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For my school project I'm trying to append a value to multiple lists at once.
An example of what I'm trying to do:
list1 = [1, 2, 3]
list2 = [4, 5, 6]
Lets say I'm trying to add a 7 onto the end of both of them. Instead of doing:
list1.append(7)
list2.append(7)
Is there a way I can do this in one line. I tried this but it doesn't work:
(list1, list2).append(7)
You can do it using a for loop. See below:
a = [1, 2, 3]
b = [3, 4, 6]
for i in (a,b):
i.append(7)
print(a)
print(b)
This gives us:
[1, 2, 3, 7]
[3, 4, 6, 7]
in this code I'm trying to delete every repeated element in the list and just make all of the elements unique and not repeated, so when I run this code give me an error:
myList = [1, 2, 4, 4, 1, 4, 2, 6, 2, 9]
repeat = 0
for i in range(len(myList)-1):
for j in range(len(myList)-1):
if myList[i]== myList[j]:
repeat+=1
if repeat>1:
del myList[j]
print("The list with unique elements only:")
print(myList)
the error which apppears is :
Traceback (most recent call last):
File "main.py", line 8, in <module>
if myList[i]== myList[j]:
IndexError: list index out of range
why is that happens and how can I solve it?
It is a really bad idea to modify an array while looping on it as you have no control on the way things are handled.
May I suggest these two solutions to your problem.
The first one is using set.
myList = [1, 2, 4, 4, 1, 4, 2, 6, 2, 9]
myList = list(set(myList))
print("The list with unique elements only:")
print(myList)
The other solution is using an other array
myList = [1, 2, 4, 4, 1, 4, 2, 6, 2, 9]
uniques = []
for number in myList:
if number not in uniques:
uniques.append(number)
print("The list with unique elements only:")
print(uniques)
You can convert list to set, it will automatically delete all of repeated elements
a = [1, 2, 4, 4, 1, 4, 2, 6, 2, 9]
unique_list = list(set(a))
print(a)
Note: We again convert set to list
What is heppening here is that you are deleting some elements in your list, making it shorter.
Since your for loops are running for the lenght of your original list, you will eventuall try to access an index that no longer exists. This will cause you to get "list index out of range"
To see this for your self, you can add a print statement, like so:
myList = [1, 2, 4, 4, 1, 4, 2, 6, 2, 9]
repeat = 0
for i in range(len(myList)-1):
for j in range(len(myList)-1):
print(i,j,len(myList))
if myList[i]== myList[j]:
repeat+=1
if repeat>1:
del myList[j]
Set data type in Python is used to remove duplicity. Whenever any iterator needs to be viewed with only the unique values in it, it can be converted into a set and that will remove all the duplicate values. For example:
lis=[2,2,3,4]
l=set(lis)
print(l)
Output:
{2, 3, 4}
It can be converted back into the list:
lis=[2,2,3,4]
l=set(lis)
print(l)
l=list(l)
print(l)
Output:
{2, 3, 4}
[2, 3, 4]
Similarly:
myList = [1, 2, 4, 4, 1, 4, 2, 6, 2, 9]
s=set(myList)
l=list(s)
print(l)
Output:
[1, 2, 4, 6, 9]
Frozen sets can also be used for this purpose. Although; elements of the frozen set remain the same after creation i.e, they can't be modified unlike the elements of the set which are mutable(can be modified).
Hope this was helpful!
This question already has an answer here:
Create index list for np.split from the list that already has number for each section
(1 answer)
Closed 3 years ago.
Let's say I've got an array [0, 1, 2, 3, 4, 5, 6, 7] and a tuple: (3, 3, 2).
I'm looking for a way to split my array to 3 array based on my tuple data:
[0, 1, 2]
[3, 4, 5]
[6, 7]
I can write a simple code like this to get what I want, however I'm looking for a correct and pythonic way to do this:
I used lists for simplicity.
a = [0, 1, 2, 3, 4, 5, 6, 7]
b = (3, 3, 2)
pointer = 0
for i in b:
lst = []
for j in range(i):
lst.append(a[pointer])
pointer += 1
print(lst)
Or this one:
a = [0, 1, 2, 3, 4, 5, 6, 7]
b = (3, 3, 2)
pointer = 0
for i in b:
lst = a[pointer:pointer+i]
pointer += i
print(lst)
Results:
[0, 1, 2]
[3, 4, 5]
[6, 7]
you can use the split method of numpy
import numpy as np
a = [0, 1, 2, 3, 4, 5, 6, 7]
b = (3, 3, 2)
c = np.split(a, np.cumsum(b)[:-1])
for r in c:
print(r)
np.split(a, b) splits a by the indices in b along a given axis(0 by default).
If you don't want to modify your input list, you can use an iterator and the itertools module.
>>> from itertools import islice
>>> a = [0, 1, 2, 3, 4, 5, 6, 7]
>>> b = (3, 3, 2)
>>> i = iter(a)
>>> [list(islice(i, x)) for x in b]
[[0, 1, 2], [3, 4, 5], [6, 7]]
In the first step you create an iterator, which starts at the first element of a. Then you iterate in a list comprehension over your numbers in b and in each step you pull accordingly many elements from the iterator and store them in your result list.
One simpler way is this:
a = [0, 1, 2, 3, 4, 5, 6, 7]
b = (3, 3, 2)
for ind in b:
print(a[:ind])
a = a[ind:]
It loops through slice sizes in b while shortening the original array every time. You can easily append the resulting slices as sublists if you need them for something else. It's almost like one of your solutions except it doesn't use any extra variables and iterates directly through elements of b.
Also, I wouldn't call variables a and b - surely not in this case where variables have clear meanings that you can express through their names. More meaningful names lessen bugs number and make code more clear, becomes a real difference with larger/more complex code. I'd call a at least in_list and b slices, but with more context this could be better.
The most "concise" syntax would be :
ex_array = [0, 1, 2, 3, 4, 5, 6, 7]
extuple = (3, 3, 2)
result = [ex_array[sum(extuple[:iii]):sum(extuple[:iii])+extuple[iii]] for iii in range(len(extuple))]
result would be a list of the expected sub-lists
Re-using the pairwise function from Compare two adjacent elements in same list, you could also:
from itertools import accumulate
from more_itertools import pairwise
a = [0, 1, 2, 3, 4, 5, 6, 7]
b = (3, 3, 2)
[a[slice(*s)] for s in pairwise(accumulate((0,)+b))]
That begin said, the np.split answer is probably faster (and easier to read).
How do I create a new list that contains sublists of ints but the way of divide it is when the next number is the minimun (or equal to the first value founded)?
For example
List1=[1,2,3,4,5,1,2,3,4,1,2,3,4,5,6]
The output that I am looking for is shown below:
Complete_List=[[1,2,3,4,5],[1,2,3,4],[1,2,3,4,5,6]]
I tried looping through the list and appending it when the value is greater than 1 . However it will not work as it doesn't create another list inside it.
Do I have to right a regex for this problem?
Some guidance would be really helpful.
Thank you
Here's something that will split a generic iterable on a given value.
def split_on_value(iterable, split_value):
iterator = iter(iterable)
outer, inner = [], [next(iterator)]
for value in iterator:
if value == split_value:
outer.append(inner)
inner = []
inner.append(value)
outer.append(inner)
return outer
value_list = [1, 2, 3, 4, 5, 1, 2, 3, 4, 1, 2, 3, 4, 5, 6]
print split_on_value(value_list, 1)
# [[1, 2, 3, 4, 5], [1, 2, 3, 4], [1, 2, 3, 4, 5, 6]]
print split_on_value(value_list, 3)
# [[1, 2], [3, 4, 5, 1, 2], [3, 4, 1, 2], [3, 4, 5, 6]]
A vanilla, straightforward, CS101 solution. Though, possibly the most efficient one, because it scans the list exactly once. It also does not assume that segments begin with 1.
fragment = []
result = []
prev = List1[0] - 1 # Preset the previous element marker
for n in List1:
if n > prev:
fragment.append(n)
else:
result.append(fragment)
fragment = [n]
prev = n
result.append(fragment)
#[[1, 2, 3, 4, 5], [1, 2, 3, 4], [1, 2, 3, 4, 5, 6]]
First you search for the 1's, or whatever your condition is, and get the indices within the list. Don't forget to append the len(list) to include the last segment.
idx = [i for i, l in enumerate(List1) if l == 1] + [len(List1)]
Optional, if you want the beginning end of the List. That is, you do not know if there will be a 1 always at index 0.
idx = [0] + idx if idx[0] != 0 else idx
Then, split the list at those indices you found.
complete_list = [List1[ind1:ind2] for ind1, ind2 in zip(idx[:-1], idx[1:])]
and the result:
[[1, 2, 3, 4, 5], [1, 2, 3, 4], [1, 2, 3, 4, 5, 6]]
You can try this to split at every instance of 1:
List1=[1,2,3,4,5,1,2,3,4,1,2,3,4,5,6]
print [map(int, list("1"+i)) for i in ''.join(map(str, List1)).split("1")][1:]
By mapping over List1 with the string function, we can then join all the numbers in the list into one large string. From there, the algorithm splits itself at each instance of one, creating a list containing the new strings of digits. from there, the code maps the integer function over a list created of the strings and appending 1 at the front of the string to make up for the lost 1 when it originally split, creating a list within a list.
I'm fairly new to Python, and besides finding it useful and rather easy to understand for the most part, there are still some things I'm unclear of, ergo this question.
Is it possible to insert specific sections of one list into a specific location in another list?
Say for example, I have a list called 'a', and in this list I have the numbers 1, 3 and 5 in this format:
a = [1, 3, 5]
I also have a list called 'b' which contains the numbers 2 and 4 in this format:
b = [2, 4]
My end goal would be for list 'a' to be changed to this:
a = [1, 2, 3, 4, 5]
As you can see, this would require me to specify using indices for both lists to combine them into one list in this custom format, and I am unsure as to how I could go about this.
I unintentionally left out a major detail, and that is the fact that I wanted to make use of the 'insert' function rather than 'append'.
Yes. Just use insert() and have the second argument be a list element.
a = [1, 3, 5]
b = [2, 4]
# The first argument is the index to insert it at.
# The second is the actual item to be inserted.
a.insert(1, b[0]) # [1, 2, 3, 5]
a.insert(3, b[1]) # [1, 2, 3, 4, 5]
Note that if you just want a sorted, you can just use this:
for item in b:
a.append(item) # Order does not matter, we will sort it later
# a is currently [1, 3, 5, 2, 4] since append() adds to the end of the list
a.sort()
# a is now [1, 2, 3, 4, 5] as sort() re-ordered it
Or if you want it even simpler, concatenate them and then sort:
a = a + b # [1, 3, 5, 2, 4]
a.sort() # [1, 2, 3, 4, 5]
Let me know if this is unclear or not what you wanted.
How does this work for you?
a = [1,2,5,6,7]
b = [3,4]
for n,k in enumerate(a):
if(k<b[0]):
a = a[0:n]+b+a[n:]
break
print(a)
using the colon to slice arrays is super useful.
If you're interleaving the lists, you can zip them using izip_longest:
>>> a = [1, 3, 5]
>>> b = [2, 4]
>>> from itertools import izip_longest
>>> [c for pair in izip_longest(a, b)
... for c in pair if c is not None]
[1, 2, 3, 4, 5]