Develop Reference

Changing elements of a list of lists [duplicate]

This question already has answers here:
How to add 1 to every element of a matrix / nested list in Python?
(4 answers)
Closed 1 year ago.
I need to add 0.05 to elements of coords array. But nothing happens.
Can you advise me where is the mistake? (I thought this would be easy but no)
coords = [[0.1,0.1,0.1],
[0.2,0.2,0.2],
[0.3,0.3,0.3]]
for i in coords:
for j in i:
j = j+0.05
print(coords)

Your solution doesn't actually modify the elements in the list by iterating over j. You could try a double list comprehension like this:
coords[:] = [j+0.05 for i in coords for j in i]
This has the advantage of editing the original object without creating a new instance of a list.

You current way will not store the results, as you did not provide a list to hold the results. I can show you two more straightforward examples:
a = [1,2,3,4]
for number in a:
number = number + 1
# check results
# nothing happens
print(a)
But if you do:
# you will get results
b = [number + 1 for number in a]
print(b)
The only difference is that you need to provide a list to hold the data.

Try this:
coords = [[0.1,0.1,0.1],
[0.2,0.2,0.2],
[0.3,0.3,0.3]]
for i in range(len(coords)):
for j in range(len(coords[i])):
coords[i][j] += .05
print(coords)

You can use lst comprehension
coords = [[0.1,0.1,0.1],
[0.2,0.2,0.2],
[0.3,0.3,0.3]]
newCoords = [[num+0.05 for num in lst] for lst in coords ]
print(newCoords)

How to create list and initialize only some of it's values in Python (in comparison to C++)

I need to create an List of size N and then initialize only N-1th and N-2th elements only. Which means if the size of the list is 5 then it should only contain elements in 3rd and 4th position.
i know how to do it in C++ but is there any way to implement it in Python?
for example: In C++
int *n = new int[5];
n[3] = 20
n[4] = 10
//and if we print the output it will show some garbage values in index 0, 1, 2 and will print 20 10 which is the values we initailized
How can i do it in python? or anything similar to this!

In python, list must be initialized with values.
Closest thing you can do:
N = 5
lst = [0] * (N-2) + [20, 10]
This:
Fills the N-2 elements of a list with default value 0
Sets the value for the last two elements
Concatenates the zeros and last two elements sub-lists of stages 1 & 2

In python,
array=[]
length=5
for i in range(length):
array.append(0)
array[3]=20
array[4]=10
Edit: As pointed out by kabanus, a more efficient way to do this would be-
array=[0]*length
Instead of the for loop.

Python slicing behavior using for loop

I am trying to use a for loop to create a series from entries in a dataframe, however I am having difficulties getting the last element of the dataframe.
import pandas as pd
a = pd.DataFrame([0,1,2,3,4,5,6,7])
a.values[-1] # returns 7
a.values[-5:-1]# returns 3,4,5,6
a.values[-5:]# returns 3,4,5,6,7
b = []
for i in range(0,len(a)):
b.append(a.values[-(i+1):-i])
I would like for 7 to be included in the list b. I realize that when i=0, a.values gives an empty array, however I'm not sure how to fix this as I can't iterate i to be blank as shown above.

Why not just manually convert negative indices to positive indices?
for i in range(0, len(a)):
b.append(a.values[(len(a) - (i+1) - 1):(len(a) - i)])
This works because python doesn't care if your upper limit in a slice is out-of-bounds, so long as the lower-limit is in bounds.
But if you're just trying to reverse the list (which it seems like you're trying to do), have you considered b = reversed(a)?

I would like for 7 to be included in the list b
import pandas as pd
a = pd.DataFrame([0,1,2,3,4,5,6,7])
b = []
for i in range(0, len(a)):
b.append(a.iloc[i])
OR, As a one-liner:
b = [a.iloc[i] for i in range(0, len(a))]

This will remove the blank and include 7 in the list:
for i in range(0,len(a)):
b.append(a.values[-i-1])

How to get mean from list that is partially empty example [0,2,4,0,4,0,0,0,0,10] [duplicate]

This question already has answers here:
Computing average of non-zero values
(3 answers)
Closed 3 years ago.
How to get mean from list that is partially empty example [0,2,4,0,4,0,0,0,0,10]
So mean should be (2+4+4+10)/4 = 5 , but statistics.mean() divides by the overall amount of numbers in the list (which in this case is 10).
So in my case I need to get mean from the list of numbers which are divisible by 3 (list b)
import random
import statistics
a = []
b = []
c = [0,2,4,0,4,0,0,0,0,10]
for x in range (10):
a.append(random.randint(0,101))
for x in a:
if x%3==0:
b.append(x)
else:
b.append(0)
average = statistics.mean(b)
print('a',a)
print('b',b)
print('Srednia',average)

You can filter out the zeros from the list using a for-loop perhaps
from statistics import mean
c = [0,2,4,0,4,0,0,0,0,10]
#Filter out 0's
non_zero_c = [item for item in c if item]
#Calculate mean from non zero numbers
print(mean(non_zero_c))
The output will be 5

In alternative to creating a list of only non-zero elements you can count them
non_zero_mean = sum(x) / sum(element != 0 for element in x)
here I'm using the "trick" that from a math point of view in Python (like in C and C++) True == 1 and False == 0.
Another alternative is
non_zero_mean = sum(x) / (len(x) - x.count(0))

Very quickly, the easiest way is to use some list comprehension to select the part of the list that interest you.
Example:
"How to get mean from list that is partially empty example [0,2,4,0,4,0,0,0,0,10] "
a = [0,2,4,0,4,0,0,0,0,10]
a_selection = [elt for elt in a if elt != 0]
m = sum(a_selection)/len(a_selection)
Same for divisible by 3:
b_selected = [elt for elt in b if elt%3==0]

Separate elements of array into two arrays [duplicate]

This question already has answers here:
How to remove items from a list while iterating?
(25 answers)
Closed 5 years ago.
I did one simple code to separate elements of an array into two new arrays : one with odd numbers and other with even numbers. So I did this:
V=[1,2,3,4,5,6]
vp=[]
vi=[]
for x in V:
if x%2==0:
vp.append(x)
V.remove(x)
else:
vi.append(x)
V.remove(x)
print (V)
print (vp)
print (vi) # sorry for the bad identation first time sharing code here
and this code give me this result:
[2,4,6]
[]
[1,3,5]
How is it happen? How am I fix this?

You shouldn't remove item while traversing an array:
V=[1,2,3,4,5,6]
vp=[]
vi=[]
for x in V:
if x%2==0:
vp.append(x)
else:
vi.append(x)

Modifying a list mid-iteration causes misbehavior (you effectively skip input elements). Don't remove from V as you go (which for long V would be expensive, each remove is O(n) making total work O(n**2)), just leave V unmodified. If necessary, clear V when you finish (a single O(n) operation), e.g. after the loop:
del V[:]

When you remove items from a list, the list gets shorter. So when you're looping over a list in a forward direction, removing items will cause the the iterator to skip forward.
To mitigate this, you can loop backwards over a list and safely remove items, since you're removing them from the end.
V = [1,2,3,4,5,6]
vp = []
vi = []
# reverse the list before iterating
for x in reversed(V):
if x % 2 == 0:
vp.append(x)
V.remove(x)
else:
vi.append(x)
V.remove(x)

Other answers that identified the problem and the fix, but here is a different way to do it using list comprehensions.
numbers = [1,2,3,4,5,6]
even = [e for e in numbers if e % 2 == 0]
odd = [e for e in numbers if e % 2 == 1]

A more concise way to make a new odd and even list from the original is to use comprehensions:
v = [1,2,3,4,5,6]
even = [number for number in v if number % 2 == 0]
odd = [number for number in v if number % 2 != 0]

Don't remove items from a list over which you are iterating. Use a copy:
V=[1,2,3,4,5,6]
vp=[]
vi=[]
for x in V[:]:
if x%2==0:
vp.append(x)
V.remove(x)
else:
vi.append(x)
V.remove(x)
print (V)
print (vp)
print (vi)
# []
# [2, 4, 6]
# [1, 3, 5]