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I have a larg list of elements:
list= [a1, b, c, b, c, d, a2, c,...a3........]
And i want to remove a specific elements from it a1, a2, a3
suppose that i can get the indexes of the elements start with a
a_indexes = [0,6, ...]
Now, i want to remove most of these elements start with a a but not all of them, i want to keep 20 of them chosen arbitrary. How can i do so ?
I know that to remove an elements from a list list_ can use:
list_.remove(list[element position])
But i am not sure how to play with the a list.
Here's a an approach that will work if I understand the question correctly.
We have a list containing numerous items. We want to remove some elements that match a certain criterion - but not all.
So:
from random import sample
li = ['a','b','a','b','a','b','a']
dc = 'a'
keep = 1 # this is how many we want to keep in the list
if (k := li.count(dc) - keep) > 0: # sanity check
il = [i for i, v in enumerate(li) if v == dc]
for i in sorted(sample(il, k), reverse=True):
li.pop(i)
print(li)
Note how the sample is sorted. This is important because we're popping elements. If we do that in no particular order then we could end up removing the wrong elements.
An example of output might be:
['b', 'b', 'a', 'b']
Suppose you have this list:
li=['d', 'a', 'c', 'a', 'g', 'b', 'f', 'a', 'c', 'g', 'e', 'f', 'e', 'g', 'b', 'b', 'c', 'e', 'a', 'd', 'g', 'd', 'd', 'a', 'c', 'e', 'a', 'c', 'f', 'a', 'b', 'a', 'a', 'f', 'b', 'd', 'd', 'b', 'f', 'a', 'd', 'g', 'd', 'b', 'e']
You can define a character to delete and a count k of how many to delete:
delete='a'
k=3
Then use random.shuffle to generate a random group of k indices to delete:
idx=[i for i,c in enumerate(li) if c==delete]
random.shuffle(idx)
idx=idx[:k]
>>> idx
[3, 7, 31]
Then delete those indices:
new_li=[e for i,e in enumerate(li) if i not in idx]
For example the original list:
['k','a','b','c','a','d','e','a','b','e','f','j','a','c','a','b']
We want to split the list into lists started with 'a' and ended with 'a', like the following:
['a','b','c','a']
['a','d','e','a']
['a','b','e','f','j','a']
['a','c','a']
The final ouput can also be a list of lists. I have tried a double for loop approach with 'a' as the condition, but this is inefficient and not pythonic.
One possible solution is using re (regex)
import re
l = ['k','a','b','c','a','d','e','a','b','e','f','j','a','c','a','b']
r = [list(f"a{_}a") for _ in re.findall("(?<=a)[^a]+(?=a)", "".join(l))]
print(r)
# [['a', 'b', 'c', 'a'], ['a', 'd', 'e', 'a'], ['a', 'b', 'e', 'f', 'j', 'a'], ['a', 'c', 'a']]
You can do this in one loop:
lst = ['k','a','b','c','a','d','e','a','b','e','f','j','a','c','a','b']
out = [[]]
for i in lst:
if i == 'a':
out[-1].append(i)
out.append([])
out[-1].append(i)
out = out[1:] if out[-1][-1] == 'a' else out[1:-1]
Also using numpy.split:
out = [ary.tolist() + ['a'] for ary in np.split(lst, np.where(np.array(lst) == 'a')[0])[1:-1]]
Output:
[['a', 'b', 'c', 'a'], ['a', 'd', 'e', 'a'], ['a', 'b', 'e', 'f', 'j', 'a'], ['a', 'c', 'a']]
Firstly you can store the indices of 'a' from the list.
oList = ['k','a','b','c','a','d','e','a','b','e','f','j','a','c','a','b']
idx_a = list()
for idx, char in enumerate(oList):
if char == 'a':
idx_a.append(idx)
Then for every consecutive indices you can get the sub-list and store it in a list
ans = [oList[idx_a[x]:idx_a[x + 1] + 1] for x in range(len(idx_a))]
You can also get more such lists if you take in-between indices also.
You can do this with a single iteration and a simple state machine:
original_list = list('kabcadeabefjacab')
multiple_lists = []
for c in original_list:
if multiple_lists:
multiple_lists[-1].append(c)
if c == 'a':
multiple_lists.append([c])
if multiple_lists[-1][-1] != 'a':
multiple_lists.pop()
print(multiple_lists)
[['a', 'b', 'c', 'a'], ['a', 'd', 'e', 'a'], ['a', 'b', 'e', 'f', 'j', 'a'], ['a', 'c', 'a']]
We can use str.split() to split the list once we str.join() it to a string, and then use a f-string to add back the stripped "a"s. Note that even if the list starts/ends with an "a", this the split list will have an empty string representing the substring before the split, so our unpacking logic that discards the first + last subsequences will still work as intended.
def split(data):
_, *subseqs, _ = "".join(data).split("a")
return [list(f"a{seq}a") for seq in subseqs]
Output:
>>> from pprint import pprint
>>> testdata = ['k','a','b','c','a','d','e','a','b','e','f','j','a','c','a','b']
>>> pprint(split(testdata))
[['a', 'b', 'c', 'a'],
['a', 'd', 'e', 'a'],
['a', 'b', 'e', 'f', 'j', 'a'],
['a', 'c', 'a']]
I want to reorder my list in a given order,
For example I have a list of ['a', 'b', 'c', 'd', 'e', 'f', 'g']
this has an index of [0,1,2,3,4,5,6] and lets say the new ordered list would have an order of [3,5,6,1,2,4,0] which would result in ['d','f','g', 'b', 'c', 'e', 'a'].
How would you result in such code?
I thought of using for loop by doing the
for i in range(Len(list))
and after that I thought go using append or creating a new list? maybe but I'm not sure if I'm approaching this right.
All you need to do is iterate the list of indexes, and use it to access the list of elements, like this:
elems = ['a', 'b', 'c', 'd', 'e', 'f', 'g']
idx = [3,5,6,1,2,4,0]
result = [elems[i] for i in idx]
print(result)
Output:
['d', 'f', 'g', 'b', 'c', 'e', 'a']
import numpy as np
my_list = ['a', 'b', 'c', 'd', 'e', 'f', 'g']
def my_function(my_list, index):
return np.take(my_list, index)
print(my_function(my_list, [3,5,6,1,2,4,0]))
Output: ['d' 'f' 'g' 'b' 'c' 'e' 'a']
I have a list list1 of 3 sublists of tuples like
[[(['A', 'B', 'A'], ['B', 'O', 'A']),
(['A', 'B', 'A'], ['B', 'A', 'O']),
(['A', 'B', 'O'], ['B', 'O', 'A']),
(['A', 'B', 'O'], ['B', 'A', 'O']),
(['A', 'B', 'A'], ['B', 'O', 'A']),
(['A', 'B', 'A'], ['B', 'A', 'O'])],
[(['A', 'B', 'A'], ['B', 'A', 'A']),
(['A', 'B', 'O'], ['B', 'A', 'A']),
(['A', 'B', 'A'], ['B', 'A', 'A'])],
[['A', 'B', 'A'], ['A', 'B', 'O']],
[['A', 'B', 'B']],
[['B', 'A', 'A']]]
Assume list2 = ['A', 'B', 'A']. My goal is to obtain a list of indices of any pairs of tuples (or a singleton set of tuple) in list1 that contain the tuple list2. I tried to use the enumerate function as follows but the result is not correct
print([i for i, j in enumerate(bigset) if ['A', 'B', 'A'] in j[0] or
['A', 'B', 'A'] == j[0] or [['A', 'B', 'A']] in j[0]])
Can anyone please help me with this problem? I'm quite stuck due to the mismatch in the different sizes of tuples of tuples appearing in list1.
Another question I have is: I want to find the total number of 3-element lists in list1. So if I do it by hand, the answer is 22. But how to do it in code? I guess we need to use two for loops?
Expected Output For list1 above with the given list2, we would get the list of indices containing list2 is [0,1,5,6,7,9,10].
Ok, so here you go
This use recursion because we don't know the depth of your list1 SO the index will be counted like this :
0,1
2,3,4,
6,7
8,
9,10,11,12
etc... (The same order you have by writing it in 1 row)
Here the result will be :
[0, 2, 8, 10, 12, 16, 18]
Now the code
def foo(l,ref):
global s
global indexes
for items in l: #if it's an element of 3 letters
if len(items)== 3 and len(items[0])==1:
if items == ref:
indexes.append(s) #save his index if it match the ref
s+= 1 #next index
else: #We need to go deeper
foo(items,ref)
return(s)
list1 = [[(['A', 'B', 'A'], ['B', 'O', 'A']),
(['A', 'B', 'A'], ['B', 'A', 'O']),
(['A', 'B', 'O'], ['B', 'O', 'A']),
(['A', 'B', 'O'], ['B', 'A', 'O']),
(['A', 'B', 'A'], ['B', 'O', 'A']),
(['A', 'B', 'A'], ['B', 'A', 'O'])],
[(['A', 'B', 'A'], ['B', 'A', 'A']),
(['A', 'B', 'O'], ['B', 'A', 'A']),
(['A', 'B', 'A'], ['B', 'A', 'A'])],
[['A', 'B', 'A'], ['A', 'B', 'O']],
[['A', 'B', 'B']],
[['B', 'A', 'A']]]
list2 = ['A', 'B', 'A']
indexes = []
s=0
count= foo(list1,list2)
print(indexes)
s is the index we are working on
count is the total amount of element (22).
Indexes is the list of index you want.
This work even if you make a list3 = [list1,list1,[list1,[list1],list1]] , you may want to try it.
Best luck to end your script now.
Would it work for your implementation if we sort out your list1 into a more friendly format first? If so, you could do that in a pretty simple way:
Go through each element of list1, if the element is itself a big list of tuples, then we want to unpack further. If the element is a tuple (so the first element of that tuple is a list), or it is itself one of your 3-element lists, then we just want to append that as it is.
nice_list = []
for i in list1:
if type(i[0]) == str or type(i[0]) == list:
# i.e. i is one of your 3-element lists, or a tuple of lists
nice_list.append(i)
else:
#If i is still a big list of other tuples, we want to unpack further
for j in i:
nice_list.append(j)
Then you could search for the indices much easier:
for i, idx in zip(nice_list, range(len(nice_list))):
if ['A', 'B', 'A'] in i:
print(idx) #Or append them to a list, whatever you wanted to do
For a not-particularly-elegant solution to your question about finding how many 3-element lists there are, yes you could use a for loop:
no_of_lists = 0
for n in nice_list:
if type(n) == tuple:
no_of_lists += len(n)
elif type(n) == list and type(n[0]) == list:
# if it is a list of lists
no_of_lists += len(n)
elif type(n) == list and type(n[0]) == str:
#if it is a 3-element list
no_of_lists_lists += 1
print('Number of 3-element lists contained:', no_of_lists)
Edit: to answer the question you asked in the comments about how the for n in nice_list part works, this just iterates through each element of the list. To explore this, try writing some code to print out nice_list[0], nice_list[1] etc, or a for loop which prints out each n so you can see what that looks like. For example, you could do:
for n in nice_list:
print(n)
to understand how that's working.
Slightly unconventional approach, due to unknown depth, and/or lack of known array flattening operation - I would try with regex:
import re
def getPos(el, arr):
el=re.escape(str(el))
el=f"(\({el})|({el}\))"
i=0
for s in re.finditer(r"\([^\)]+\)", str(arr)):
if(re.match(el,s.group(0))):
yield i
i+=1
Which yields:
>>> print(list(getPos(list2, list1)))
[0, 1, 4, 5, 6, 8, 9]
(Which I believe is the actual result you want).
This question already has answers here:
How do I clone a list so that it doesn't change unexpectedly after assignment?
(24 answers)
Closed 6 years ago.
I am using a recursive function on a list of lists with an accumulator, but instead of creating the right list of lists, it makes a list with duplicates of the last item inserted in the accumulator. I recreated the problem in a much simpler recursive function and list. This function takes a list of lists and makes 2 copies and reverses them n times.
def recursauto(x, n, instSet):
#Base Case
if(n==0):
instSet.append(x)
print(x) #Print to see what SHOULD be added
else:
temp = [x]*(2) # Make a list with 2 copies of the original list
for i in range(len(temp)):
temp[i][i] = temp[i][i][::-1] # Make the list backwards
for i in range(2):
recursauto(temp[i], n-1, instSet) #Input each element
MyList = [] #Empyt list for accumulator
print("Correct output:")
recursauto([["A", "l", "e", "x"], ["A", "b", "d", "a", "l", "a", "h"]], 2, MyList)
print("Wrong printed list:")
for i in MyList:
print(i) #Print what is in the accumulator
The output comes out wrong and the accumulator does not have the right things that were put into it.
Correct output:
[['A', 'l', 'e', 'x'], ['A', 'b', 'd', 'a', 'l', 'a', 'h']]
[['A', 'l', 'e', 'x'], ['A', 'b', 'd', 'a', 'l', 'a', 'h']]
[['x', 'e', 'l', 'A'], ['h', 'a', 'l', 'a', 'd', 'b', 'A']]
[['x', 'e', 'l', 'A'], ['h', 'a', 'l', 'a', 'd', 'b', 'A']]
Wrong printed list:
[['x', 'e', 'l', 'A'], ['h', 'a', 'l', 'a', 'd', 'b', 'A']]
[['x', 'e', 'l', 'A'], ['h', 'a', 'l', 'a', 'd', 'b', 'A']]
[['x', 'e', 'l', 'A'], ['h', 'a', 'l', 'a', 'd', 'b', 'A']]
[['x', 'e', 'l', 'A'], ['h', 'a', 'l', 'a', 'd', 'b', 'A']]
I know that there is an easier way to do this, but the function I'm actually making requires recursion. Like I said, this is just a simplified recreation of the problem I am having.
temp = [x]*(2)
The above line does not create a list of two copies of the original list; in fact, it just stores a reference to the same original list twice. If you want a distinct copy of x, try using the list copy constructor like temp = [list(x), list(x)] or alternatively the shallow copy method [x.copy() x.copy()].
See the below example.
>>> ls = ['a']
>>> dup = [ls] * 2
>>> dup
[['a'], ['a']]
>>> ls.append('b')
>>> dup
[['a', 'b'], ['a', 'b']]