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I am trying to create a python program to shuffle an array so that the horizontal and vertical rows never have a repeat number.
Input: [1,2,3,4]
Output:
1 2 3 4
2 3 4 1
3 4 1 2
4 1 2 3
My program calculates the shifting of each element correctly, but when it appends the list to the output list, the output list only has repeat copies of the last item in the list.
def numbers(list_of_numbers):
finalValues = [list_of_numbers]
#print(list_of_numbers)
for i in range(1,len(list_of_numbers)):
print("Last array of final: ", finalValues[-1])
tempArray = finalValues[-1]
print("Temp Array: ",tempArray)
temp = tempArray[0]
for j in range(0,len(list_of_numbers)-1):
tempArray[j] = tempArray[j+1]
tempArray[-1] = temp
finalValues.append(tempArray)
print("Final Values: ",finalValues)
return finalValues
numbers([1,2,3,4])
Program Output
[[4, 1, 2, 3], [4, 1, 2, 3], [4, 1, 2, 3], [4, 1, 2, 3]]
Correct Output
[[1,2,3,4], [2,3,4,1], [3,4,1,2], [4,1,2,3]]
The problem comes from the line:
tempArray = finalValues[-1]
You don't create a copy of the previous list, but only a new name to refer to it. After that, all changes you make to tempArray are actually changes to this list, and when you finally do:
finalValues.append(tempArray)
you just add another reference to this same list in finalValues.
In the end, finalValues contains 4 references to the same list, which you can access with finalValues[0], finalValues[1]...
What you need is to create a new list by copying the previous one. One way to do it is to use a slice:
tempArray = finalValues[-1][:]
You can find other ways to close or copy a list in this question
And so, the complete code gives the expected output:
Last array of final: [1, 2, 3, 4]
Temp Array: [1, 2, 3, 4]
Final Values: [[1, 2, 3, 4], [2, 3, 4, 1]]
Last array of final: [2, 3, 4, 1]
Temp Array: [2, 3, 4, 1]
Final Values: [[1, 2, 3, 4], [2, 3, 4, 1], [3, 4, 1, 2]]
Last array of final: [3, 4, 1, 2]
Temp Array: [3, 4, 1, 2]
Final Values: [[1, 2, 3, 4], [2, 3, 4, 1], [3, 4, 1, 2], [4, 1, 2, 3]]
[[1, 2, 3, 4], [2, 3, 4, 1], [3, 4, 1, 2], [4, 1, 2, 3]]
Thierry has provided a very comprehensive explanation of why your code doesn't work as you expect. As such it is the best answer to your question.I have added my answer just as an example of you you can code this in a less complex way .
create the 2d list with the first index as list of numbers. for each iteration take the last index of temp and slice from index 1 to the end then add on index 0.
then return the list
def numbers(list_of_numbers):
temp = [list_of_numbers]
for _ in range(1, len(list_of_numbers)):
temp.append(temp[-1][1:] + temp[-1][0:1])
return temp
print(numbers([1,2,3,4]))
OUTPUT
[[1, 2, 3, 4], [2, 3, 4, 1], [3, 4, 1, 2], [4, 1, 2, 3]]
The problems is in shallow assignment of arrays. You should make deep copy, to really clone arrays, to make them independent.
I did it in your own code. There are a few changes of your code:
import copy that it have been added to first row.
Three usages of copy.deepcopy function instead of =(simple assignment).
import copy
def numbers(list_of_numbers):
finalValues = copy.deepcopy([list_of_numbers])
#print(list_of_numbers)
for i in range(1,len(list_of_numbers)):
print("Last array of final: ", finalValues[-1])
tempArray = copy.deepcopy(finalValues[-1])
print("Temp Array: ",tempArray)
temp = tempArray[0]
for j in range(0,len(list_of_numbers)-1):
tempArray[j] = tempArray[j+1]
tempArray[-1] = temp
finalValues.append(copy.deepcopy(tempArray))
print("Final Values: ",finalValues)
return finalValues
numbers([1,2,3,4])
Program Output
[[4, 1, 2, 3], [4, 1, 2, 3], [4, 1, 2, 3], [4, 1, 2, 3]]
Program Output
[[1,2,3,4], [2,3,4,1], [3,4,1,2], [4,1,2,3]]
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]
Let's say
>>> a = [1,2,3,4,5]
And I want an output like
>>> b
[[1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 3, 4, 5]]
Here is my code:
class Child(object):
def get_lines(self):
a = [1,2,3,4,5]
b=[]
c=[]
j=0
for i in a:
print i
b.append(i)
print b
c.insert(j,b)
j=j+1
print c
son= Child()
son.get_lines()
When I print list b in loop, it gives:
1
[1]
2
[1, 2]
3
[1, 2, 3]
4
[1, 2, 3, 4]
5
[1, 2, 3, 4, 5]
and the output is:
[[1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5]]
Where do I make wrong in the code?
b is always the same list object (note I've changed print c to return c):
>>> map(id, Child().get_lines())
...
[49021616, 49021616, 49021616, 49021616, 49021616]
c contains five references to the same list. I think what you want is:
class Child(object):
def get_lines(self):
a = [1, 2, 3, 4, 5]
return [a[:x+1] for x in range(len(a))]
This makes a shallow copy of part (or all) of a for each step:
>>> Child().get_lines()
[[1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 3, 4, 5]]
Replace:
c.insert(j,b)
with:
c.append(b[:])
and try again.
You need to make a copy of b. Otherwise you add the same b again and again resulting in the full list at all indices. 'b[:]' copies the list.
This solution does the same but is a bit shorter:
a = [1, 2, 3, 4, 5]
c = []
for i in range(1, len(a) + 1):
c.append(a[:i])
now c is:
[[1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 3, 4, 5]]
a[:i] slices the list a from the beginning to excluding the index i.
In this case it does a[:1], a[:2] and so on. a[:1] makes a new list [1],
a[:2] a new list [1, 2,] and so on. Using append() is simpler and insert().
Another alternative is a list comprehension:
a = [1, 2, 3, 4, 5]
[a[:i] for i in range(1, len(a) + 1)]
also results in:
[[1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 3, 4, 5]]
Here in this case you are not supposed to append value to the list object (b).
since list is mutable, it will refer to the exact same object in memory until a reassign occurs.
>>> b=[]
>>> b.append(1)
>>> id(b)
4337935424
>>> b.append(2)
>>> id(b)
4337935424
>>> b.append(3)
>>> id(b)
4337935424
>>> b = [1, 2, 3, 4]
>>> id(b)
4337942608
so that in your code c will make five references to the same list.
It will instruct a new object and then (to the extent possible) inserts references into it to the objects found in the original.
>>> class Child(object):
...
... def get_lines(self):
... a = [1, 2, 3, 4, 5]
... return map(lambda x: a[:x+1], range(len(a)))
...
>>> Child().get_lines()
[[1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 3, 4, 5]]
You insert same b five times in list c. As a list actually contains references to objects and not copies, you have 5 times a reference to a same b list in which you have added successively 1, 2, 3, 4, 5. So the result.
You must instead add copies of list b :
def get_lines(self):
a = [1,2,3,4,5]
b=[]
c=[]
j=0
for i in a:
print i
b.append(i)
print b
c.insert(j,b[:]) # forces insertion of a copy
j=j+1
print c
In the loop, value of b persists since it's a mutable object. Hence when you print c, the last value of b is shown.
Instead of using b as temporary variable, you can directly use a as follows:
class Child(object):
def get_lines(self):
a = [1,2,3,4,5]
b = []
for index, element in enumerate(a, 1):
b.append(x[:index])
return b
son= Child()
son.get_lines()
This question already has answers here:
How to group a list of tuples/objects by similar index/attribute in python?
(3 answers)
Closed 8 years ago.
I see splitting-a-list-of-arbitrary-size-into-only-roughly-n-equal-parts. How about not-equal splitting? I have list having items with some attribute (value which can be retrieved for running same function against every item), how to split items having same attribute to be new list e.g. new sublist? Something lambda-related could work here?
Simple example could be:
list = [1, 1, 1, 2, 3, 3, 3, 3, 4, 4]
After fancy operation we could have:
list = [[1, 1, 1], [2], [3, 3, 3, 3], [4, 4]]
>>> L = [1, 1, 1, 2, 3, 3, 3, 3, 4, 4]
>>> [list(g) for i, g in itertools.groupby(L)]
[[1, 1, 1], [2], [3, 3, 3, 3], [4, 4]]
>>> L2 = ['apple', 'aardvark', 'banana', 'coconut', 'crow']
>>> [list(g) for i, g in itertools.groupby(L2, operator.itemgetter(0))]
[['apple', 'aardvark'], ['banana'], ['coconut', 'crow']]
You should use the itertools.groupby function from the standard library.
This function groups the elements in the iterable it receives (by default using the identity function, i.e., checking consequent elements for equality), and for each streak of grouped elements, it reutrns a 2-tuple consisting of the streak representative (the element itself), and an iterator of the elements within the streak.
Indeed:
l = [1, 1, 1, 2, 3, 3, 3, 3, 4, 4]
list(list(k[1]) for k in groupby(l))
>>> [[1, 1, 1], [2], [3, 3, 3, 3], [4, 4]]
P.S. you should avoid using list as a variable name, as it would conflict with the built-in type/function.
Here's a pretty simple roll your own solution. If the 'attribute' in question is simply the value of the item, there are more straightforward approaches.
def split_into_sublists(data_list, sizes_list):
if sum(sizes_list) != len(data_list):
raise ValueError
count = 0
output = []
for size in sizes_list:
output.append(data_list[count:count+size])
count += size
return output
if __name__ == '__main__':
data_list = [1, 1, 1, 2, 3, 3, 3, 3, 4, 4]
sizes_list = [3,1,4,2]
list2 = [[1, 1, 1], [2], [3, 3, 3, 3], [4, 4]]
print(split_into_sublists(data_list, sizes_list) == list2) # True
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.