [5, 9] position
How i have write the program to get the output as "2 numbers of 9s are present at 5 and 9 position".
You can use the list comprehension. enumerate return the element (num) of list and the index of element (idx) and if the element (num) is equal to 9 then the positions list will contain the idx as a string. The indexing starts from 1 (not 0) (start=1). You can get the number of 9s with len(positions) and you can join the element with " and ".join(positions).
Code:
seq = [1, 2, 3, 4, 9, 6, 8, 13, 9, 12, 19]
positions = [str(idx) for idx, num in enumerate(seq, start=1) if num == 9]
print("{} numbers of 9s are present at {} positions".format(len(positions), " and ".join(positions)))
Output:
>>> python3 test.py
2 numbers of 9s are present at 5 and 9 positions
I guess it is the most elegant way to solve your issue.
your_list = [4,5,6,7,9,6,7,9]
indexes = []
for i, num in enumerate(your_list, start=1):
if num == 9:
indexes.append(i)
def output_list(str_list):
last_value = str_list[-1]
first_indexes = str_list[:-1]
first_indexes_str = ", ".join(first_indexes)
if len(first_indexes) == 0:
return f'{len(indexes)} numbers of 9s are present at {last_value} position'
return f'{len(indexes)} numbers of 9s are present at {first_indexes_str} and {last_value} position'
print(output_list(indexes))
first empty list of the index list.
loop over the list and get the index by using (enumerate).
if the num is nine append to the indexes list.
after that print by using the f'...' and print how much find and where.
and add the function that parses the data as wanted
Well, first make a for loop in your list, taking each element, then compare this element to 9 to see if it's a 9 or not. compt variable will count "9" encounters.
compt = 0
seq = [1, 2, 3, 4, 9, 6, 8, 13, 9, 12, 19]
for k in seq:
if(k == 9):
compt += 1
print(compt)
If you also want the index / position of nines founds, then do the same, looping on index, and keep an index list :
compt = 0
indexlist = []
seq = [1, 2, 3, 4, 9, 6, 8, 13, 9, 12, 19]
for k in range(0, len(seq)):
if(seq[k] == 9):
compt += 1
indexlist.append(k)
print(str(compt) + " nines founds, in following positions : " + str(indexlist))
You can use locate present in more_itertools module
from more_itertools import locate
arr = [1, 2, 3, 4, 9, 6, 8, 13, 9, 12, 19]
indices = list(locate(arr, lambda a: a==9))
You can add +1 to each element in the indices to get the position
seq = [1, 2, 3, 4, 9, 6, 8, 13, 9, 12, 19]
positions = [i + 1 for i in range(len(seq)) if seq[i] == 9]
print(f"{len(positions)} numbers of 9s are present at {positions} position")
So I have a list li = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13] and I only want to print out elements that are a part of an arithmetic sequence 6n - 5 (1st, 7th and 13th).
How can I do that if I have a list with n elements?
You could simply say
print([x for x in li if x % 6 == 1])
or, alternatively, if you just want the sequence and don't want to bother about creating li in the first place,
print([6*n-5 for n in range(1, (13+5)//6+1)])
Use the code:
li = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
new=[]
for i in li:
if int((i+5)/6)==((i+5)/6):
#You can also use
#if ((i+5)/6).is_integer():
new.append(i)
I tried to make it as easy as possible.
Hope it helps :)
li = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
for n in range(1,len(li)+1): #choose n such that it is length of the list since it cant have more values than the number of values in the list.
for i in li:
if (6*n - 5) == i:
print(i)
Hope this helps
Thanks
Michael
From what I understand, you want the elements whose positions are generated by the sequence. Hence, you want elements from an array of length n whose index is from the sequence function 6x-5.
NOTE: I am assuming you are using 1-based indexing, which means, when you say 1st element in your list, you intend to get 1 and not 2.
n = 13
li = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
# generate the sequence till n
seq = []
x = 1
while 6*x-5 <= n:
seq.append(6*x-5)
x += 1
# Print the elements from the sequence:
for idx in seq:
print(li[idx-1])
# If you want to store it in another list:
li2 = [li[idx-1] for idx in seq]
Below is a more generic and efficient way for above code:
n = 13
li = list(range(1, n+1)) # More easy to write
# More efficient way is to create a generator function
def get_seq(n):
x = 1
while 6*x-5 <= n:
yield 6*x-5
x += 1
# Get the generator object
seq = get_seq(n)
# Print the elements from the sequence:
for idx in seq:
print(li[idx-1])
# Want to store it in another list:
seq = get_seq(n) # Don't forget to get a new generator object.
li2 = [li[idx-1] for idx in seq]
Output for both snippets:
1
7
13
Hope the answer helps, and remove confusion for others as well ;)
You can simply generate the sequence for any n.
for example:
n = 10
print([ 6*x - 5 for x in range(1,n)])
output:
[1, 7, 13, 19, 25, 31, 37, 43, 49]
>>> [Finished in 0.2s]
But if you just want to filter your existing list li:
li = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
print([ x for x in li if x % 6 == 1 ])
output:
[1, 7, 13]
>>>
[Finished in 0.3s]
li[1] is 2, li[7] is 8 and the element with the index 13 is out of range.
li = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
for num, i in enumerate(li):
if ((num + 5)/6).is_integer():
print(i)
# 2
# 8
If you want to start with the index 1 add start=1 to the enumerate() function.
for num, i in enumerate(li, start=1):
if ((num + 5)/6).is_integer():
print(i)
# 1
# 7
# 13
I got a list like these:
List1: [1, 5, 9, 1, 5, 9, 15, 21, 29, 1, 5, 9, 15]
I want a new list, which should content the highest number, before it starts again with 1.
List_new: [9, 29, 15]
I tried this:
List_new = []
for i in range(len(List1)):
j = List1[i]
if j + 1 == '1':
List_new += [j]
else:
continue
print(j)
But I got an empty list back.
Simply with built-in only libs:
from itertools import groupby
result = [max(group) for r, group in groupby(your_list, lambda x: x == 1) if not r]
def max_of_sublists(megalist):
maxitem = 0
for item in megalist:
if item == 1 and maxitem:
yield maxitem
maxitem = 0
if maxitem < item:
maxitem = item
yield maxitem
biglist=[1, 5, 9, 1, 5, 9, 15, 21, 29, 1, 5, 9, 15]
print([x for x in max_of_sublists(biglist)])
Your code has a few issues. Here's a version that works.
list1 = [1, 5, 9, 1, 5, 9, 15, 21, 29, 1, 5, 9, 15]
list2 = []
for i in range(len(list1)-1):
if list1[i+1] == 1:
list2.append(list1[i])
list2.append(list1[-1]) # adds the last element
This outputs:
>>> list2
[9, 29, 15]
Here is a simple for loop that will answer your question:
List_new = [List1[0]] # initialize with first element
for i in List1[1:]: # simply iterate over list elements, not indices
if i != 1 and i > List_new[-1]:
List_new[-1] = i # current element is the new maximum
elif i == 1:
List_new.append(i) # encountered a 1, start looking for new maximum
See inline comments for explanations.
This problem can be implemented in a one liner using python modules as in the very elegant solution suggested by Andrey. However, if you would like to follow on the logic, check out this solution.
def max_values_between_ones(numbers):
max_values = []
max_value = None
for i in range(len(numbers)):
if numbers[i] == 1:
if max_value != None:
max_values.append(max_value)
max_value = None
# max_value is None when they were no values != 1 before this 1
else:
if max_value != None:
# this part was missing in your code, to get the max value
# you should be comparing the current value with the max value so far
max_value = max(numbers[i], max_value)
else:
# set max_value to any not 1 value
max_value = numbers[i]
# if the list didn't end with 1, add the last max_value
if max_value != None:
max_values.append(max_value)
return max_values
numbers = [1, 5, 9, 1, 5, 9, 15, 21, 29, 1, 5, 9, 15]
max_values = max_values_between_ones(numbers)
print(max_values)
>> [9, 29, 15]
Like this:
l = [1, 5, 9, 1, 5, 9, 15, 21, 29, 1, 5, 9, 15]
pos = [item for item in range(0, len(l)) if l[item] == 1]
new_list = []
for n in range(len(pos)):
if n != len(pos) - 1:
new_list.append(l[pos[n]:pos[n+1]])
else:
new_list.append(l[pos[n]:])
print map(lambda x: max(x), new_list)
List1 = [1, 5, 9, 1, 5, 9, 15, 21, 29, 1, 5, 9, 15]
maxi = 0
List2 = []
for i in range(0,len(List1)):
if maxi < List1[i]:
maxi = List1[i]
if (i == len(List1)-1 or List1[i] == 1) and maxi > 1:
List2.append(maxi)
maxi = 0
print List2
I am trying to write a program that takes an input number T(number of test cases), and then asks for the numbers N.
This is my code:
T = int(raw_input())
L = [int(raw_input()) for i in range(T)]
L1 = []
for i in range(0,L[i]):
if (i%3 == 0 or i%5 ==0):
L1.append(i)
print L1
Input: 2 10 20
Output: [0, 3, 5, 6, 9, 10, 12, 15, 18]
I would like the output to be of the following format:
[[0, 3, 5, 6, 9], [0, 3, 5, 6, 9, 10, 12, 15, 18]]
Here [0, 3, 5, 6, 9] is the list that has elements with both multiples of 3 and 5 for number 10
[0, 3, 5, 6, 9, 10, 12, 15, 18] is the list that has elements with both multiples of 3 and 5 for number 20
I am new to python. kindly let me know how I should proceed on this.
The following will produce a list of lists containing all the multiples of 3 and 5 that are less than the given number.
L = [10,20]
L1 = []
for i in L:
L2 = [] # initialize a new list
for j in range(i):
if not (j%3 and j%5): # use falsy values and DeMorgan's Law
L2.append(j) # append to this list
if L2: # use this if you don't want to keep empty lists
L1.append(L2)
>>> L1
[[0, 3, 5, 6, 9], [0, 3, 5, 6, 9, 10, 12, 15, 18]]
I think what you want is splitting a list by input values. Hope it helps
num = int(raw_input())
upperBounds= [int(raw_input()) for i in range(num)]
res= []
for upperBound in upperBounds:
res.append([i for i in range(0,upperBound) if not (i % 3 and i % 5)])
output:
2
10
20
[[0, 3, 5, 6, 9], [0, 3, 5, 6, 9, 10, 12, 15, 18]]
This can be easily done by applying appropriate logic:
if the element at index 0 we have to iterate from 0 to that element
else we have to iterate form L[index-1] to L[index]
T = int(raw_input())
L = [int(raw_input()) for i in range(T)]
L1 = []
for j in xrange(len(L)):
temp = []
get = 0 if not j else L[j-1]
# if j==0:
# get = 0
# else:
# get = L[j-1]
for i in range(get, L[j]):
if (i%3 == 0 or i%5 ==0):
temp.append(i)
L1.append(temp)
print L1
>>> [[0, 3, 5, 6, 9], [10, 12, 15, 18]]
Or a more Pythonic and compacted version may look like:
T = int(raw_input())
L = [int(raw_input()) for i in range(T)]
L1 = []
for j in xrange(len(L)):
get = 0 if not j else L[j-1]
L1.append([i for i in range(get, L[j]) if (i%3 == 0 or i%5 ==0)])
print L1
You can simply generate a list of multiples with range(l,u,s) with l the lower bounds, u the upper bound and d the step.
Now if we want to generate multiples of i for a given range, we can use the following function:
def multiples(factor, lower, upper) :
return set(range(lower+(factor-lower)%factor,upper,factor))
We thus manipulate the lower bound as lower+(factor-lower)%factor in order to search - in constant time - the first multiple that is greater than or equal to lower.
Next we need to multiples of 3 and 5:
def multiples35(lower, upper):
return sorted(list(multiples(3,lower,upper)|multiples(5,lower,upper)))
Now we only need to iterate over the list of values and generate the list of multiples for each two numbers:
def func(B):
return [multiples35(0,upper) for upper in B]
Or as full code:
import sets
def multiples(factor, lower, upper) :
return set(range(lower+(factor-lower)%factor,upper,factor))
def multiples35(lower, upper):
return sorted(list(multiples(3,lower,upper)|multiples(5,lower,upper)))
def func(B):
return [multiples35(0,upper) for upper in B]
The main function reads then:
T = int(raw_input())
B = [int(raw_input()) for i in range(T)]
print func(B)
I'm trying to run selection sort in python, this is the code that I'm using
def main(list):
input_array = [12, 9, 13, 7, 3, 19, 6, 5]
output_array = selection_sort(input_array)
print(output_array)
def selection_sort(param):
for i in range(0, (len(param) - 1)):
min = i
for j in range(i + 1, len(param)):
if param[min] < param[j]:
min = j
if min != i:
temp = param[i]
param[i] = param[min]
param[min] = temp
return param
The output that I get is
Process finished with exit code 0
I'm using PyCharm as the idea, if that's of any consequence.
input_array should be a list, you are making it a set
input_array = [12, 9, 13, 7, 3, 19, 6, 5]
don't use the variable name list, it is the name for the built-in list
don't use min as a variable name, it is the name for the built-in function min
you don't need an argument for your main method here
you are not calling your main method, call it after the definition of selection_sort
change the line
minimum, i, j, temp = 0
to
minimum, i, j, temp = 0, 0, 0, 0
This will sort in ascending order:
def selection_sort(my_list):
for j in range(len(my_list)):
for i in range(j, len(my_list)):
if my_list[i] < my_list[j]:
my_list[j], my_list[i] = my_list[i], my_list[j]
return my_list
def main():
input_array = [12, 9, 13, 7, 3, 19, 6, 5]
output_array = selection_sort(input_array)
print(output_array)
[3, 5, 6, 7, 9, 12, 13, 19]