multiply a list of numbers with 10 - python-3.x

I have a list of numbers t. I want to multiply the numbers in the list with 10. Why does this not work?:
for i in t
i = i*10
Why do I have to do this?:
for i in range(len(t)):
t[i] = t[i]*10

Well, it doesn't work because that's no the correct syntax. You could, however, clean things up a bit with a list comprehension:
t = [x * 10 for x in t]

It doesn't work that way in Python.
The index variable (your i) of the for .. in construct is just a variable. At the start of each pass through the loop, Python assigns the corresponding value from the in sequence to the loop's index variable. It's as if you had a lot of copies of the loop body, and before each copy you put an assignment statement i = t[0], i = t[1], and so on. The index variable does not remember that it was assigned from the in sequence (your t). The index variable is not an equivalent name (an "alias") for the corresponding value of the in sequence. Changing the index variable does not affect the in sequence.
python -c 't = [1, 2, 3]
for i in t:
i = 1
print t'
[1, 2, 3]
But you're not wrong to wonder whether it's an alias! Another language that is frequently compared with Python does work that way (quoting manual page "perlsyn"):
If any element of LIST is an lvalue, you can modify it by modifying VAR inside the loop. Conversely, if any element of LIST is NOT an lvalue, any attempt to modify that element will fail. In other words, the foreach loop index variable is an implicit alias for each item in the list that you're looping over.
So:
perl -e '#t = (1, 2, 3); foreach $i (#t) { $i = 1; } print #t'
111

Related

The best way of iterating through an array whose length changes in Python

I am implementing an algorithm which might affect the size of some array, and I need to iterate through the entire array. Basically a 'for x in arrayname' would not work because it does not update if the contents of arrayname are changed in the loop. I came up with an ugly solution which is shown in the following example:
test = np.array([1,2,3])
N = len(test)
ii=0
while ii < N:
N = len(test)
print(test[ii])
if test[ii] ==2:
test = np.append(test,4)
ii+=1
I am wondering whether a cleaner solution exists.
Thanks in advance!
Assuming all the elements are going to be added at the end and no elements are being deleted you could store the new elements in a separate list:
master_list = [1,2,3]
curr_elems = master_list
while len(curr_elems) > 0: # keep looping over new elements added
new_elems = []
for item in curr_elems: # loop over the current list of elements, initially the list but then all the added elements on second run etc
if should_add_element(item):
new_elems.append(generate_new_element(item))
master_list.extend(new_elems) # add all the new elements to our master list
curr_elems = new_elems # and prep to iterate over the new elements for next iteration of the while loop
The while loop seems the best solution. As the condition is re-evaluated at each iteration, you don’t need to reset the length of the list in the loop, you can do it inside the condition:
import random
l = [1, 2, 3, 4, 5]
i = 0
while i < len(l):
if random.choice([True, False]):
del l[i]
else:
i += 1
print(f'{l=}')
This example gives a blueprint for a more complex algorithm. Of course, in this simple case, it could be coded more simply with a filter, or like this:
l = [1, 2, 3, 4, 5]
[x for x in l if random.choice([True, False])]
You might want to check this related post for more creative solutions: How to remove items from a list while iterating?

If condition working differently for same value in python

I am trying to write a function which will return True or False if the given number is not greater than 2.
So simple, but the if condition is returning different outputs for same value '2'. The code I used is:
The code I used is:
ele_list = [1,2,3,2]
for i in ele_list:
if not i>2:
print(i,False)
ele_list.remove(i)
print(ele_list)
The ouput I am receiving is:
1 False
[2, 3, 2]
2 False
[3, 2]
I am confused to see that the first 2 in the list is passing through the if condition but the second 2 in the list is not passing through the condition. Please help me figure out this..
Removing elements from the list you're looping over is generally a bad idea.
What's happening here is that when you're removing an element, you're changing the length of the array, and therefor changing what elements are located at what indexes as well as changing the "goal" of the forloop.
Lets have a look at the following example:
ele_list = [4,3,2,1]
for elem in ele_list:
print(elem)
ele_list.remove(elem)
In the first iteration of the loop elem is the value 4 which is located at index 0. Then you're removing from the array the first value equal to elem. In other words the value 4 at index 0 is now removed. This shifts which element is stored at what index. Before the removal ele_list[0] would be equal to 4, however after the removal ele_list[0] will equal 3, since 3 is the value that prior to the removal was stored at index 1.
Now when the loop continues to the second iteration the index that the loop "looks at" is incremented by 1. So the variable elem will now be the value of ele_list[1] which in the updated list (after the removal of the value 4 in the previous iteration) is equal to 2. Then you're (same as before) removing the value at index 1 from the list, so now the length of the list just 2 elements.
When the loops is about to start the third iteration it checks to see if the new index (in this case 2) is smaller than the length of the list. Which its not, since 2 is not smaller than 2. So the loop ends.
The simplest solutions is to create a new copy of the array and loop over the copy instead. This can easily be done using the slice syntax: ele_list[:]
ele_list = [1,2,3,2]
for elem in ele_list[:]:
if not elem > 2:
print(elem, False)
ele_list.remove(elem)
print(ele_list)
the problem is that you're modifying your list as you're iterating over it, as mentioned in #Olian04's answer.
it sounds like what you really want to do, however, is only keep values that are > 2. this is really easy using a list comprehension:
filtereds_vals = [v for v in ele_list if v > 2]
if you merely want a function that gives you True for numbers greater than 2 and False for others, you can do something like this:
def gt_2(lst):
return [v > 2 for v in lst]
or, finally, if you want to find out if any of the values is > 2 just do:
def any_gt_2(lst):
return any(v > 2 for v in lst)
I think the problem here is how the remove function interacts with the for function.
See the documentation, read the "note" part:
https://docs.python.org/3.7/reference/compound_stmts.html?highlight=while#grammar-token-for-stmt
This can lead to nasty bugs that can be avoided by making a temporary copy using a slice of the whole sequence
A possible solution, as suggested into the documentation:
ele_list = [1,2,3,2]
for i in ele_list[:]:
if not i>2:
print(i,False)
ele_list.remove(i)
print(ele_list)
"""
1 False
[2, 3, 2]
2 False
[3, 2]
2 False
[3]
"""

python3 functional programming: Accumulating items from a different list into an initial value

I have some code that performs the following operation, however I was wondering if there was a more efficient and understandable way to do this. I am thinking that there might be something in itertools or such that might be designed to perform this type of operation.
So I have a list of integers the represents changes in the number of items from one period to the next.
x = [0, 1, 2, 1, 3, 1, 1]
Then I need a function to create a second list that accumulates the total number of items from one period to the next. This is like an accumulate function, but with elements from another list instead of from the same list.
So I can start off with an initial value y = 3.
The first value in the list y = [3]. The I would take the second
element in x and add it to the list, so that means 3+1 = 4. Note that I take the second element because we already know the first element of y. So the updated value of y is [3, 4]. Then the next iteration is 4+2 = 6. And so forth.
The code that I have looks like this:
def func():
x = [0, 1, 2, 1, 3, 1, 1]
y = [3]
for k,v in enumerate(x):
y.append(y[i] + x[i])
return y
Any ideas?
If I understand you correctly, you do what what itertools.accumulate does, but you want to add an initial value too. You can do that pretty easily in a couple ways.
The easiest might be to simply write a list comprehension around the accumulate call, adding the initial value to each output item:
y = [3 + val for val in itertools.accumulate(x)]
Another option would be to prefix the x list with the initial value, then skip it when accumulate includes it as the first value in the output:
acc = itertools.accumulate([3] + x)
next(acc) # discard the extra 3 at the start of the output.
y = list(acc)
Two things I think that need to be fixed:
1st the condition for the for loop. I'm not sure where you are getting the k,v from, maybe you got an example using zip (which allows you to iterate through 2 lists at once), but in any case, you want to iterate through lists x and y using their index, one approach is:
for i in range(len(x)):
2nd, using the first append as an example, since you are adding the 2nd element (index 1) of x to the 1st element (index 0) of y, you want to use a staggered approach with your indices. This will also lead to revising the for loop condition above (I'm trying to go through this step by step) since the first element of x (0) will not be getting used:
for i in range(1, len(x)):
That change will keep you from getting an index out of range error. Next for the staggered add:
for i in range(1, len(x)):
y.append(y[i-1] + x[i])
return y
So going back to the first append example. The for loop starts at index 1 where x = 1, and y has no value. To create a value for y[1] you append the sum of y at index 0 to x at index 1 giving you 4. The loop continues until you've exhausted the values in x, returning accumulated values in list y.

Switching positions of two strings within a list

I have another question that I'd like input on, of course no direct answers just something to point me in the right direction!
I have a string of numbers ex. 1234567890 and I want 1 & 0 to change places (0 and 9) and for '2345' & '6789' to change places. For a final result of '0678923451'.
First things I did was convert the string into a list with:
ex. original = '1234567890'
original = list(original)
original = ['0', '1', '2' etc....]
Now, I get you need to pull the first and last out, so I assigned
x = original[0]
and
y = original[9]
So: x, y = y, x (which gets me the result I'm looking for)
But how do I input that back into the original list?
Thanks!
The fact that you 'pulled' the data from the list in variables x and y doesn't help at all, since those variables have no connection anymore with the items from the list. But why don't you swap them directly:
original[0], original[9] = original[9], original[0]
You can use the slicing operator in a similar manner to swap the inner parts of the list.
But, there is no need to create a list from the original string. Instead, you can use the slicing operator to achieve the result you want. Note that you cannot swap the string elements as you did with lists, since in Python strings are immutable. However, you can do the following:
>>> a = "1234567890"
>>> a[9] + a[5:9] + a[1:5] + a[0]
'0678923451'
>>>

How do you modify a variable that's a value in a dictionary when calling that variable by its key?

n = 3
d = {'x':n}
d['x'] += 1
print(n)
When I run it, I get
3
How do I make n = 4?
You can't do this, at least, not in any simple way.
The issue is very similar when you're just dealing with two variables bound to the same object. If you rebind one of them with an assignment, you will not see the new value through the other variable:
a = 3
b = a
a += 1 # binds a to a new integer, 4, since integers are immutable
print(b) # prints 3, not 4
One exception is if you are not binding a new value to the variable, but instead modifying a mutable object in-place. For instance, if instead of 1 you has a one-element list [1], you could replace the single value without creating a new list:
a = [3]
b = a
a[0] += 1 # doesn't rebind a, just mutates the list it points to
print(b[0]) # prints 4, since b still points to the same list as a
So, for your dictionary example you could take a similar approach and have n and your dictionary value be a list or other container object that you modify in-place.
Alternatively, you could store the variable name "n" in your dictionary and then rather than replacing it in your other code, you could use for a lookup in the globals dict:
n = 3
d = {"x": "n"} # note, the dictionary value is the string "n", not the variable n's value
globals()[d["x"]] += 1
print(n) # this actually does print 4, as you wanted
This is very awkward, of course, and only works when n is a global variable (you can't use the nominally equivalent call to locals in a function, as modifying the dictionary returned by locals doesn't change the local variables). I would not recommend this approach, but I wanted to show it can be done, if only badly.
You could use a class to contain the data values to enable additions. Basically you are creating a mutable object which acts as an integer.
It is a work around, but lets you accomplish what you want.
Note, that you probably need to override a few more Python operators to get full coverage:
class MyInt(object):
val = 0
def __init__(self,val):
self.val = val
def __iadd__(self,val):
self.val = self.val + val
def __repr__(self):
return repr(self.val)
n = MyInt(3)
print(n)
d = {'x':n}
d['x'] += 1
print(n)

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