Develop Reference

adding rows based on values of other rows

I have a list (in a dataframe) that looks like this:
oddnum = [1, 3, 5, 7, 9, 11, 23]
I want to create a new list that looks like this:
newlist = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 23]
I want to test if the distance between two numbers is 2 (if oddnum[index+1]-oddnum[index] == 2)
If the distance is 2, then I want to add the number following oddnum[index] and create a new list (oddnum[index] + 1)
If the distance is greater than two, keep the list as is
I keep getting key error because (I think) the list runs out of [index] and [index+1] no longer exists once it reaches the end of the list. How do I do this?

To pass errors, the best method is to use try and except conditions. Here's my code:
oddnum = [1, 3, 5, 7, 9, 11, 23]
res = [] # The new list
for i in range(len(oddnum)):
res.append(oddnum[i]) # Append the first value by default
try: # Tries to run the code
if oddnum[i] + 2 == oddnum[i+1]: res.append(oddnum[i]+1) # Appends if the condition is met
except: pass # Passes on exception (in our case KeyError)
print(res)

oddnum = [1, 3, 5, 7, 9, 11, 23]
new_list = []
for pos, num in enumerate(oddnum):
new_list.append(num)
try:
if num-oddnum[pos+1] in [2, -2]:
new_list.append(num+1)
except:
pass
print(new_list)
Use try: except: to prevent exceptions popping up and ignore it

Returning a list of the ordered elements with a conition in while loop structure

I am trying to solve a assignment where are 13 lights and starting from 1, light is turned off at every 5th light, when the count reaches 13, start from 1st item again. The function should return the order of lights turned off. In this case, for a list of 13 items, the return list would be [5, 10, 2, 8, 1, 9, 4, 13, 12, 3, 7, 11, 6]. Also, turned off lights would not count again.
So the way I was going to approach this problem was to have a list named turnedon, which is [1,2,3,4,5,6,7,8,9,10,11,12,13] and an empty list called orderoff and append to this list whenever a light gets turned off in the turnedon list. So while the turnedon is not empty, iterate through the turnedon list and append the light getting turned off and remove that turnedoff light from the turnedon list, if that makes sense. I cannot figure out what should go into the while loop though. Any idea would be really appreciated.
def orderoff():
n=13
turnedon=[]
for n in range(1,n+1):
turnedon.append(n)
orderoff=[]
while turneon !=[]:

This problem is equivalent to the well-known Josephus problem, in which n prisoners stand in a circle, and they are killed in a sequence where each time, the next person to be killed is k steps around the circle from the previous person; the steps are only counted over the remaining prisoners. A sample solution in Python can be found on the Rosetta code website, which I've adapted slightly below:
def josephus(n, k):
p = list(range(1, n+1))
i = 0
seq = []
while p:
i = (i+k-1) % len(p)
seq.append(p.pop(i))
return seq
Example:
>>> josephus(13, 5)
[5, 10, 2, 8, 1, 9, 4, 13, 12, 3, 7, 11, 6]

This works, but the results are different from yours:
>>> pos = 0
>>> result = []
>>> while len(result) < 13 :
... pos += 5
... pos %= 13
... if pos not in result :
... result.append(pos)
...
>>> result = [i+1 for i in result] # make it 1-based, not 0-based
>>> result
[6, 11, 3, 8, 13, 5, 10, 2, 7, 12, 4, 9, 1]
>>>

I think a more optimal solution would be to use a loop, add the displacement each time, and use modules to keep the number in range
def orderoff(lights_num,step):
turnd_off=[]
num =0
for i in range(max):
num =((num+step-1)%lights_num)+1
turnd_off.append(num)
return turnd_off
print(orderoff(13))

Handling lists in Python

Suppose I have a list:
a = [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89]
I want to write a program that prints out all the elements of the list that are less than 10.
Actually its pretty simple I got this program but, I need to do it in a single line and I've no idea how to do that. Need some help with this.

print [x for x in a if x < 10]
Take a further look at lambda functions, I feel this is what you are looking for.

So in order to print something out of a list that is less than 10 In the same line, first you need to create a list:
numbers = []
loop through every single element of the list
for i in a:
And then you need a If statement to check if the element is less than 10
if i < 10:
Append the number to the list
numbers.append(str(i))
Join the results together:
result = " ".join(numbers)
And lastly printing it out
print(result)
And if you combine everything together, this is what you should get:
a = [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89]
numbers = []
for i in a:
if i < 10:
numbers.append(str(i))
result = " ".join(numbers)
print(result)
The result should be:
1 1 2 3 5 8

Dynamically update variables used in a For statement in python

I want to update the list used in the for loop statement as it is being appended with new values within the loop.I know this code wont update the list but I am not able to find a way to do so. I hope to required output might explain it better
INPUT
list=[1,2,3]
count=0
for x in list:
if int(len(list))>10:
break
else:
count=count+1
x=x+10
if x>3:
list=list+[x]
print "list = %r" %list
print "count = %r" %count
OUTPUT
list = [1, 2, 3, 11, 12, 13]
count = 3
Required OUTPUT
list = [1, 2, 3, 11, 12, 13, 21, 22, 23, 31]
count = 10

I suggest you use a while loop instead of a for, which saves the need of the break keyword :
l=[1,2,3]
count = len(l) # There are already len(l) elements in the list
pos = 0
while len(l) < 10:
count += 1
x = l[pos]
l.append(x+10)
pos += 1
print "list = %r" %l
print "count = %r" %count
Which gives the output :
list = [1, 2, 3, 11, 12, 13, 21, 22, 23, 31]
count = 10
Also, you can notice that i renamed the list variable to l, to prevent any confusion between the type list and the variable itself.
I used append to add elements at the end of the list.
Hope it'll be helpful.

two problems here:
first: the if x>3: test. (Why did you include that?)
second when you have a for statement, it loops on the values present initially only.
This should work:
l = [1,2,3]
count = 0
while len(l) < 10:
l.append(l[count]+10)
count += 1

Creating a new array of numbers based on an existing string array

I am trying to create a list of values that correlate to a string by comparing each character of my string to that of my "alpha_list". This is for encoding procedure so that the numerical values can be added later.
I keep getting multiple errors from numerous different ways i have tried to make this happen.
import string
alpha_list = " ABCDEFGHIJKLMNOPQRSTUVWXYZ"
ints = "HELLO WORLD"
myotherlist = []
for idx, val in enumerate(ints):
myotherlist[idx] = alpha_list.index(val)
print(myotherlist)
Right now this is my current error reading
Traceback (most recent call last):
File "C:/Users/Derek/Desktop/Python/test2.py", line 11, in <module>
myotherlist[idx] = alpha_list.index(val)
IndexError: list assignment index out of range
I am pretty new to python so if I am making a ridiculously obvious mistake please feel free to criticize.
The print(myotherlist) output that i am looking for should look something like this:
[8, 5, 12, 12, 15, 0, 23, 15, 18, 12, 4]

Just use append:
for val in ints:
myotherlist.append(alpha_list.index(val))
print(myotherlist)
myotherlist is an empty list so you cannot access using myotherlist[idx] as there is no element 0 etc..
Or just use a list comprehension:
my_other_list = [alpha_list.index(val) for val in ints]
Or a functional approach using map:
map(alpha_list.index,ints))
Both output:
In [7]: [alpha_list.index(val) for val in ints]
Out[7]: [8, 5, 12, 12, 15, 0, 23, 15, 18, 12, 4]
In [8]: map(alpha_list.index,ints)
Out[8]: [8, 5, 12, 12, 15, 0, 23, 15, 18, 12, 4]

import string - don't use that a bunch of books say its better to use the built in str
myotherlist[idx] = alpha_list.index(val) is why you are getting the error. This is saying 'Go to idx index and put alpha_list.index(val) there, but since the list is empty it cannot do that.
So if you replace
for idx, val in enumerate(ints):
myotherlist[idx] = alpha_list.index(val)
with
for letter in ints: #iterates over the 'HELLO WORLD' string
index_to_append = alpha_list.index(letter)
myotherlist.append(index_to_append)
you will get the expected result!
If there is something not clear please let me know!