We Keep Coding

Extract words from random strings

Below I have some strings in a list:
some_list = ['a','l','p','p','l','l','i','i','r',i','r','a','a']
Now I want to take the word april from this list. There are only two april in this list. So I want to take that two april from this list and append them to another extract list.
So the extract list should look something like this:
extract = ['aprilapril']
or
extract = ['a','p','r','i','l','a','p','r','i','l']
I tried many times trying to get the everything in extract in order, but I still can't seems to get it.
But I know I can just do this
a_count = some_list.count('a')
p_count = some_list.count('p')
r_count = some_list.count('r')
i_count = some_list.count('i')
l_count = some_list.count('l')
total_count = [a_count,p_count,r_count,i_count,l_count]
smallest_count = min(total_count)
extract = ['april' * smallest_count]
Which I wouldn't be here If I just use the code above.
Because I made some rules for solving this problem
Each of the characters (a,p,r,i and l) are some magical code elements, these code elements can't be created out of thin air; they are some unique code elements, that has some uniquw identifier, like a secrete number that is associated with them. So you don't know how to create this magical code elements, the only way to get the code elements is to extract them to a list.
Each of the characters (a,p,r,i and l) must be in order. Imagine they are some kind of chains, they will only work if they are together. Meaning that we got to put p next to and in front of a, and l must come last.
These important code elements are some kind of top secrete stuff, so if you want to get it, the only way is to extract them to a list.
Below are some examples of a incorrect way to do this: (breaking the rules)
import re
word = 'april'
some_list = ['aaaaaaappppppprrrrrriiiiiilll']
regex = "".join(f"({c}+)" for c in word)
match = re.match(regex, text)
if match:
lowest_amount = min(len(g) for g in match.groups())
print(word * lowest_amount)
else:
print("no match")
from collections import Counter
def count_recurrence(kernel, string):
# we need to count both strings
kernel_counter = Counter(kernel)
string_counter = Counter(string)
effective_counter = {
k: int(string_counter.get(k, 0)/v)
for k, v in kernel_counter.items()
}
min_recurring_count = min(effective_counter.values())
return kernel * min_recurring_count
This might sounds really stupid, but this is actually a hard problem (well for me). I originally designed this problem for myself to practice python, but it turns out to be way harder than I thought. I just want to see how other people solve this problem.
If anyone out there know how to solve this ridiculous problem, please help me out, I am just a fourteen-year-old trying to do python. Thank you very much.

I'm not sure what do you mean by "cannot copy nor delete the magical codes" - if you want to put them in your output list you will need to "copy" them somehow.
And btw your example code (a_count = some_list.count('a') etc) won't work since count will always return zero.
That said, a possible solution is
worklist = [c for c in some_list[0]]
extract = []
fail = False
while not fail:
lastpos = -1
tempextract = []
for magic in magics:
if magic in worklist:
pos = worklist.index(magic, lastpos+1)
tempextract.append(worklist.pop(pos))
lastpos = pos-1
else:
fail = True
break
else:
extract.append(tempextract)
Alternatively, if you don't want to pop the elements when you find them, you may compute the positions of all the occurences of the first element (the "a"), and set lastpos to each of those positions at the beginning of each iteration

May not be the most efficient way, although code works and is more explicit to understand the program logic:
some_list = ['aaaaaaappppppprrrrrriiiiiilll']
word = 'april'
extract = []
remove = []
string = some_list[0]
for x in range(len(some_list[0])//len(word)): #maximum number of times `word` can appear in `some_list[0]`
pointer = i = 0
while i<len(word):
j=0
while j<(len(string)-pointer):
if string[pointer:][j] == word[i]:
extract.append(word[i])
remove.append(pointer+j)
i+=1
pointer = j+1
break
j+=1
if i==len(word):
for r_i,r in enumerate(remove):
string = string[:r-r_i] + string[r-r_i+1:]
remove = []
elif j==(len(string)-pointer):
break
print(extract,string)

Extracting multiple data from a single list

I working on a text file that contains multiple information. I converted it into a list in python and right now I'm trying to separate the different data into different lists. The data is presented as following:
CODE/ DESCRIPTION/ Unity/ Value1/ Value2/ Value3/ Value4 and then repeat, an example would be:
P03133 Auxiliar helper un 203.02 417.54 437.22 675.80
My approach to it until now has been:
Creating lists to storage each information:
codes = []
description = []
unity = []
cost = []
Through loops finding a code, based on the code's structure, and using the code's index as base to find the remaining values.
Finding a code's easy, it's a distinct type of information amongst the other data.
For the remaining values I made a loop to find the next value that is numeric after a code. That way I can delimitate the rest of the indexes:
The unity would be the code's index + index until isnumeric - 1, hence it's the first information prior to the first numeric value in each line.
The cost would be the code's index + index until isnumeric + 2, the third value is the only one I need to store.
The description is a little harder, the number of elements that compose it varies across the list. So I used slicing starting at code's index + 1 and ending at index until isnumeric - 2.
for i, carc in enumerate(txtl):
if carc[0] == "P" and carc[1].isnumeric():
codes.append(carc)
j = 0
while not txtl[i+j].isnumeric():
j = j + 1
description.append(" ".join(txtl[i+1:i+j-2]))
unity.append(txtl[i+j-1])
cost.append(txtl[i+j])
I'm facing some problems with this approach, although there will always be more elements to the list after a code I'm getting the error:
while not txtl[i+j].isnumeric():
txtl[i+j] list index out of range.
Accepting any solution to debug my code or even new solutions to problem.
OBS: I'm also going to have to do this to a really similar data font, but the code would be just a sequence of 7 numbers, thus harder to find amongst the other data. Any solution that includes this facet is also appreciated!

A slight addition to your code should resolve this:
while i+j < len(txtl) and not txtl[i+j].isnumeric():
j += 1
The first condition fails when out of bounds, so the second one doesn't get checked.
Also, please use a list of dict items instead of 4 different lists, fe:
thelist = []
thelist.append({'codes': 69, 'description': 'random text', 'unity': 'whatever', 'cost': 'your life'})
In this way you always have the correct values together in the list, and you don't need to keep track of where you are with indexes or other black magic...
EDIT after comment interactions:
Ok, so in this case you split the line you are processing on the space character, and then process the words in the line.
from pprint import pprint # just for pretty printing
textl = 'P03133 Auxiliar helper un 203.02 417.54 437.22 675.80'
the_list = []
def handle_line(textl: str):
description = ''
unity = None
values = []
for word in textl.split()[1:]:
# it splits on space characters by default
# you can ignore the first item in the list, as this will always be the code
# str.isnumeric() doesn't work with floats, only integers. See https://stackoverflow.com/a/23639915/9267296
if not word.replace(',', '').replace('.', '').isnumeric():
if len(description) == 0:
description = word
else:
description = f'{description} {word}' # I like f-strings
elif not unity:
# if unity is still None, that means it has not been set yet
unity = word
else:
values.append(word)
return {'code': textl.split()[0], 'description': description, 'unity': unity, 'values': values}
the_list.append(handle_line(textl))
pprint(the_list)
str.isnumeric() doesn't work with floats, only integers. See https://stackoverflow.com/a/23639915/9267296

Why I have IndexError when program makes succesful first step?

I tried to make sorter that deletes duplicates of IP's in first list and saves it into a file, but after first succesful round it gives me IndexError: list index out of range.
I've expected normal sorting process, but it doesn't works
Code:
ip1 = open('hosts', 'r')
ip2 = open('rotten', 'r')
ipList1 = [line.strip().split('\n') for line in ip1]
ipList2 = [line.strip().split('\n') for line in ip2]
for i in range(len(ipList1)):
for a in range(len(ipList2)):
if(ipList1[i] == ipList2[a]):
print('match')
del(ipList1[i])
del(ipList2[a])
i -= 1
a -= 1
c = open('end', 'w')
for d in range(len(ipList1)):
c.write(str(ipList1[d]) + '\n')
c.close()

You're deleting from the list while iterating over it, that's why you're getting an IndexError.
This could be easier done with sets:
with open('hosts') as ip1, open('rotten') as ip2:
ipList1 = set(line.strip().split('\n') for line in ip1)
ipList2 = set(line.strip().split('\n') for line in ip2)
good = ipList1 - ipList2
with open('end', 'w') as c:
for d in good:
c.write(d + '\n')

You changed lists in a fly. For expression gets a list with, as an example, of 5 elements length, after the first iteration you remove 4, so in the second iteration for tried to extract the second element but now it does not exist.
If necessary save the ordering you can use generator expression:
ips = [ip for ip in ipList1 if ip not in set(list2)]
If doesn't, just use sets expression.

You should never modify a list that you are currently iterating over.
A fix would be just to make a third list that saves the non duplicates. Another way would be to just use sets and subtract them from each other although I do know if you like duplicates in one list itself. Also, the way you are doing it right now a duplicate is only found if its at the same index.
ip2 = open('rotten', 'r')
ipList1 = [line.strip().replace('\n', '') for line in ip1]
ipList2 = [line.strip().replace('\n', '') for line in ip2]
ip1.close()
ip2.close()
newlist = []
for v in ip1:
if v not in ip2:
newlist.append(v)
c = open('end', 'w')
c.write('\n'.join(newlist))
c.close()

Other answers focus on deleting from a container while iterating over it. While that’s generally a bad idea, it’s not the crux of the problem here because you have (unpythonically) set up the for loops to use a sequence of indices, and so you aren’t strictly speaking iterating over the lists themselves anyway.
No, the problem here is that i-=1 and a-=1 have no effect: when a for loop begins a new iteration, it doesn’t work off of the previous value of the index. It just takes the next value that it was always destined to take, from the iterator that you established at the beginning (in your case, the output of range())

Assign variable if list index out of range python error

How to pass a string to a variable if an index error is found? Consider the code:
for l1, l2 in zip(open('file1.list'), open ('file2.list')):
a=fasta1[int(l1)]
b=fasta2[int(l2)]
alignments = pairwise2.align.globalxx(a,b)
top_aln = alignments[0]
aln_a, aln_b, score, begin, end = top_aln
print aln_a+'\n'+aln_b
outfast1 = aln_a
outfast2 = aln_b
A number of these functions must be imported (pairwise2 align),
but the file.lists are single column text files with one sequence id (text and numbers) per line, that are used to extract from the fasta1 and fasta2 text files.
Basically, I want to try: each list command ( a=fasta1[int(l1)]) and if there is no error (the id is in range), do as normal (assign variables a and b for that iteration), but if NOT, assign the 'a' variable some placeholder text like 'GGG':
for l1, l2 in zip(open('file1.list'), open ('file2.list')):
try:
a=fasta1[int(l1)]
except IndexError,e:
a="GGG"
continue
try:
b=fasta2[int(l2)]
except (IndexError):
b="CCC"
continue
This code doesn't quite work (when integrated with above code), which isn't surprising given my lack of python prowess, but I don't quite know why. I actually get no text output, despite the print calls... Am I thinking about this right? If there is NO error in the index, I just want it to go on and do the pairwise alignment (with the first a and b variables) and then print some text to stdout.
Any ideas?

Python's conditional (aka ternary) expressions can one-line this for you. They're often criticized for lack of readability, but I think this example reads well enough.
a = fasta1[int(l1)] if int(l1) < len(fasta1) else "GGG"

You don't need continue, because it will skip that iteration of the loop. Consider the following:
for l1, l2 in zip(open('file1.list'), open ('file2.list')):
a = 'GGG'
b = 'CCC'
try:
a = fasta1[int(l1)]
b = fasta2[int(l2)]
except IndexError:
pass

Transposition Cipher in Python

Im currently trying to code a transposition cipher in python. however i have reached a point where im stuck.
my code:
key = "german"
length = len(key)
plaintext = "if your happy and you know it clap your hands, clap your hands"
Formatted = "".join(plaintext.split()).replace(",","")
split = split_text(formatted,length)
def split_text(formatted,length):
return [formatted[i:i + length] for i in range(0, len(formatted), length)]
def encrypt():
i use that to count the length of the string, i then use the length to determine how many columns to create within the program. So it would create this:
GERMAN
IFYOUR
HAPPYA
NDYOUK
NOWITC
LAPYOU
RHANDS
CLAPYO
URHAND
S
this is know where im stuck. as i want to get the program to create a string by combining the columns together. so it would combine each column to create:
IHNNLRCUSFADOAHLRYPYWPAAH .....
i know i would need a loop of some sort but unsure how i would tell the program to create such a string.
thanks

you can use slices of the string to get each letter of the string in steps of 6 (length)
print(formatted[0::length])
#output:
ihnnlrcus
Then just loop through all the possible start indices in range(length) and link them all together:
def encrypt(formatted,length):
return "".join([formatted[i::length] for i in range(length)])
note that this doesn't actually use split_text, it would take formatted directly:
print(encrypt(formatted,length))
the problem with using the split_text you then cannot make use of tools like zip since they stop when the first iterator stops (so because the last group only has one character in it you only get the one group from zip(*split))
for i in zip("stuff that is important","a"):
print(i)
#output:
("s","a")
#nothing else, since one of the iterators finished.
In order to use something like that you would have to redefine the way zip works by allowing some of the iterators to finish and continue until all of them are done:
def myzip(*iterators):
iterators = tuple(iter(it) for it in iterators)
while True: #broken when none of iterators still have items in them
group = []
for it in iterators:
try:
group.append(next(it))
except StopIteration:
pass
if group:
yield group
else:
return #none of the iterators still had items in them
then you can use this to process the split up data like this:
encrypted_data = ''.join(''.join(x) for x in myzip(*split))