Develop Reference

How can I convert a result into a list of variables that I can use as an input?

I was able to come up with these two parts, but I'm having trouble linking them.
Part 1 - This accepts a filter which is listed as 'project = status = blocked'. This will list all issue codes that match the filter and separate them line by line. Is it necessary to convert the results into a list? I'm also wondering if it converts the entire result into one massive string or if each line is a string.
issues_in_project = jira.search_issues(
'project = status = Blocked'
)
issueList = list(issues_in_project)
search_results = '\n'.join(map(str, issueList))
print(search_results)
Part 2 - Right now, the jira.issue will only accept an issue code one at a time. I would like to use the list generated from Part 1 to keep running the code below for each and every issue code in the result. I'm having trouble linking these two parts.
issue = jira.issue(##Issue Code goes here##)
print(issue.fields.project.name)
print(issue.fields.summary + " - " + issue.fields.status.statusCategory.name)
print("Description: " + issue.fields.description)
print("Reporter: " + issue.fields.reporter.displayName)
print("Created on: " + issue.fields.created)

Part 1
'project = status = Blocked' is not a valid JQL. So first of all, you will not get a valid result from calling jira.search_issues('project = status = Blocked').
The result of jira.search_issues() is basically a list of jira.resources.Issue objects and not a list of string or lines of string. To be correct, I should say the result of jira.search_issues() is of type jira.client.ResultList, which is a subclass of python's list.
Part 2
You already have all the required data in issues_in_project if your JQL is correct. Therefore, you can loop through the list and use the relevant information of each JIRA issue. For your information, jira.issue() returns exactly one jira.resources.Issue object (if the issue key exists).
Example
... # initialize jira
issues_in_project = jira.search_issues('status = Blocked')
for issue in issues_in_project:
print(issue.key)
print(issue.fields.summary)

Deleting/Removing element from a list when comparing to another list Python

So I have a good one. I'm trying to build two lists (ku_coins and bin_coins) of crypto tickers from two different exchanges, but I don't want to double up, so if it appears on both exchanges I want to remove it from ku_coins.
A slight complication occurs as Kucoin symbols come in as AION-BTC, while Binance symbols come in as AIONBTC, but it's no problem.
So firstly, I create the two lists of symbols, which runs fine, no problem. What I then try and do is loop through the Kucoin symbols and convert them to the Binance style symbol, so AIONBTC instead of AION-BTC. Then if it appears in the Binance list I want to remove it from the Kucoin list. However, it appears to randomly refuse to remove a handful of symbols that match the requirement. For example AION.
It removes the majority of doubled up symbols but in AIONs case for example it just won't delete it.
If I just do print(i) after this loop:
for i in ku_coins:
if str(i[:-4] + 'BTC') in bin_coins:
It will happily print AION-BTC as one of the symbols, as it fits the requirement perfectly. However, when I stick the ku_coins.remove(i) command in before printing, it suddenly decideds not to print AION suggesting it doesn't match the requirements. And it's doing my head in. Obviously the remove command is causing the problem, but I can't for the life of me figure out why. Any help really appreciated.
import requests
import json
ku_dict = json.loads(requests.get('https://api.kucoin.com/api/v1/market/allTickers').text)
ku_syms = ku_dict['data']['ticker']
ku_coins = []
for x in range(0, len(ku_syms)):
if ku_syms[x]['symbol'][-3:] == 'BTC':
ku_coins.append(ku_syms[x]['symbol'])
bin_syms = json.loads(requests.get('https://www.binance.com/api/v3/ticker/bookTicker').text)
bin_coins = []
for i in bin_syms:
if i['symbol'][-3:] == 'BTC':
bin_coins.append(i['symbol'])
ku_coins.sort()
bin_coins.sort()
for i in ku_coins:
if str(i[:-4] + 'BTC') in bin_coins:
ku_coins.remove(i)

#top bantz, #Fourier has already mentioned that you shouldn't modify a list you're iterating over. What you can do in this case is to create a copy of ku_coins first then iterate over that, and then remove the element from the original ku_coins that matches your if condition. See below:
ku_coins.sort()
bin_coins.sort()
# Create a copy
ku_coins_ = ku_coins[:]
# Then iterate over that copy
for i in ku_coins_:
if str(i[:-4] + 'BTC') in bin_coins:
ku_coins.remove(i)

How about modifying the code to:
while ku_coins:
i = ku_coins.pop()
if str(i[:-4] + 'BTC') in bin_coins:
pass
else:
# do something
the pop() method removes i from the ku_coins list
pop()

How to do this recursion method to print out every 3 letters from a string on a separate line?

I'm making a method that takes a string, and it outputs parts of the strings on separate line according to a window.
For example:
I want to output every 3 letters of my string on separate line.
Input : "Advantage"
Output:
Adv
ant
age
Input2: "23141515"
Output:
231
141
515
My code:
def print_method(input):
mywindow = 3
start_index = input[0]
if(start_index == input[len(input)-1]):
exit()
print(input[1:mywindow])
printmethod(input[mywindow:])
However I get a runtime error.... Can someone help?

I think this is what you're trying to get. Here's what I changed:
Renamed input to input_str. input is a keyword in Python, so it's not good to use for a variable name.
Added the missing _ in the recursive call to print_method
Print from 0:mywindow instead of 1:mywindow (which would skip the first character). When you start at 0, you can also just say :mywindow to get the same result.
Change the exit statement (was that sys.exit?) to be a return instead (probably what is wanted) and change the if condition to be to return once an empty string is given as the input. The last string printed might not be of length 3; if you want this, you could use instead if len(input_str) < 3: return
def print_method(input_str):
mywindow = 3
if not input_str: # or you could do if len(input_str) == 0
return
print(input_str[:mywindow])
print_method(input_str[mywindow:])

edit sry missed the title: if that is not a learning example for recursion you shouldn't use recursion cause it is less efficient and slices the list more often.
def chunked_print (string,window=3):
for i in range(0,len(string) // window + 1): print(string[i*window:(i+1)*window])
This will work if the window size doesn't divide the string length, but print an empty line if it does. You can modify that according to your needs

Compare two lists in python and obtain non-equality

This piece of code in theory have to compare two lists which have the ID of a tweet, and in this comparison if it already exists in screen printing , otherwise not.
But I print all or not being listed.
Any suggestions to compare these two lists of ID's and if not the ID of the first list in the second then print it ?
Sorry for the little efficient code . ( and my English )
What I seek is actually not do RT ( retweet ) repeatedly when I already have . I use Tweepy library , I read the timeline , and make the tweet RT I did not do RT
def analizarRT():
timeline = []
temp = []
RT = []
fileRT = openFile('rt.txt')
for status in api.user_timeline('cnn', count='6'):
timeline.append(status)
for i in range(6):
temp.append(timeline[i].id)
for a in range(6):
for b in range(6):
if str(temp[a]) == fileRT[b]:
pass
else:
RT.append(temp[a])
for i in RT:
print i
Solved add this function !
def estaElemento(tweetId, arreglo):
encontrado = False
for a in range(len(arreglo)):
if str(tweetId) == arreglo[a].strip():
encontrado = True
break
return encontrado

Its a simple program, don't complicate it. As per your comments, there are two lists:)
1. timeline
2. fileRT
Now, you want to compare the id's in both these lists. Before you do that, you must know the nature of these two lists.
I mean, what is the type of data in the lists?
Is it
list of strings? or
list of objects? or
list of integers?
So, find out that, debug it, or use print statements in your code. Or please add these details in your question. So, you can give a perfect answer.
Mean while, try this:
if timeline.id == fileRT.id should work.
Edited:
def analizarRT():
timeline = []
fileRT = openFile('rt.txt')
for status in api.user_timeline('cnn', count='6'):
timeline.append(status)
for i in range(6):
for b in range(6):
if timeline[i].id == fileRT[b].id:
pass
else:
newlist.append(timeline[i].id)
print newlist
As per your question, you want to obtain them, right?. I have appended them in a newlist. Now you can say print newlist to see the items

your else statement is associated with the for statement, you probably need to add one more indent to make it work on the if statement.

Serial Key Generation and Validation

I'm toying around with writing creating a serial code generator/validator, but I can't seem to get how to do a proper check.
Here's my generator code:
# Serial generator
# Create sequences from which random.choice can choose
Sequence_A = 'ABCDEF'
Sequence_B = 'UVWQYZ'
Sequence_C = 'NOPQRS'
Sequence_D = 'MARTIN'
import random
# Generate a series of random numbers and Letters to later concatenate into a pass code
First = str(random.randint(1,5))
Second = str(random.choice(Sequence_A))
Third = str(random.randint(6,9))
Fourth = str(random.choice(Sequence_B))
Fifth = str(random.randint(0,2))
Sixth = str(random.choice(Sequence_C))
Seventh = str(random.randint(7,8))
Eighth = str(random.choice(Sequence_D))
Ninth = str(random.randint(3,5))
serial = First+Second+Third+Fourth+Fifth+Sixth+Seventh+Eighth+Ninth
print serial
I'd like to make a universal check so that my validation code will accept any key generated by this.
My intuition was to create checks like this:
serial_check = raw_input("Please enter your serial code: ")
# create a control object for while loop
control = True
# Break up user input into list that can be analyzed individually
serial_list = list(serial_check)
while control:
if serial_list[0] == range(1,5):
pass
elif serial_list[0] != range(1,5):
control = False
if serial_list[1] == random.choice('ABCDEF'):
pass
elif serial_list[1] != random.choice('ABCDEF'):
control = False
# and so on until the final, where, if valid, I would print that the key is valid.
if control == False:
print "Invalid Serial Code"
I'm well aware that the second type of check won't work at all, but it's a place holder because I've got no idea how to check that.
But I thought the method for checking numbers would work, but it doesn't either.

The expression `range(1, 5)' creates a list of numbers from 1 to 4. So in your first test, you're asking whether the first character in your serial number is equal to that list:
"1" == [1, 2, 3, 4]
Probably not...
What you probably want to know is whether a digit is in the range (i.e. from 1 to 5, I assume, not 1 to 4).
Your other hurdle is that the first character of the serial is a string, not an integer, so you would want to take the int() of the first character. But that will raise an exception if it's not a digit. So you must first test to make sure it's a digit:
if serial_list[0].isdigit() and int(serial_list[0]) in range(1, 6):
Don't worry, if it's not a digit, Python won't even try to evaluate the part after and. This is called short-circuiting.
However, I would not recommend doing it this way. Instead, simply check to make sure it is at least "1" and no more than "5", like this:
if "1" <= serial_list <= "5":
You can do the same thing with each of your tests, varying only what you're checking.
Also, you don't need to convert the serial number to a list. serial_check is a string and accessing strings by index is perfectly acceptable.
And finally, there's this pattern going on in your code:
if thing == other:
pass
elif thing != other:
(do something)
First, because the conditions you are testing are logical opposites, you don't need elif thing != other -- you can just say else, which means "whatever wasn't matched by any if condition."
if thing == other:
pass
else:
(do something)
But if you're just going to pass when the condition is met, why not just test the opposite condition to begin with? You clearly know how to write it 'cause you were putting it in the elif. Put it right in the if instead!
if thing != other:
(do something)
Yes, each of your if statements can easily be cut in half. In the example I gave you for checking the character range, probably the easiest way to do it is using not:
if not ("1" <= serial_list <= "5"):

Regarding your python, I'm guessing that when your wrote this:
if serial_list[0] == range(1,5):
You probably meant this:
if 1 <= serial_list[0] <= 5:
And when you wrote this:
if serial_list[1] == random.choice('ABCDEF'):
You probably meant this:
if serial_list[1] in 'ABCDEF':
There are various other problems with your code, but I'm sure you'll improve it as you learn python.
At a higher level, you seem to be trying to build something like a software activation code generator/validator. You should know that just generating a string of pseudo-random characters and later checking that each is in range is an extremely weak form of validation. If you want to prevent forgeries, I would suggest learning about HMAC (if you're validating on a secure server) or public key cryptography (if you're validating on a user's computer) and incorporating that into your design. There are libraries available for python that can handle either approach.