Related
I'm solving the problem.
You have to leave the last list, but the previous list is printed out again.
def merge(xs,ys):
# xs, ys, ss = xs, ys, []
xs, ys, ss = xs[:], ys[:], []
while xs!=[] and ys!=[]:
if xs[0] <= ys[0]:
ss.append(xs[0])
xs.remove(xs[0])
else:
ss.append(ys[0])
ys.remove(ys[0])
ss.extend(xs)
ss.extend(ys)
return ss
accumulator = []
remain = []
def merge2R(xss):
if len(xss)% 2 != 0 :
OExcept = len(xss)-1
remain.append((xss[OExcept]))
xss.remove(xss[OExcept])
if xss != []:
accumulator.append(merge(xss[0],xss[1]))
xss.remove(xss[0])
xss.remove(xss[0])
return merge2R(xss)
else:
return accumulator + remain
The result comes out like this.
How can I fix it?
>>> merge2R([[2],[1,3],[4,6,7],[5,8],[9]])
[[1, 2, 3], [4, 5, 6, 7, 8], [1, 2, 3], [4, 5, 6, 7, 8], [9]]
Desired result value:
>>> merge2R([[2],[1,3],[4,6,7],[5,8]])
[[1,2,3], [4,5,6,7,8]]
>>> merge2R([[2],[1,3],[4,6,7],[5,8],[9]])
[[1,2,3], [4,5,6,7,8], [9]]
Your code was working, you just needed to reset accumulator and remain
def merge2R(xss):
# Declare use of globals
global remain
global accumulator
if len(xss) % 2 != 0:
OExcept = len(xss)-1
remain.append((xss[OExcept]))
xss.remove(xss[OExcept])
if xss != []:
accumulator.append(merge(xss[0], xss[1]))
xss.remove(xss[0])
xss.remove(xss[0])
return merge2R(xss)
else:
x = accumulator + remain
# Must reset accumulator and remain
accumulator = []
remain = []
return x
Because you initialise both arrays as empty, then append to them:
# Remain
remain.append((xss[OExcept]))
# Accumulator
accumulator.append(merge(xss[0], xss[1]))
After you have finished with the data in those arrays (at end of function) you need to discard it:
accumulator = []
remain = []
The result of not discarding these arrays is evident when calling the function with the same argument multiple times:
print(merge2R([[2], [1, 3]]))
print(merge2R([[2], [1, 3]]))
print(merge2R([[2], [1, 3]]))
print(merge2R([[2], [1, 3]]))
print(merge2R([[2], [1, 3]]))
[[1, 2, 3]]
[[1, 2, 3], [1, 2, 3]]
[[1, 2, 3], [1, 2, 3], [1, 2, 3]]
[[1, 2, 3], [1, 2, 3], [1, 2, 3], [1, 2, 3]]
[[1, 2, 3], [1, 2, 3], [1, 2, 3], [1, 2, 3], [1, 2, 3]]
How can I get product of nested list :
[[-1, 3, 1],[6, 1, 2],[4, 3, 1],[0, 1, 1]]
to:
[-3,12,12,1]
where eg : -1 * 3 * 1 = -3 and so on.
This is my current solution :
for i in range(len(array2)):
for j in range(len(array2[i])):
prod = array2[i][j] * array2[i][j + 1] * array2[i][j + 2]
print(prod)
I'm getting the following error :
IndexError: list index out of range
You might look at operator.mul combined with functools.reduce for this to make it short and very clear and totally avoid indexing (which is often the source of small errors):
from operator import mul
from functools import reduce
l = [[-1, 3, 1],[6, 1, 2],[4, 3, 1],[0, 1, 1]]
[reduce(mul, s) for s in l]
# [-3, 12, 12, 0]
Edit based on comment
If you want to ignore zeros, you can simply filter them out (this assumes that you don't have rows of all zeros, in which case it's not clear what the answer would be):
[reduce(mul, filter(None, s)) for s in l]
# [-3, 12, 12, 1]
Your index is out of bounds when you get j+1 and j+2, the correct way is to stop at len-2
array2 = [[-1, 3, 1],
[6, 1, 2],
[4, 3, 1],
[0, 1, 1]]
for i in range(len(array2)):
for j in range(len(array2[i])-2): # Check here the range
prod = array2[i][j] * array2[i][j + 1] * array2[i][j + 2]
print(prod)
Try the following function:
input_list = [[-1, 3, 1], [6, 1, 2], [4, 3, 1], [0, 1, 1]]
def compute_multiplication(nested_list):
for l in nested_list:
res = 1
for element in l:
res *= element
print(res)
With the final line being:
compute_multiplication(input_list)
This provides the following output:
-3
12
12
0
my_list = [[-1,3,1],[6,1,2],[4,3,1],[0,1,1]]
products = []
for sub_list in my_list:
partial_prod = 1
for item in sub_list:
partial_prod = partial_prod * item
products.append(partial_prod)
A solution using numpy.prod() :
import numpy
matrix = [
[-1, 3, 1],
[6, 1, 2],
[4, 3, 1],
[0, 1, 1]
]
for row in matrix:
prod = numpy.prod(row)
print(prod)
Assumption is that there are only three entries per row : a list comprehension should suffice. Multiply each item by the other in each row :
matrix = [
[-1, 3, 1],
[6, 1, 2],
[4, 3, 1],
[0, 1, 1]
]
first, make a condition to change zero values to 1 (this is based on the condition, that 0 be ignored and only the result of non zeros be returned)
filtered = [[ent if ent!= 0 else 1 for ent in row ] for row in matrix]
multiply each entry by the other
res = [start*middle*end for start,middle,end in filtered]
#[-3, 12, 12, 1]
if the number of entries is more than three, then #mark meyer's solution is apt for this
I have an array such as
data = [
[1, 0],
[2, 0],
[3, 1],
[4, 1],
[5, 1],
[6, 0],
[7, 0]]
and I want the result to be
verified_data = [[1, 0], [2, 0], [3, 1]]
So how can I remove the last elements if they are 0, and also if last N elements are same (except the first 1). What is the proper way to achieve this? Use of numpy is also fine.
Editing as I have written a solution even if it looks ugly:
def verify_data(data):
rev_data = reversed(data)
for i, row in list(enumerate(rev_data )):
if row[1] == 0:
del data[- 1]
else:
break
rev_data = reversed(data)
last_same_data = None
for i, row in list(enumerate(rev_data)):
if not last_same_data:
last_same_data = row[1]
continue
if last_same_data == row[1]:
del data[-1]
else:
break
return data
I've split removing trailing zeros and removing trailing duplicates into two functions. Using the list[-n] indices to avoid explicit index tracking.
In [20]: def remove_trailing_duplicates(dat):
...: key=dat[-1][1]
...: while (len(dat)>1) and (dat[-2][1]==key):
...: dat.pop() # Remove the last item.
...: key=dat[-1][1] # Reset key to last item.
In [21]: def remove_trailing_zeros(dat):
# len(dat)>0 can give an empty list, >1 leaves at least the first item
...: while len(dat)>0 and dat[-1][1]==0:
dat.pop()
In [22]: data = [
...: [1, 0],
...: [2, 0],
...: [3, 1],
...: [4, 1],
...: [5, 1],
...: [6, 0],
...: [7, 0]]
In [23]: remove_trailing_zeros(data)
In [24]: data
Out [24]: [[1, 0], [2, 0], [3, 1], [4, 1], [5, 1]]
In [25]: remove_trailing_duplicates(data)
In [26]: data
Out[26]: [[1, 0], [2, 0], [3, 1]]
This works with the data you used in the question and checks for only one item left in the duplicates function. What would you want for ALL data items being [n, 0]? An empty list or the first item remaining?
HTH
Is there a method similar to str.replace which can do the following:
>> replace(sequence=[0,1,3], old=[0,1], new=[1,2])
[1,2,3]
It should really act like str.replace : replacing a "piece" of a sequence by another sequence, not map elements of "old" with "new" 's ones.
Thanks :)
No, I'm afraid there is no built-in function that does this, however you can create your own!
The steps are really easy, we just need to slide a window over the list where the width of the window is the len(old). At each position, we check if the window == to old and if it is, we slice before the window, insert new and concatenate the rest of the list on after - this can be done simply be assigning directly to the old slice as pointed out by #OmarEinea.
def replace(seq, old, new):
seq = seq[:]
w = len(old)
i = 0
while i < len(seq) - w + 1:
if seq[i:i+w] == old:
seq[i:i+w] = new
i += len(new)
else:
i += 1
return seq
and some tests show it works:
>>> replace([0, 1, 3], [0, 1], [1, 2])
[1, 2, 3]
>>> replace([0, 1, 3, 0], [0, 1], [1, 2])
[1, 2, 3, 0]
>>> replace([0, 1, 3, 0, 1], [0, 1], [7, 8])
[7, 8, 3, 7, 8]
>>> replace([1, 2, 3, 4, 5], [1, 2, 3], [1, 1, 2, 3])
[1, 1, 2, 3, 4, 5]
>>> replace([1, 2, 1, 2], [1, 2], [3])
[3, 3]
As pointed out by #user2357112, using a for-loop leads to re-evaluating replaced sections of the list, so I updated the answer to use a while instead.
I tried this but before using this method read this about eval() by Ned :
import re
import ast
def replace(sequence, old, new):
sequence = str(sequence)
replace_s=str(str(old).replace('[', '').replace(']', ''))
if '.' in replace_s:
replace_ss=list(replace_s)
for j,i in enumerate(replace_ss):
if i=='.':
try:
replace_ss[0]=r"\b"+ replace_ss[0]
replace_ss[j]=r".\b"
except IndexError:
pass
replace_s="".join(replace_ss)
else:
replace_s = r"\b" + replace_s + r"\b"
final_ = str(new).replace('[', '').replace(']', '')
return ast.literal_eval(re.sub(replace_s, final_, sequence))
print(replace([0, 1, 3], [0, 1], [1, 2]))
output:
[1, 2, 3]
I'm trying to create a pair of functions that, given a list of "starting" numbers, will recursively add to each index position up to a defined maximum value (much in the same way that a odometer works in a car--each counter wheel increasing to 9 before resetting to 1 and carrying over onto the next wheel).
The code looks like this:
number_list = []
def counter(start, i, max_count):
if start[len(start)-1-i] < max_count:
start[len(start)-1-i] += 1
return(start, i, max_count)
else:
for j in range (len(start)):
if start[len(start)-1-i-j] == max_count:
start[len(start)-1-i-j] = 1
else:
start[len(start)-1-i-j] += 1
return(start, i, max_count)
def all_values(fresh_start, i, max_count):
number_list.append(fresh_start)
new_values = counter(fresh_start,i,max_count)
if new_values != None:
all_values(*new_values)
When I run all_values([1,1,1],0,3) and print number_list, though, I get:
[[1, 1, 1], [1, 1, 1], [1, 1, 1], [1, 1, 1], [1, 1, 1], [1, 1, 1], [1, 1, 1], [1, 1, 1],
[1, 1, 1], [1, 1, 1], [1, 1, 1], [1, 1, 1], [1, 1, 1], [1, 1, 1], [1, 1, 1], [1, 1, 1],
[1, 1, 1], [1, 1, 1], [1, 1, 1], [1, 1, 1], [1, 1, 1], [1, 1, 1], [1, 1, 1], [1, 1, 1],
[1, 1, 1], [1, 1, 1], [1, 1, 1]]
Which is unfortunate. Doubly so knowing that if I replace the first line of all_values with
print(fresh_start)
I get exactly what I'm after:
[1, 1, 1]
[1, 1, 2]
[1, 1, 3]
[1, 2, 1]
[1, 2, 2]
[1, 2, 3]
[1, 3, 1]
[1, 3, 2]
[1, 3, 3]
[2, 1, 1]
[2, 1, 2]
[2, 1, 3]
[2, 2, 1]
[2, 2, 2]
[2, 2, 3]
[2, 3, 1]
[2, 3, 2]
[2, 3, 3]
[3, 1, 1]
[3, 1, 2]
[3, 1, 3]
[3, 2, 1]
[3, 2, 2]
[3, 2, 3]
[3, 3, 1]
[3, 3, 2]
[3, 3, 3]
I have already tried making a copy of fresh_start (by way of temp = fresh_start) and appending that instead, but with no change in the output.
Can anyone offer any insight as to what I might do to fix my code? Feedback on how the problem could be simplified would be welcome as well.
Thanks a lot!
temp = fresh_start
does not make a copy. Appending doesn't make copies, assignment doesn't make copies, and pretty much anything that doesn't say it makes a copy doesn't make a copy. If you want a copy, slice it:
fresh_start[:]
is a copy.
Try the following in the Python interpreter:
>>> a = [1,1,1]
>>> b = []
>>> b.append(a)
>>> b.append(a)
>>> b.append(a)
>>> b
[[1, 1, 1], [1, 1, 1], [1, 1, 1]]
>>> b[2][2] = 2
>>> b
[[1, 1, 2], [1, 1, 2], [1, 1, 2]]
This is a simplified version of what's happening in your code. But why is it happening?
b.append(a) isn't actually making a copy of a and stuffing it into the array at b. It's making a reference to a. It's like a bookmark in a web browser: when you open a webpage using a bookmark, you expect to see the webpage as it is now, not as it was when you bookmarked it. But that also means that if you have multiple bookmarks to the same page, and that page changes, you'll see the changed version no matter which bookmark you follow.
It's the same story with temp = a, and for that matter, a = [1,1,1]. temp and a are "bookmarks" to a particular array which happens to contain three ones. And b in the example above, is a bookmark to an array... which contains three bookmarks to that same array that contains three ones.
So what you do is create a new array and copy in the elements of the old array. The quickest way to do that is to take an array slice containing the whole array, as user2357112 demonstrated:
>>> a = [1,1,1]
>>> b = []
>>> b.append(a[:])
>>> b.append(a[:])
>>> b.append(a[:])
>>> b
[[1, 1, 1], [1, 1, 1], [1, 1, 1]]
>>> b[2][2] = 2
>>> b
[[1, 1, 1], [1, 1, 1], [1, 1, 2]]
Much better.
When I look at the desired output I can't help but think about using one of the numpy grid data production functions.
import numpy
first_column, second_column, third_column = numpy.mgrid[1:4,1:4,1:4]
numpy.dstack((first_column.flatten(),second_column.flatten(),third_column.flatten()))
Out[23]:
array([[[1, 1, 1],
[1, 1, 2],
[1, 1, 3],
[1, 2, 1],
[1, 2, 2],
[1, 2, 3],
[1, 3, 1],
[1, 3, 2],
[1, 3, 3],
[2, 1, 1],
[2, 1, 2],
[2, 1, 3],
[2, 2, 1],
[2, 2, 2],
[2, 2, 3],
[2, 3, 1],
[2, 3, 2],
[2, 3, 3],
[3, 1, 1],
[3, 1, 2],
[3, 1, 3],
[3, 2, 1],
[3, 2, 2],
[3, 2, 3],
[3, 3, 1],
[3, 3, 2],
[3, 3, 3]]])
Of course, the utility of this particular approach might depend on the variety of input you need to deal with, but I suspect this could be an interesting way to build the data and numpy is pretty fast for this kind of thing. Presumably if your input list has more elements you could have more min:max arguments fed into mgrid[] and then unpack / stack in a similar fashion.
Here is a simplified version of your program, which works. Comments will follow.
number_list = []
def _adjust_counter_value(counter, n, max_count):
"""
We want the counter to go from 1 to max_count, then start over at 1.
This function adds n to the counter and then returns a tuple:
(new_counter_value, carry_to_next_counter)
"""
assert max_count >= 1
assert 1 <= counter <= max_count
# Counter is in closed range: [1, max_count]
# Subtract 1 so expected value is in closed range [0, max_count - 1]
x = counter - 1 + n
carry, x = divmod(x, max_count)
# Add 1 so expected value is in closed range [1, max_count]
counter = x + 1
return (counter, carry)
def increment_counter(start, i, max_count):
last = len(start) - 1 - i
copy = start[:] # make a copy of the start
add = 1 # start by adding 1 to index
for i_cur in range(last, -1, -1):
copy[i_cur], add = _adjust_counter_value(copy[i_cur], add, max_count)
if 0 == add:
return (copy, i, max_count)
else:
# if we have a carry out of the 0th position, we are done with the sequence
return None
def all_values(fresh_start, i, max_count):
number_list.append(fresh_start)
new_values = increment_counter(fresh_start,i,max_count)
if new_values != None:
all_values(*new_values)
all_values([1,1,1],0,3)
import itertools as it
correct = [list(tup) for tup in it.product(range(1,4), range(1,4), range(1,4))]
assert number_list == correct
Since you want the counters to go from 1 through max_count inclusive, it's a little bit tricky to update each counter. Your original solution was to use several if statements, but here I have made a helper function that uses divmod() to compute each new digit. This lets us add any increment to any digit and will find the correct carry out of the digit.
Your original program never changed the value of i so my revised one doesn't either. You could simplify the program further by getting rid of i and just having increment_counter() always go to the last position.
If you run a for loop to the end without calling break or return, the else: case will then run if there is one present. Here I added an else: case to handle a carry out of the 0th place in the list. If there is a carry out of the 0th place, that means we have reached the end of the counter sequence. In this case we return None.
Your original program is kind of tricky. It has two explicit return statements in counter() and an implicit return at the end of the sequence. It does return None to signal that the recursion can stop, but the way it does it is too tricky for my taste. I recommend using an explicit return None as I showed.
Note that Python has a module itertools that includes a way to generate a counter series like this. I used it to check that the result is correct.
I'm sure you are writing this to learn about recursion, but be advised that Python isn't the best language for recursive solutions like this one. Python has a relatively shallow recursion stack, and does not automatically turn tail recursion into an iterative loop, so this could cause a stack overflow inside Python if your recursive calls nest enough times. The best solution in Python would be to use itertools.product() as I did to just directly generate the desired counter sequence.
Since your generated sequence is a list of lists, and itertools.product() produces tuples, I used a list comprehension to convert each tuple into a list, so the end result is a list of lists, and we can simply use the Python == operator to compare them.