I've found answers that tackle the question of how to multiply with the element value, but what concerns me is how to make copies of the element itself. I have:
a = [1, 2, 3]
x = 3
b = []
I tried:
b.append(a * x)
But that gives me:
[1, 2, 3, 1, 2, 3, 1, 2, 3]
and I need:
b = [[1, 2, 3], [1, 2, 3], [1, 2, 3]]
Please note I'm a programming novice. Thanks!
If you need to copy the list and not a reference to the list, you can't use *.
b = [a[:] for i in range(x)]
(a[:] creates a copy of the list.)
Related
Suppose I have a two lists:
a = [1,2,3,4,5]
b = [float]
My requirement is to just get the value of the 'a' list in 'b', i.e.
b = [1,2,3,4,5, float]
So, if we either append of insert, that will again create list of list, which I don't want.
Any suggestions
Assume your input is:
a = [1,2,3,4,5]
b = ['float']
b{:] = a+b
print (b)
The output of this will be:
[1, 2, 3, 4, 5, 'float']
If your input is:
a = [1,2,3,4,5]
b = [float]
then your output will be:
[1, 2, 3, 4, 5, <class 'float'>]
this may be a bit clumsy, but it does what you want:
a = [1, 2, 3, 4, 5]
b = ["float"]
print(id(b))
for n in reversed(a):
b.insert(0, n)
print(b) # [1, 2, 3, 4, 5, 'float']
print(id(b))
note that the id does not change - no new list is created.
slice-assignment would also work:
b0 = b[0]
b[:len(a)] = a
b.append(b0)
if you don't mind working with a deque instead of a list you could also do this:
from collections import deque
a = [1, 2, 3, 4, 5]
b = deque(["float"])
b.extendleft(reversed(a)) # deque([1, 2, 3, 4, 5, 'float'])
you can use:
a.__add__(b)
output:
[1, 2, 3, 4, 5, <class 'float'>]
I had followed the book and can sum lists in lists by column but one of the test cases is missing variables in the list and I'm unable to move forward because I keep getting an index error.
The first initial_list works as it should giving [3,6,9]
The second one though should apparently give me [3,4,9,4]
list_initial = [[1, 2, 3], [1, 2, 3],[1, 2, 3 ]]
list_initial = [[1, 2, 3], [1], [1, 2, 3, 4]]
def column_sums(list_initial):
column = 0
list_new = []
while column < len(list_initial):
total = sum(row[column] for row in list_initial )
list_new.append(total)
column = column + 1
print(list_new)
column_sums(list_initial)
You can effectively "transpose" your data so that rows become columns, and then use itertools.zip_longest with a fillvalue of 0, to sum across them, eg:
from itertools import zip_longest
list_initial = [[1, 2, 3], [1], [1, 2, 3, 4]]
summed = [sum(col) for col in zip_longest(*list_initial, fillvalue=0)]
# [3, 4, 6, 4]
I'm just starting with Python and trying to find a general solution to transform a list of lists [1, [2, 3, 4], 5 ] to a list [[1,2,5], [1,3,5], [1,4,5]] in Python.
I've tried creating some dynamic lists but not getting what i want, not even for this simple list in the example. Any help will be greatly appreciated.
inter_l = []
aba = []
v = [1, [2, 3], 4, 5, 6]
g = globals()
for elem in v:
if isinstance(elem, (list,)):
l_ln = len(elem)
indx = v.index(elem)
for i in range(0, l_ln):
g['depth_{0}'.format(i)] = [elem[i]]
inter_l.append(list(g['depth_{0}'.format(i)]))
else:
aba.append(elem)
t = aba.extend(inter_l)
w = aba.extend(inter_l)
print(v)
print(aba)
print(inter_l)
[1, [2, 3], 4, 5, 6]
[1, 4, 5, 6, [2], [3], [2], [3]]
[[2], [3]]
The easiest way would be to leverage itertools.product function, but since it expects iterables as its inputs, the input would have to be transformed a little. One way to achieve this would be something like this:
transformed = [e if isinstance(e, list) else [e] for e in v]
which converts all non-list elements into lists and then pass this transformed input to product:
list(itertools.product(*transformed))
Note, that * in front of transformed expands transformed list into positional arguments, so that instead of a single argument of type list, a list of its elements is passed instead.
The entire pipeline looks something like this:
>>> v = [1, [2, 3, 4], 5]
>>> t = [e if isinstance(e, list) else [e] for e in v]
>>> list(itertools.product(*t))
[(1, 2, 5), (1, 3, 5), (1, 4, 5)]
I need to generate a list of lists in that special way:
[3, 1, 4] -> [[1, 2, 3], [1], [1, 2, 3, 4]]
That means that every list in a list of lists must be in range of the given list values. I've tried smth like:
L = [3, 1, 4]
q = [i for i in L]
print(list([x] for x in range(y for y in q)))
But it return a TypeError: generator cannot be interpreted as an integer
That all has to be a single generator expression.
Using a list comprehension.
Try:
L = [3, 1, 4]
print([list(range(1, i+1)) for i in L])
Output:
[[1, 2, 3], [1], [1, 2, 3, 4]]
Let us say we have a list of integers:
list = [6, 4, 1, 4, 4, 4, 4, 4, 2, 1]
I now wrote a function which returns another list with all the integers from the list above without repeats.
def no_repeats(s):
new_list = []
for number in s:
if new_list.count(number) < 1:
new_list.append(number)
return(new_list)
The new_list returns [6, 4, 1, 2] which is good! My question is how I would now write two similar functions:
A function clean(s) which does not return a new list like the function above, but changes the original list by deleting all the numbers that repeat. Thus, the result has to be the same and the function must not include "return" or create a new list. It must only clean the original list.
A function double(s) which, again, changes the original list (does not return a new list!) but this time, by doubling every number in the original list. Thus, double(list) should change the original list above to:
[6, 6, 4, 4, 1, 1, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 2, 2, 1, 1]
Thank you for all the help!
Removing duplicates inplace without preserving the order:
def no_repeats(L):
L[:] = set(L)
There are several variations possible (preserve order, support non-hashable items, support item that do not define total ordering) e.g., to preserve order:
from collections import OrderedDict
def no_repeats(L):
L[:] = OrderedDict.fromkeys(L)
To double each element's value inplace:
def double(L):
for i in range(len(L)):
L[i] *= 2
To duplicate each element:
def duplicate_elements(L):
L[:] = [x for x in L for _ in range(2)]
>>> def clean(s):
... s[:] = [s[i] for i in range(len(s)) if s[i] not in s[:i]]
...
>>> st = [1, 2, 3, 2, 1]
>>> clean(st)
>>> st
[1, 2, 3]
>>> def double(s):
... s[:] = [s[i//3] for i in range(3*len(s)) if i % 3]
...
>>> st = [1, 2, 3, 2, 1]
>>> double(st)
>>> st
[1, 1, 2, 2, 3, 3, 2, 2, 1, 1]
neither is particularly efficient nor pythonic, yet do address the OP question
def double(s):
... s[:] = [s[i//2] for i in range(2*len(s))]
will also do the trick, with a little less obsfucation