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Fixes to questions 2, 9, 18, 31, 35, 44, 52, 56, 66 & 78 #106

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@Renelvon Renelvon commented Mar 14, 2020

Fixes #98
Fixes #101

I have added a few more fixes and simplifications, let me know what you think.

I ran all the generators; some of the output is due to older commits.

@rougier
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rougier commented Mar 18, 2020

Thanks. There's a new format where only file has to be changed. Could you use it ? Also, it would be better to have one PR per proposed changed (would make discussion easier).

@Renelvon
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I only updated the .ktx file manually, the rest was done by running the generator in the last commit. Should I only submit changes in the .ktx file? I will close the PR and resubmit one PR per fix once you clarify. :)

@@ -72,8 +72,8 @@ print(Z)
`hint: reshape`

```python
nz = np.nonzero([1,2,0,0,4,0])
print(nz)
Z = np.arange(9).reshape((3, 3))
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Why this change ?

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The current solution corresponds to question 10.

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My bad... Thanks for fixing it!

B = np.ones(3)*2
C = np.ones(3)*3
A = np.ones(3)
B = np.full(3, 2)
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I prefer the version with ones (in order to keep it simple)

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Okay, I will rebase and drop commit. Should the C matrix stay as well?

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ok


```python
Z = np.random.random((10,2))
X,Y = Z[:,0], Z[:,1]
R = np.sqrt(X**2+Y**2)
R = np.hypot(Y,X)
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Y,X -> X,Y (for consistency)

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The order is such so that it's consistent with the order of arguments in the arctan2 invocation. It's my personal mnemonic, but I will change it if you insist.

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I would prefer yes.

@@ -673,10 +673,10 @@ print(F)
# Author: Nadav Horesh

w,h = 16,16
I = np.random.randint(0,2,(h,w,3)).astype(np.ubyte)
I = np.random.randint(0,2,(h,w,3))
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Why did you remove the cast to ubyte ?

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In a sufficiently long array, the expected number of different "colors" should be 2**3 == 8. For the solution to work, F must be a matrix of at least 24-bit integers, but this is not what happens on my machine: I[..., 1] * 256 has type np.uint16, and the same happens with I[..., 0] * 256 * 256, leading to overflow. Writing (I[..., 0] * 256) * 256 doesn't solve the problem. A cast to a sufficient wide type has to happen at some point, or the initial cast has to go.

>>> import numpy as np
>>> w, h = 16, 16
>>> I = np.random.randint(0, 2, (h, w, 3))
>>> I2 = I.astype(np.ubyte)
>>> F = I[...,0]*256*256 + I[...,1]*256 +I[...,2]
>>> F2 = I2[...,0]*256*256 + I2[...,1]*256 +I2[...,2]
>>> print(len(np.unique(F)), len(np.unique(F2)))
8 4
>>> print(F.dtype, F2.dtype)
int64 uint16

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ok

@@ -826,9 +826,9 @@ np.negative(Z, out=Z)
def distance(P0, P1, p):
T = P1 - P0
L = (T**2).sum(axis=1)
U = -((P0[:,0]-p[...,0])*T[:,0] + (P0[:,1]-p[...,1])*T[:,1]) / L
U = ((P0 - p)*T).sum(axis=1) / L
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Did you check this is correct ?

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?

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No.9's solution is wrong in 100_Numpy_exercises_with_hints_with_solutions.md file
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