I checked out
gradient descent using python and numpy
but it didn't solve my problem.
I'm trying to get familiar with image-processing and I want to generate a few test arrays to mess around with in Python.
Is there a method (like np.arange) to create a m x n array where the inner entries form some type of gradient?
I did an example of a naive method for generating the desired output.
Excuse my generality of the term gradient, I'm using it in it's simple meaning as smooth transition in color.
#!/usr/bin/python
import numpy as np
import matplotlib.pyplot as plt
#Set up parameters
m = 15
n = 10
A_placeholder = np.zeros((m,n))
V_m = np.arange(0,m).astype(np.float32)
V_n = np.arange(0,n).astype(np.float32)
#Iterate through combinations
for i in range(m):
m_i = V_m[i]
for j in range(n):
n_j = V_n[j]
A_placeholder[i,j] = m_i * n_j #Some combination
#Relabel
A_gradient = A_placeholder
A_placeholder = None
#Print data
print A_gradient
#[[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[ 0. 1. 2. 3. 4. 5. 6. 7. 8. 9.]
[ 0. 2. 4. 6. 8. 10. 12. 14. 16. 18.]
[ 0. 3. 6. 9. 12. 15. 18. 21. 24. 27.]
[ 0. 4. 8. 12. 16. 20. 24. 28. 32. 36.]
[ 0. 5. 10. 15. 20. 25. 30. 35. 40. 45.]
[ 0. 6. 12. 18. 24. 30. 36. 42. 48. 54.]
[ 0. 7. 14. 21. 28. 35. 42. 49. 56. 63.]
[ 0. 8. 16. 24. 32. 40. 48. 56. 64. 72.]
[ 0. 9. 18. 27. 36. 45. 54. 63. 72. 81.]
[ 0. 10. 20. 30. 40. 50. 60. 70. 80. 90.]
[ 0. 11. 22. 33. 44. 55. 66. 77. 88. 99.]
[ 0. 12. 24. 36. 48. 60. 72. 84. 96. 108.]
[ 0. 13. 26. 39. 52. 65. 78. 91. 104. 117.]
[ 0. 14. 28. 42. 56. 70. 84. 98. 112. 126.]]
#Show Image
plt.imshow(A_gradient)
plt.show()
I've tried np.gradient but it didn't give me the desired output.
#print np.gradient(np.array([V_m,V_n]))
#Traceback (most recent call last):
# File "Untitled.py", line 19, in <module>
# print np.gradient(np.array([V_m,V_n]))
# File "/Users/Mu/anaconda/lib/python2.7/site-packages/numpy/lib/function_base.py", line 1458, in gradient
# out[slice1] = (y[slice2] - y[slice3])
#ValueError: operands could not be broadcast together with shapes (10,) (15,)
A_placeholder[i,j] = m_i * n_j
Any operation like that can be expressed in numpy using broadcasting
A = np.arange(m)[:, None] * np.arange(n)[None, :]
Related
I have the following output for a
[ 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27.
29. 31. 33. 35. 37. 39. 41. 43. 45. 47. 97. 99. 101. 103.
105. 107. 109. 111. 113. 115. 117. 119. 121. 123. 125. 127. 129. 131.
133. 135. 137. 139. 141. 143.]
I want to reshape it to the below
[[1. 3. 5. 7. 9. 11. 13. 15.]
[17. 19. 21. 23. 25. 27. 29. 31.]
[33. 35. 37. 39. 41. 43. 45. 47.]
[97. 99. 101. 103. 105. 107. 109. 111.]
[113. 115. 117. 119. 121. 123. 125. 127.]
[129. 131. 133. 135. 137. 139. 141. 143.]]
I tried to use a.resize(6, 8), but it gives me this error: "resize only works on single-segment arrays"
Also, when I am trying to use a.reshape(6, 8), it gives me the same array.
I don't understand what is the reason for that as I have tested another array and worked well.
try a.reshape((8, 6))
notice the double parentheses
a = np.array([1., 3., 5., 7., 9., 11., 13., 15., 17., 19., 21., 23., 25., 27.,
29., 31., 33., 35., 37., 39., 41., 43., 45., 47., 97., 99., 101., 103.,
105., 107., 109., 111., 113., 115., 117., 119., 121., 123., 125., 127., 129., 131.,
133., 135., 137., 139., 141., 143.])
print(a.reshape((8, 6)))
out:
[[ 1. 3. 5. 7. 9. 11.]
[ 13. 15. 17. 19. 21. 23.]
[ 25. 27. 29. 31. 33. 35.]
[ 37. 39. 41. 43. 45. 47.]
[ 97. 99. 101. 103. 105. 107.]
[109. 111. 113. 115. 117. 119.]
[121. 123. 125. 127. 129. 131.]
[133. 135. 137. 139. 141. 143.]]
Process finished with exit code 0
do notice that for the output you requested, the dimensions should be
a.reshape((6,8))
out:
[[ 1. 3. 5. 7. 9. 11. 13. 15.]
[ 17. 19. 21. 23. 25. 27. 29. 31.]
[ 33. 35. 37. 39. 41. 43. 45. 47.]
[ 97. 99. 101. 103. 105. 107. 109. 111.]
[113. 115. 117. 119. 121. 123. 125. 127.]
[129. 131. 133. 135. 137. 139. 141. 143.]]
Process finished with exit code 0
you can read about NumPy's reshape here: reshape documentation
Try
b = a.reshape((8,6))
and keep in mind 2 things, for future use of similar methods:
the reshape method takes a tuple as input, in that case (8,6) , calling b = a.reshape(8,6) gives 2 int arguments to the method instead of the tuple it expects. always pay attention to the expected values. you can investigate that by just hovering over a function in pycharm and most editors.
in numpy, many methods do not manipulate the given object but rather return a new value for you to use.
it is healthy to always check for that in documentation, in order to avoid catastrophic heartbreaks, trust me.
I have this file which has an array of data written to it:
[[[ 32. 28. 28. ... 24. 24. 24.]
[ 30. 29. 29. ... 24. 24. 24.]
[ 29. 29. 28. ... 24. 24. 24.]
...
[137. 138. 129. ... 34. 34. 34.]
[140. 139. 128. ... 31. 34. 34.]
[136. 135. 122. ... 30. 30. 33.]]
[[ 40. 40. 40. ... 33. 33. 33.]
[ 38. 38. 37. ... 33. 33. 33.]
[ 37. 37. 37. ... 33. 33. 33.]
...
[140. 137. 132. ... 41. 43. 42.]
[139. 136. 129. ... 42. 43. 43.]
[140. 139. 133. ... 40. 42. 43.]]
[[ 10. 8. 7. ... 4. 4. 4.]
[ 8. 7. 7. ... 4. 4. 4.]
[ 7. 6. 6. ... 4. 4. 4.]
...
[101. 103. 94. ... 12. 13. 13.]
[105. 104. 92. ... 12. 13. 13.]
[ 99. 99. 99. ... 9. 10. 11.]]]
I do not know how to read from this file and use it within my code. Any help would be great! I have this within my code so far:
# Read and pre-process input images
n, c, h, w = net.inputs[input_blob].shape
images = np.ndarray(shape=(n, c, h, w))
for i in range(n):
image = cv2.imread(args.input[i])
if image.shape[:-1] != (h, w):
log.warning("Image {} is resized from {} to {}".format(args.input[i], image.shape[:-1], (h, w)))
image = cv2.resize(image, (w, h))
# Swapping Red and Blue channels
#image[:, :, [0, 2]] = image[:, :, [2, 0]]
# Change data layout from HWC to CHW
image = image.transpose((2, 0, 1))
images[i] = image
eoim = image
eoim16 = eoim.astype(np.float16)
val = []
preprocessed_image_path = 'C:/Users/Owner/Desktop/Ubotica/IOD/cloud_detect/'
formated_image_file = "output_patch_fp"
f = open(preprocessed_image_path + "/" + formated_image_file + ".txt", 'r')
val = f
print(f)
print(val)
# divide by 255 to get value in range 0->1 if necessary (depends on input pixel format)
if(eoim16.max()>1.0):
eoim16 = np.divide(eoim16,255)
print(eoim16)
#f.close()
#print(val)
#val = np.reshape(val, (3,512,512))
eoim16 = np.ndarray(shape=(c, h, w))
#res = val
# calling the instance method using the object cloudDetector
res = cloudDetector.infer(eoim16)
res = res[out_blob]
But when I try to print out val and f (just to see if the data matches and is actually being read within my code nothing appears. Is there any way to solve this so that my array reads into val and I can use the data within my code? Much appreciated!
Try using the eval function. It takes strings and interprets them as Python code.
a = eval(fileData)
print(a)
In Tensorflow, SAME padding aims to produce a same sized output as the input, given a stride = 1, by padding the input with zeros as appropriate. For an odd sized kernel, for example like 5x5, it puts the center of the kernel (2,2) onto the first pixel of the input (0,0) and starts to convolve. Both in the x and y coordinates, 2 pixels of zero padding is needed then.
What if an even kernel, for example a 6x6 is used instead? It won't have a pixel's center as its actual center. How does VALID padding handle this? For example according to Image convolution with even-sized kernel the convention in the general image processing literature is to place one more pixel before the zero, like -3 -2 -1 0 1 2 in this case. Three pixel will be hit in the padding area. I refered to the Tensorflow documents for this, but could not find a clarifying answer.
Like you say, the documentation does not seem to specify it clearly. Looking at the source of the 2D convolution kernel (conv_ops.cc), a comment explains:
// Total padding on rows and cols is
// Pr = (R' - 1) * S + (Kr - 1) * Dr + 1 - R
// Pc = (C' - 1) * S + (Kc - 1) * Dc + 1 - C
// where (R', C') are output dimensions, (R, C) are input dimensions, S
// is stride, (Dr, Dc) are dilations, (Kr, Kc) are filter dimensions.
// We pad Pr/2 on the left and Pr - Pr/2 on the right, Pc/2 on the top
// and Pc - Pc/2 on the bottom. When Pr or Pc is odd, this means
// we pad more on the right and bottom than on the top and left.
So it seems you would get one extra padding at the right column and bottom row with even-sized kernels. We can look at one example:
import tensorflow as tf
input_ = tf.ones((1, 10, 10, 1), dtype=tf.float32)
kernel = tf.ones((6, 6, 1, 1), dtype=tf.float32)
conv = tf.nn.conv2d(input_, kernel, [1, 1, 1, 1], 'SAME')
with tf.Session() as sess:
print(sess.run(conv)[0, :, :, 0])
Output:
[[16. 20. 24. 24. 24. 24. 24. 20. 16. 12.]
[20. 25. 30. 30. 30. 30. 30. 25. 20. 15.]
[24. 30. 36. 36. 36. 36. 36. 30. 24. 18.]
[24. 30. 36. 36. 36. 36. 36. 30. 24. 18.]
[24. 30. 36. 36. 36. 36. 36. 30. 24. 18.]
[24. 30. 36. 36. 36. 36. 36. 30. 24. 18.]
[24. 30. 36. 36. 36. 36. 36. 30. 24. 18.]
[20. 25. 30. 30. 30. 30. 30. 25. 20. 15.]
[16. 20. 24. 24. 24. 24. 24. 20. 16. 12.]
[12. 15. 18. 18. 18. 18. 18. 15. 12. 9.]]
Indeed, it does look like extra zeros are added to the right and bottom sides.
I found this usefull article on polyfit which works pretty good:
http://www.emilkhatib.com/analyzing-trends-in-data-with-pandas/
import numpy as np
coefficients, residuals, _, _, _ = np.polyfit(range(len(selected.index)),selected,1,full=True)
mse = residuals[0]/(len(selected.index))
nrmse = np.sqrt(mse)/(selected.max() - selected.min())
print('Slope ' + str(coefficients[0]))
print('NRMSE: ' + str(nrmse))
now i would like to use this on a rolling base..
def test(input_list, i):
if sum(~np.isnan(x) for x in input_list) < 2:
return np.NaN
print(input_list)
coefficients, residuals, _, _, _ = np.polyfit(range(len(input_list)),input_list,1,full=True)
mse = residuals[0]/(len(input_list))
nrmse = np.sqrt(mse)/(input_list.max() - input_list.min())
print('Slope ' + str(coefficients[0]))
print('NRMSE: ' + str(nrmse))
a = coefficients[0]*i + coefficients[1]
return a
df['pred'] = df['abs'].rolling(window=2, min_periods=1, center=False).apply(lambda x: test(x, base1.index))
but i wont get it working :)
i get
IndexError: index 0 is out of bounds for axis 0 with size 0 instead of correct results :)
anybody got an idea? thanks! e.
****EDIT1****
sorry, i missed posting a concrete example...
i managed to get the function working, by transforming the numpy array in a df.
but somehow residuals is empty
import quandl
import MySQLdb
import pandas as pd
import numpy as np
import sys
import matplotlib.pyplot as plt
def test(input_list, i):
if sum(~np.isnan(x) for x in input_list) < 2:
return np.NaN
abc = pd.DataFrame(input_list)
coefficients, residuals, _, _, _ = np.polyfit(range(len(abc)),abc[0],1,full=True)
#residuals is empty... why?
a = coefficients[0]*len(abc) + coefficients[1]
return a
df = quandl.get("WIKI/GOOGL")
df = df.ix[:, ['High', 'Low', 'Close']]
#reseit index for calc
#base1['DateTime'] = base1.index
#base1.index = range(len(base1))
df['close_pred'] = df['Close'].rolling(window=15, min_periods=2, center=False).apply(lambda x: test(x, 0))
print(df.head(30).to_string())
Residuals are empty just for 1st iteration see little modified code and answer
def test(data):
if sum(~np.isnan(x) for x in data) < 2:
return np.NaN
df = pd.DataFrame(data)
coefficients, residuals, _, _, _ = np.polyfit(range(len(data)),df[0],1,full=True)
#if residuals.size == 0:
# residuals = [0]
print(coefficients[-2], residuals, data)
return coefficients[-2]
and answer
df_xx['pred'] = df_xx[0].rolling(window=5, min_periods=2, center=False).apply(lambda y: test(y))
0.9999999999999998 [] [0. 1.]
1.0 [4.29279946e-34] [0. 1. 2.]
1.0000000000000002 [3.62112419e-33] [0. 1. 2. 3.]
0.9999999999999999 [8.77574736e-31] [0. 1. 2. 3. 4.]
0.9999999999999999 [1.25461096e-30] [1. 2. 3. 4. 5.]
0.9999999999999999 [2.93468782e-30] [2. 3. 4. 5. 6.]
0.9999999999999997 [1.38665176e-30] [3. 4. 5. 6. 7.]
0.9999999999999997 [2.18347839e-30] [4. 5. 6. 7. 8.]
0.9999999999999999 [6.21693422e-30] [5. 6. 7. 8. 9.]
1.0 [1.07025673e-29] [ 6. 7. 8. 9. 10.]
1.0000000000000002 [1.4374879e-29] [ 7. 8. 9. 10. 11.]
0.9999999999999997 [1.14542951e-29] [ 8. 9. 10. 11. 12.]
1.0000000000000004 [9.73226454e-30] [ 9. 10. 11. 12. 13.]
0.9999999999999997 [1.99069506e-29] [10. 11. 12. 13. 14.]
0.9999999999999997 [1.09437894e-29] [11. 12. 13. 14. 15.]
1.0 [3.60983058e-29] [12. 13. 14. 15. 16.]
1.0000000000000002 [1.90967258e-29] [13. 14. 15. 16. 17.]
1.0000000000000002 [3.13030715e-29] [14. 15. 16. 17. 18.]
1.0 [1.25806434e-29] [15. 16. 17. 18. 19.]
simple code below fix it
if residuals.size == 0:
residuals = [0]
Suppose I have some nearest neighbor classifier. For a new observation it computes the distance between the new observation and all observations in the "known" data set. It returns the class label of the observation, that has the smallest distance to the new observation.
import numpy as np
known_obs = np.random.randint(0, 10, 40).reshape(8, 5)
new_obs = np.random.randint(0, 10, 80).reshape(16, 5)
labels = np.random.randint(0, 2, 8).reshape(8, )
def my_dist(x1, known_obs, axis=0):
return (np.square(np.linalg.norm(x1 - known_obs, axis=axis)))
def nn_classifier(n, known_obs, labels, axis=1, distance=my_dist):
return labels[np.argmin(distance(n, known_obs, axis=axis))]
def classify_batch(new_obs, known_obs, labels, classifier=nn_classifier, distance=my_dist):
return [classifier(n, known_obs, labels, distance=distance) for n in new_obs]
print(classify_batch(new_obs, known_obs, labels, nn_classifier, my_dist))
For performance reasons I would like to avoid the for loop in the classify_batch function. Is there a way to use numpy operations to apply the nn_classifier function to each row of new_obs?
I already tried apply_along_axis but as often mentioned it is convenient but not fast.
The key to avoiding the loop is to express the action on the (16,8) array of 'distances'. The labels[] and argmin steps just cloud the issue.
If I set labels = np.arange(8), then this
arr = np.array([my_dist(n, known_obs, axis=1) for n in new_obs])
print(arr)
print(np.argmin(arr, axis=1))
produces the same thing. It still has a list comprehension, but we are closer to 'source'.
[[ 32. 115. 22. 116. 162. 86. 161. 117.]
[ 106. 31. 142. 164. 92. 106. 45. 103.]
[ 44. 135. 94. 18. 94. 50. 87. 135.]
[ 11. 92. 57. 67. 79. 43. 118. 106.]
[ 40. 67. 126. 98. 50. 74. 75. 175.]
[ 78. 61. 120. 148. 102. 128. 67. 191.]
[ 51. 48. 57. 133. 125. 35. 110. 14.]
[ 47. 28. 93. 91. 63. 49. 32. 88.]
[ 61. 86. 23. 141. 159. 85. 146. 22.]
[ 131. 70. 155. 149. 129. 127. 44. 138.]
[ 97. 138. 87. 117. 223. 77. 130. 122.]
[ 151. 78. 211. 161. 131. 115. 46. 164.]
[ 13. 50. 31. 69. 59. 43. 80. 40.]
[ 131. 108. 157. 161. 207. 85. 102. 146.]
[ 39. 106. 67. 23. 61. 67. 70. 88.]
[ 54. 51. 74. 68. 42. 86. 35. 65.]]
[2 1 3 0 0 1 7 1 7 6 5 6 0 5 3 6]
With
print((new_obs[:,None,:] - known_obs[None,:,:]).shape)
I get a (16,8,5) array. So can I apply the linalg.norm on the last axis?
This seems to do the trick
np.square(np.linalg.norm(diff, axis=-1))
So together:
diff = (new_obs[:,None,:] - known_obs[None,:,:])
dist = np.square(np.linalg.norm(diff, axis=-1))
idx = np.argmin(dist, axis=1)
print(idx)