So basically I'm trying to check if a bunch of strings in a list called list9000 contain an "#" sign. What I want is to empty the list once it has 6 elements, but before clearing it to do a for loop checking the elements for any "#" signs. I've tried using del and other emptying techniques, but it just doesn't seem to work. Here's my work so far:
if( len(list9000) == 6):
# print(list9000)
i = list9000.count("#")
if(i>1):
amount9000 = amount9000 - i + 1
numWrong = numWrong - i + 1
list9000[:] = []
list9000.append(line)
This is just a snippet of my code. There are about 300 lines of other code. I am reading a text file, in which I add the lines of text in the file to my list. If I could solve this problem, I would be basically done with my project!
Edit: I've tried using del list9000[:], but it doesn't work.
**Update: ** I have printed out the length of the list, and it doesn't seem to be 6 most of the time, but rather increased by 6 every time.
You can just reassign list9000 to a new empty list object. But be careful, Python creates reference if you are assigning mutable objects. A simple list9000[:] = [] should do the job to clear your list.
if( len(list9000) == 6):
# print(list9000)
i = 0
for item in list9000:
if("#" in item):
i += i + 1
if(i>1):
amount9000 = amount9000 - i + 1
numWrong = numWrong - i + 1
list9000[:] = []
Related
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)
I'm trying to create variables within these loops but python gives me a syntax error. tme and empty both give me this problem. I'm trying to get python to read an excel sheet and put the values in a list. The variables I'm trying to create are supposed to help me find the empty spaces in python and remove the value from another list I created so that the values I obtain correspond to the correct input and the inputs that have no value in excel are removed from that list. It gives an index error too if there's a fix for that I'd appreciate it, but I can figure that out if the variables work correctly.
span = list()
print(span)
price = list() #creates a list with all the values before giving index error
for q in range(1,chart.nrows):
for i in range (1,chart.ncol):
L = (chart.cell_value(q,i))
if L != '': #all values that exist will be appended to the list
price.append(L)
else: #find x and y parts of space, have that popped from span list
moth = chart.cell_value(q,0)
m = str(moth)
y = year = str(chart.cell_value(0,i)
tme = m + ' ' + y
empty = span.index(tme)
span.pop(empty)
Just from a syntax point of view you were missing a closing bracket in
y = year = str(chart.cell_value(0,i))
I am writing a simple secret santa script that selects a "GiftReceiver" and a "GiftGiver" from a list. Two lists and an empty dataframe to be populated are produced:
import pandas as pd
import random
santaslist_receivers = ['Rudolf',
'Blitzen',
'Prancer',
'Dasher',
'Vixen',
'Comet'
]
santaslist_givers = santaslist_receivers
finalDataFrame = pd.DataFrame(columns = ['GiftGiver','GiftReceiver'])
I then have a while loop that selects random elements from each list to pick a gift giver and receiver, then remove from the respective list:
while len(santaslist_receivers) > 0:
print (len(santaslist_receivers)) #Used for testing.
gift_receiver = random.choice(santaslist_receivers)
santaslist_receivers.remove(gift_receiver)
print (len(santaslist_receivers)) #Used for testing.
gift_giver = random.choice(santaslist_givers)
while gift_giver == gift_receiver: #While loop ensures that gift_giver != gift_receiver
gift_giver = random.choice(santaslist_givers)
santaslist_givers.remove(gift_giver)
dummyDF = pd.DataFrame({'GiftGiver':gift_giver,'GiftReceiver':gift_receiver}, index = [0])
finalDataFrame = finalDataFrame.append(dummyDF)
The final dataframe only contains three elements instead of six:
print(finalDataframe)
returns
GiftGiver GiftReceiver
0 Dasher Prancer
0 Comet Vixen
0 Rudolf Blitzen
I have inserted two print lines within the while loop to investigate. These print the length of the list santaslist_receivers before and after the removal of an element. The expected return is to see original list length on the first print, then minus 1 on the second print, then the same length again on the first print of the next iteration of the while loop, then so on. Specifically I expect:
6,5,5,4,4,3,3... and so on.
What is returned is
6,5,4,3,2,1
Which is consistent with the DataFrame having only 3 rows, but I do not see the cause of this.
What is the error in my code or my approach?
You can solve it by simply changing this line
santaslist_givers = santaslist_receivers
to
santaslist_givers = list(santaslist_receivers)
Python variables are pointers essentially so they refer to the same list , ie santaslist_givers and santaslist_receivers were accessing the same location in memory in your implementation . To make them different use a list function
And for some extra information , you can refer copy.deepcopy
You should make an explicit copy of your list here
santaslist_givers = santaslist_receivers
there are multiple options for doing this as explained in this question.
In this case I would recommend (if you have Python >= 3.3):
santaslist_givers = santaslist_receivers.copy()
If you are on an older version of Python, the typical way to do it is:
santaslist_givers = santaslist_receivers[:]
I have an if loop in which I am trying to;
(1) Create a dataframe from a filepath.
(2) Format this dataframe
(3) Add that dataframe to a dictionary that is a property of an instance of a class.
Here is my code defining the class and the method:
class myClass:
def __init__(self, name, filepathlist):
self.name = name
self.filepathlist = filepathlist
def formatData(self):
i = 0
self.dataframeDict = {}
if i < (len(self.filepathlist) - 1):
DFRAW = pd.read_csv(self.filepathlist[i], header = 9) #Row 9 is the row that is not blank (all blank auto-skipped)
DFRAW['DateTime'], DFRAW['dummycol1'] = DFRAW[' ;W;W;W;W'].str.split(';', 1).str
DFRAW['Col1'], DFRAW['dummycol2'] = DFRAW['dummycol1'].str.split(';', 1).str
DFRAW['Col2'], DFRAW['dummycol3'] = DFRAW['dummycol2'].str.split(';', 1).str
DFRAW['Col3'], DFRAW['Col4'] = DFRAW['dummycol3'].str.split(';', 1).str
DFRAW = DFRAW.drop([' ;W;W;W;W', 'dummycol1', 'dummycol2', 'dummycol3'], axis = 1)
dictIndex = self.filepathlist[i][39:44]
self.dataframeDict.update({dictIndex: DFRAW})
i = i + 1
Then I create an instance of the class and run the method:
filepathlist = ['filepath1','filepath2']
myINST = myClass('Mydataname', filepathlist)
myINST.formatData()
I then expect myINST.dataframeDict to have two dataframes as per the 2 input filepaths and thus 2 iterations of the if loop. However only 1 is present.
What is the error in my code or my approach?
It is hard to tell whether this will completely solve your problem, because no dummy data is provided. You will, however, get one step closer to your solution if you replace if i < (len(self.filepathlist) - 1): with while i < (len(self.filepathlist) - 1):.
You are currently just checking if i=0 is smaller than len(self.filepathlist)-1. If so, then the if-block is executed once. What you are actually looking for is a loop that keeps on iterating, as long as i is smaller than len(self.filepathlist)-1. This is done with while-loops.
You need to change your condition to for i in range(len(self.filepathlist)):
(Also, remove the assignment of i as the for loop does it automatically. For the same reason, you should also remove the line which increments i).
If you want to use a while loop, change the if line to while i < len(self.filepathlist):.
Notice that there's no -1. This is because you're using < instead of <=. If you want to use -1, then you also need the <= as this will ensure the loop runs the correct number of times.
I need the following code to finish quicker without threads or multiprocessing. If anyone knows of any tricks that would be greatly appreciated. maybe for i in enumerate() or changing the list to a string before calculating, I'm not sure.
For the example below, I have attempted to recreate the variables using a random sequence, however this has rendered some of the conditions inside the loop useless ... which is ok for this example, it just means the 'true' application for the code will take slightly longer.
Currently on my i7, the example below (which will mostly bypass some of its conditions) completes in 1 second, I would like to get this down as much as possible.
import random
import time
import collections
import cProfile
def random_string(length=7):
"""Return a random string of given length"""
return "".join([chr(random.randint(65, 90)) for i in range(length)])
LIST_LEN = 18400
original = [[random_string() for i in range(LIST_LEN)] for j in range(6)]
LIST_LEN = 5
SufxList = [random_string() for i in range(LIST_LEN)]
LIST_LEN = 28
TerminateHook = [random_string() for i in range(LIST_LEN)]
#^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Exclude above from benchmark
ListVar = original[:]
for b in range(len(ListVar)):
for c in range(len(ListVar[b])):
#If its an int ... remove
try:
int(ListVar[b][c].replace(' ', ''))
ListVar[b][c] = ''
except: pass
#if any second sufxList delete
for d in range(len(SufxList)):
if ListVar[b][c].find(SufxList[d]) != -1: ListVar[b][c] = ''
for d in range(len(TerminateHook)):
if ListVar[b][c].find(TerminateHook[d]) != -1: ListVar[b][c] = ''
#remove all '' from list
while '' in ListVar[b]: ListVar[b].remove('')
print(ListVar[b])
ListVar = original[:]
That makes a shallow copy of ListVar, so your changes to the second level lists are going to affect the original also. Are you sure that is what you want? Much better would be to build the new modified list from scratch.
for b in range(len(ListVar)):
for c in range(len(ListVar[b])):
Yuck: whenever possible iterate directly over lists.
#If its an int ... remove
try:
int(ListVar[b][c].replace(' ', ''))
ListVar[b][c] = ''
except: pass
You want to ignore spaces in the middle of numbers? That doesn't sound right. If the numbers can be negative you may want to use the try..except but if they are only positive just use .isdigit().
#if any second sufxList delete
for d in range(len(SufxList)):
if ListVar[b][c].find(SufxList[d]) != -1: ListVar[b][c] = ''
Is that just bad naming? SufxList implies you are looking for suffixes, if so just use .endswith() (and note that you can pass a tuple in to avoid the loop). If you really do want to find the the suffix is anywhere in the string use the in operator.
for d in range(len(TerminateHook)):
if ListVar[b][c].find(TerminateHook[d]) != -1: ListVar[b][c] = ''
Again use the in operator. Also any() is useful here.
#remove all '' from list
while '' in ListVar[b]: ListVar[b].remove('')
and that while is O(n^2) i.e. it will be slow. You could use a list comprehension instead to strip out the blanks, but better just to build clean lists to begin with.
print(ListVar[b])
I think maybe your indentation was wrong on that print.
Putting these suggestions together gives something like:
suffixes = tuple(SufxList)
newListVar = []
for row in original:
newRow = []
newListVar.append(newRow)
for value in row:
if (not value.isdigit() and
not value.endswith(suffixes) and
not any(th in value for th in TerminateHook)):
newRow.append(value)
print(newRow)