We Keep Coding

How to define a constant function defined in intervals in python?

I want to define a simple function which assumes different constant value (y=[1,4,2,3]) for defined intervals.
I implement it in this way:
import numpy as np
def f(x):
if (x>=0 and x<=1900):
return 1
if (x>1900 and x<=3600):
return 4
if (x>3600 and x<=5400):
return 2
if (x>5400 and x<=7200):
return 3
x=np.linspace(0,7200,1000)
y=f(x)
However, when I run the script, an error appears:
"ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()"
Do you know how to fix this?

The reason is that your function is only applicable to a single element rather than vectorization. np.vectorize is a general solution, but its performance is poor. For the example here, you can use np.searchsorted to vectorize:
>>> np.array([1, 1, 4, 2, 3])[np.searchsorted([0, 1900, 3600, 5400, 7200], x)]
array([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, 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, 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, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4,
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4,
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4,
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4,
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4,
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4,
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4,
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4,
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4,
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4,
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,
3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,
3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,
3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,
3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,
3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,
3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,
3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,
3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,
3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,
3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,
3, 3, 3, 3, 3, 3, 3, 3, 3, 3])

x is not what you think it is.
Try print(x) and see what it is actually looking like. It is first up a list and inside the list contains
[ 0. 7.20720721 14.41441441 21.62162162 28.82882883 .... 7192.79279279 7200. ]
I am unsure what you are trying to achieve but x is an array and not a single value, therefor you either need to loop over it or point to the exact index you want to test.

Numpy, duplicating data in different order

I have data for example
>>> a = np.array([1, 2, 3, 4])
>>> b = np.array([3, 4, 5, 6])
I want to duplicate each item in each vector to the value of the length of the vector. So the results can are
>>> a2 = np.array([1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4])
>>> b2 = np.array([3, 4, 5, 6, 3, 4, 5, 6, 3, 4, 5, 6, 3, 4, 5, 6])
Using np.tile(b, len(b)) can output b2. However, how can I get a2?

The two replications are a bit different. The first one can be obtained with .repeat(..) [numpy-doc]:
>>> a.repeat(len(a))
array([1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4])
the second one with .tile(..) [numpy-doc]:
>>> np.tile(b, len(b))
array([3, 4, 5, 6, 3, 4, 5, 6, 3, 4, 5, 6, 3, 4, 5, 6])

You can do both in one go using np.meshgrid
A,B = map(np.ravel,np.meshgrid(a,b,indexing='ij'))
A
# array([1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4])
B
# array([3, 4, 5, 6, 3, 4, 5, 6, 3, 4, 5, 6, 3, 4, 5, 6])

Converting a heatmap image in the JPEG format to categories in python

I have a massive heatmap photo in the JPEG format and I'd like to convert the entries to 5 different categories based on the color of each one, and store the final result in three 2D-numpy arrays (One for section A, one for section B, one for section C).
The categories are as follows:
Red: 5,
Dark brown: 4,
Light brown: 3,
Light orange: 2,
White: 1
You can see the cropped photo that shows only a few top rows in the JPEG format in the following image:
How can I do this in python?

I would recommend you install and use OpenCV to help with the processing. This also provides a Python library to work with.
The following works for the heatmap you have given. You will need to modify it accordingly for your massive heatmap.
It first crops the image into 3, one for A, B and C respectively. Next it extracts the pixels for each category colour, converts them first to white and then resizes each to a 16 x 16 greyscale image. Next it converts non zero values to the current category value and adds this to your array.
import numpy as np
import cv2
def parse_image(img):
# Start with all zeros in a 16x16 array
data = np.zeros((16, 16), dtype=int)
# List holding min and max BGR values for each category
categories = [
(5, (0, 0, 250), (5, 5, 255)), # Red
(4, (5, 60, 152), (12, 78, 160)), # Dark brown
(3, (9, 105, 225), (18, 120, 240)), # Light brown
(2, (220, 235, 250), (230, 245, 255)), # Light orange
(1, (250, 250, 250), (255, 255, 255)), # White
]
for category, bgr_min, bgr_max in categories:
# Extract pixels in the required range and convert them to 255
mask = cv2.inRange(img, bgr_min, bgr_max)
image_cat = cv2.bitwise_or(img, np.full(img.shape, 255, dtype=np.uint8), mask=mask)
# Convert the image into greyscale
image_grey = cv2.cvtColor(image_cat, cv2.COLOR_BGR2GRAY)
# Resize the image to 16x16
values = cv2.resize(image_grey, (16, 16))
# Convert non black values into the current category value
values[values > 0] = category
# Add the values to the data array
data = data + values
return data
# Load the heatmap
image_src = cv2.imread("heatmap.jpg")
cv2.imshow("Source", image_src)
# Crop into 3 sub images
starty = 28
cropx = 234
cropy = 184
images = []
for number, startx in enumerate([30, 303, 572], start=1):
images.append(image_src[starty:starty+cropy,startx:startx+cropx])
# Parse A, B and C
abc = [parse_image(img) for img in images]
print abc
So for the heatmap you provided you would get the following output (which could be reshaped into a single array):
[array([[2, 2, 5, 3, 5, 3, 5, 4, 2, 5, 2, 2, 5, 2, 2, 2],
[2, 2, 2, 3, 2, 2, 4, 3, 2, 4, 2, 2, 5, 2, 2, 2],
[2, 2, 2, 2, 2, 2, 3, 2, 3, 3, 2, 2, 5, 2, 2, 2],
[4, 3, 2, 5, 4, 2, 5, 5, 5, 5, 2, 2, 5, 2, 2, 2],
[2, 3, 2, 4, 2, 2, 5, 3, 3, 5, 2, 2, 5, 2, 2, 2],
[3, 4, 3, 5, 4, 2, 5, 5, 5, 5, 2, 2, 5, 2, 2, 2],
[4, 4, 5, 5, 4, 2, 5, 5, 5, 5, 2, 2, 5, 2, 2, 2],
[5, 5, 5, 5, 5, 3, 2, 2, 5, 2, 2, 2, 5, 3, 5, 5],
[3, 4, 5, 4, 5, 2, 5, 5, 5, 5, 2, 2, 5, 2, 3, 5],
[3, 3, 5, 5, 4, 2, 5, 5, 3, 5, 2, 2, 5, 2, 2, 3],
[4, 4, 5, 5, 5, 5, 2, 2, 5, 2, 2, 2, 5, 2, 2, 5],
[3, 2, 2, 2, 3, 2, 4, 5, 5, 3, 2, 2, 5, 2, 2, 5],
[1, 1, 2, 2, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 2],
[2, 2, 2, 2, 2, 0, 2, 2, 2, 2, 2, 2, 5, 2, 2, 2],
[4, 3, 2, 5, 5, 2, 5, 5, 2, 5, 2, 2, 5, 2, 2, 2],
[2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 5, 2, 2, 2]]), array([[2, 3, 3, 3, 2, 2, 2, 2, 0, 3, 2, 3, 2, 2, 2, 3],
[2, 3, 2, 3, 2, 3, 3, 2, 0, 3, 2, 2, 2, 2, 2, 3],
[2, 2, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 2, 2, 2],
[2, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 2, 2, 2, 2],
[2, 3, 2, 3, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 0, 3],
[2, 2, 2, 3, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2],
[2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2],
[2, 2, 3, 3, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2],
[3, 3, 4, 3, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 2, 3],
[3, 3, 4, 3, 0, 3, 3, 2, 2, 2, 2, 3, 2, 2, 2, 3],
[2, 0, 3, 3, 0, 2, 3, 2, 0, 2, 3, 3, 2, 2, 2, 2],
[2, 3, 2, 3, 3, 2, 3, 2, 0, 2, 3, 3, 2, 2, 2, 2],
[1, 1, 2, 1, 0, 2, 1, 1, 3, 1, 2, 2, 1, 2, 2, 0],
[2, 2, 2, 2, 2, 2, 0, 2, 2, 2, 2, 2, 2, 2, 2, 2],
[3, 2, 2, 3, 2, 2, 3, 2, 3, 3, 2, 2, 2, 2, 2, 3],
[2, 2, 2, 2, 2, 2, 3, 2, 3, 2, 2, 2, 2, 2, 2, 2]]), array([[2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2],
[2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 3, 3, 2],
[2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2],
[2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2],
[2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2],
[2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2],
[2, 2, 0, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2],
[2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 5],
[2, 2, 5, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2],
[2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 3, 2],
[2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2],
[2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2],
[1, 1, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 3, 2, 2],
[2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 2],
[2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2],
[2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2]])]
Note, working with JPG images means there will be artefacts in image, so colours will not have exact values. This is why there are min and max values. Also normally you will see RGB values. But here you will be working with BGR values instead. You will need to tweak the min max values to ensure you do not get any 0 values.

Find index of non-equal neighbors (adjacent elements) of the list

I would like to create a list that:
iterates through each value of List1
populates List2 with the index number where the value differs from the previous index number
print List1 yields:
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 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, 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, 1,11, 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, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 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4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3]
Here is the code I have so far:
List2 = []
ArraySize = len(List1)
index=0
for c in range (1,ArraySize):
if ArraySize[c] != ArraySize[c-1]:
c=List2[index]
index=index+1
This provides the error:
if ArraySize[c] != ArraySize[c-1]:
TypeError: 'int' object has no attribute '__getitem__'
I don't want to alter List_1 at all, because I use it to produce a graph later in the script.
List2 for these values in List1 should end up looking like:
[93,307,526,692,1106,1856,2043]
Any help is greatly appreciated! Thank you!

As the comment on your question says, you are using ArraySize in the wrong place, and assigning c in a way that doesn't do anything useful, you aren't updating List2.
And there's no need to keep a counter for the index when you can use enumerate() (also: What does enumerate mean? ) to do that for you, e.g.
List1 = [0, 0, ...]
List2 = []
for index, value in enumerate(List1):
if index > 0 and List1[index-1] != value:
List2.append(index)
Output is
> List2
=> [93, 218, 219, 307, 526, 692, 1106, 1856, 2043]
Try online at http://repl.it

In order to achieve this, you may use zip() and enumerate() along with conditional list comprehension expression as:
>>> [n for n, (i, j) in enumerate(zip(my_list, my_list[1:]), 1) if i!=j]
[93, 218, 219, 307, 526, 692, 1106, 1856, 2043]
Issue with you code:
You are assigning ArraySize = len(List1) which will hold int value. Then later you are doing ArraySize[c] i.e accessing it like a list. But since it hold integer value, you are getting the error:
TypeError: 'int' object has no attribute '__getitem__'
# Function called by list when you ^
# try to access the items based on index

Is there a built-In function in python to count the number of 1's in a number? [closed]

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.
Closed 10 years ago.
Please suggest an efficient algorithm if no such library function exists.

something like str(number).count("1") ?

Something like:
for baseconv in (int, oct, hex, bin):
as_text = str(baseconv(123456789))
print as_text, as_text.count('1')

For the number of '1's in the binary representation, an 8 bit lookup table is quite efficient. eg. for 32 bit numbers
>>> tab=[0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 4, 5, 5, 6, 5, 6, 6, 7, 5, 6, 6, 7, 6, 7, 7, 8]
>>> N=123456789
>>> tab[N>>24]+tab[N>>16&255]+tab[N>>8&255]+tab[N&255]
16