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When we have repeated measurements on an experimental unit, typically these units cannot be considered 'independent' and need to be modeled in a way that we get valid estimates for our standard errors.
When I compare the intervals obtained by computing the marginal means for the treatment using a mixed model (treating the unit as a random effect) and in the other case, first averaging over the unit and THEN runnning a simple linear model on the averaged responses, I get the exact same uncertainty intervals.
How do we incorporate the uncertainty of the measurements of the unit, into the uncertainty of what we think our treatments look like?
In order to really propogate all the uncertainty, shouldn't we see what the treatment looks like, averaged over "all possible measurements" on a unit?
``` r
library(dplyr)
#>
#> Attaching package: 'dplyr'
#> The following objects are masked from 'package:stats':
#>
#> filter, lag
#> The following objects are masked from 'package:base':
#>
#> intersect, setdiff, setequal, union
library(emmeans)
library(lme4)
#> Loading required package: Matrix
library(ggplot2)
tmp <- structure(list(treatment = c("A", "A", "A", "A", "A", "A", "A",
"A", "A", "A", "A", "A", "B", "B", "B", "B", "B", "B", "B", "B",
"B", "B", "B", "B"), response = c(151.27333548, 162.3933313,
159.2199999, 159.16666725, 210.82, 204.18666667, 196.97333333,
194.54666667, 154.18666667, 194.99333333, 193.48, 191.71333333,
124.1, 109.32666667, 105.32, 102.22, 110.83333333, 114.66666667,
110.54, 107.82, 105.62000069, 79.79999821, 77.58666557, 75.78666928
), experimental_unit = c("A-1", "A-1", "A-1", "A-1", "A-2", "A-2",
"A-2", "A-2", "A-3", "A-3", "A-3", "A-3", "B-1", "B-1", "B-1",
"B-1", "B-2", "B-2", "B-2", "B-2", "B-3", "B-3", "B-3", "B-3"
)), row.names = c(NA, -24L), class = c("tbl_df", "tbl", "data.frame"
))
### Option 1 - Treat the experimental unit as a random effect since there are
### 4 repeat observations for the same unit
lme4::lmer(response ~ treatment + (1 | experimental_unit), data = tmp) %>%
emmeans::emmeans(., ~ treatment) %>%
as.data.frame()
#> treatment emmean SE df lower.CL upper.CL
#> 1 A 181.0794 10.83359 4 151.00058 211.1583
#> 2 B 101.9683 10.83359 4 71.88947 132.0472
#ggplot(.,aes(treatment, emmean)) +
#geom_pointrange(aes(ymin = lower.CL, ymax = upper.CL))
### Option 2 - instead of treating the unit as random effect, we average over the
### 4 repeat observations, and run a simple linear model
tmp %>%
group_by(experimental_unit) %>%
summarise(mean_response = mean(response)) %>%
mutate(treatment = c(rep("A", 3), rep("B", 3))) %>%
lm(mean_response ~ treatment, data = .) %>%
emmeans::emmeans(., ~ treatment) %>%
as.data.frame()
#> treatment emmean SE df lower.CL upper.CL
#> 1 A 181.0794 10.83359 4 151.00058 211.1583
#> 2 B 101.9683 10.83359 4 71.88947 132.0472
#ggplot(., aes(treatment, emmean)) +
#geom_pointrange(aes(ymin = lower.CL, ymax = upper.CL))
### Whether we include a random effect for the unit, or average over it and THEN model it, we find no difference in the
### marginal means for the treatments
### How do we incoporate the variation of the repeat measurments to the marginal means of the treatments?
### Do we then ignore the variation in the 'subsamples' and simply average over them PRIOR to modeling?
<sup>Created on 2021-07-31 by the [reprex package](https://reprex.tidyverse.org) (v2.0.0)</sup>
emmeans() does take into account the errors of random effects. This is what I get when I remove the complex sequences of pipes:
> mmod = lme4::lmer(response ~ treatment + (1 | experimental_unit), data = tmp)
> emmeans(mmod, "treatment")
treatment emmean SE df lower.CL upper.CL
A 181 10.8 4 151.0 211
B 102 10.8 4 71.9 132
Degrees-of-freedom method: kenward-roger
Confidence level used: 0.95
This is as shown. If I fit a fixed-effects model that accounts for experimental units as a fixed effect, I get:
> fmod = lm(response ~ treatment + experimental_unit, data = tmp)
> emmeans(fmod, "treatment")
NOTE: A nesting structure was detected in the fitted model:
experimental_unit %in% treatment
treatment emmean SE df lower.CL upper.CL
A 181 3.25 18 174.2 188
B 102 3.25 18 95.1 109
Results are averaged over the levels of: experimental_unit
Confidence level used: 0.95
The SEs of the latter results are considerably lower, and that is because the random variations in experimental_unit are modeled as fixed variations.
Apparently the piping you did accounts for the variation of the random effects and includes those in the EMMs. I think that is because you did things separately for each experimental unit and somehow combined those results. I'm not very comfortable with a sequence of pipes that is 7 steps long, and I don't understand why that results in just one set of means.
I recommend against the as.data.frame() at the end. That zaps out annotations that can be helpful in understanding what you have. If you are doing that to get more digits precision, I'll claim that those are digits you don't need, it just exaggerates the precision you are entitled to claim.
Notes on some follow-up comments
Subsequently, I am convinced that what we see in the piped operations in the second part of the OP doe indeed comprise computing the mean of each EU, then analyzing those.
Let's look at that in the context of the formal model. We have (sorry MathJax doesn't work on stackoverflow, but I'll leave the markup there anyway)
$$ Y_{ijk} = \mu + \tau_i + U_{ij} + E_{ijk} $$
where $Y_{ijk}$ is the kth response measurement on the ith treatment and jth EU in the ith treatment, and the rhs terms represent respectively the overall mean, the (fixed) treatment effects, the (random) EU effects, and the (random) error effects. We assume the random effects are all mutually independent. With a balanced design, the EMMs are just the marginal means:
$$ \bar Y_{i..} = \mu + \tau_i + \bar U_{i.} + \bar E_{i..} $$
where a '.' subscript means we averaged over that subscript. If there are n EUs per treatment and m measurements on each EU, we get that
$$ Var(\bar Y_{i..} = \sigma^2_U / n + \sigma^2_E / mn $$
Now, if we aggregate the data on EUs ahead of time, we are starting with
$$ \bar Y_{ij.} = \mu + U_{ij} + \bar E_{ij.} $$
However, if we then compute marginal means by averaging over j, we get exactly the same thing as we did before with $\bar Y_{i..}$, and the variance is exactly as already shown. That is why it doesn't matter if we aggregated first or not.
Spotify Codes are little barcodes that allow you to share songs, artists, users, playlists, etc.
They encode information in the different heights of the "bars". There are 8 discrete heights that the 23 bars can be, which means 8^23 different possible barcodes.
Spotify generates barcodes based on their URI schema. This URI spotify:playlist:37i9dQZF1DXcBWIGoYBM5M gets mapped to this barcode:
The URI has a lot more information (62^22) in it than the code. How would you map the URI to the barcode? It seems like you can't simply encode the URI directly. For more background, see my "answer" to this question: https://stackoverflow.com/a/62120952/10703868
The patent explains the general process, this is what I have found.
This is a more recent patent
When using the Spotify code generator the website makes a request to https://scannables.scdn.co/uri/plain/[format]/[background-color-in-hex]/[code-color-in-text]/[size]/[spotify-URI].
Using Burp Suite, when scanning a code through Spotify the app sends a request to Spotify's API: https://spclient.wg.spotify.com/scannable-id/id/[CODE]?format=json where [CODE] is the media reference that you were looking for. This request can be made through python but only with the [TOKEN] that was generated through the app as this is the only way to get the correct scope. The app token expires in about half an hour.
import requests
head={
"X-Client-Id": "58bd3c95768941ea9eb4350aaa033eb3",
"Accept-Encoding": "gzip, deflate",
"Connection": "close",
"App-Platform": "iOS",
"Accept": "*/*",
"User-Agent": "Spotify/8.5.68 iOS/13.4 (iPhone9,3)",
"Accept-Language": "en",
"Authorization": "Bearer [TOKEN]",
"Spotify-App-Version": "8.5.68"}
response = requests.get('https://spclient.wg.spotify.com:443/scannable-id/id/26560102031?format=json', headers=head)
print(response)
print(response.json())
Which returns:
<Response [200]>
{'target': 'spotify:playlist:37i9dQZF1DXcBWIGoYBM5M'}
So 26560102031 is the media reference for your playlist.
The patent states that the code is first detected and then possibly converted into 63 bits using a Gray table. For example 361354354471425226605 is encoded into 010 101 001 010 111 110 010 111 110 110 100 001 110 011 111 011 011 101 101 000 111.
However the code sent to the API is 6875667268, I'm unsure how the media reference is generated but this is the number used in the lookup table.
The reference contains the integers 0-9 compared to the gray table of 0-7 implying that an algorithm using normal binary has been used. The patent talks about using a convolutional code and then the Viterbi algorithm for error correction, so this may be the output from that. Something that is impossible to recreate whithout the states I believe. However I'd be interested if you can interpret the patent any better.
This media reference is 10 digits however others have 11 or 12.
Here are two more examples of the raw distances, the gray table binary and then the media reference:
1.
022673352171662032460
000 011 011 101 100 010 010 111 011 001 100 001 101 101 011 000 010 011 110 101 000
67775490487
2.
574146602473467556050
111 100 110 001 110 101 101 000 011 110 100 010 110 101 100 111 111 101 000 111 000
57639171874
edit:
Some extra info:
There are some posts online describing how you can encode any text such as spotify:playlist:HelloWorld into a code however this no longer works.
I also discovered through the proxy that you can use the domain to fetch the album art of a track above the code. This suggests a closer integration of Spotify's API and this scannables url than previously thought. As it not only stores the URIs and their codes but can also validate URIs and return updated album art.
https://scannables.scdn.co/uri/800/spotify%3Atrack%3A0J8oh5MAMyUPRIgflnjwmB
Your suspicion was correct - they're using a lookup table. For all of the fun technical details, the relevant patent is available here: https://data.epo.org/publication-server/rest/v1.0/publication-dates/20190220/patents/EP3444755NWA1/document.pdf
Very interesting discussion. Always been attracted to barcodes so I had to take a look. I did some analysis of the barcodes alone (didn't access the API for the media refs) and think I have the basic encoding process figured out. However, based on the two examples above, I'm not convinced I have the mapping from media ref to 37-bit vector correct (i.e. it works in case 2 but not case 1). At any rate, if you have a few more pairs, that last part should be simple to work out. Let me know.
For those who want to figure this out, don't read the spoilers below!
It turns out that the basic process outlined in the patent is correct, but lacking in details. I'll summarize below using the example above. I actually analyzed this in reverse which is why I think the code description is basically correct except for step (1), i.e. I generated 45 barcodes and all of them matched had this code.
1. Map the media reference as integer to 37 bit vector.
Something like write number in base 2, with lowest significant bit
on the left and zero-padding on right if necessary.
57639171874 -> 0100010011101111111100011101011010110
2. Calculate CRC-8-CCITT, i.e. generator x^8 + x^2 + x + 1
The following steps are needed to calculate the 8 CRC bits:
Pad with 3 bits on the right:
01000100 11101111 11110001 11010110 10110000
Reverse bytes:
00100010 11110111 10001111 01101011 00001101
Calculate CRC as normal (highest order degree on the left):
-> 11001100
Reverse CRC:
-> 00110011
Invert check:
-> 11001100
Finally append to step 1 result:
01000100 11101111 11110001 11010110 10110110 01100
3. Convolutionally encode the 45 bits using the common generator
polynomials (1011011, 1111001) in binary with puncture pattern
110110 (or 101, 110 on each stream). The result of step 2 is
encoded using tail-biting, meaning we begin the shift register
in the state of the last 6 bits of the 45 long input vector.
Prepend stream with last 6 bits of data:
001100 01000100 11101111 11110001 11010110 10110110 01100
Encode using first generator:
(a) 100011100111110100110011110100000010001001011
Encode using 2nd generator:
(b) 110011100010110110110100101101011100110011011
Interleave bits (abab...):
11010000111111000010111011110011010011110001...
1010111001110001000101011000010110000111001111
Puncture every third bit:
111000111100101111101110111001011100110000100100011100110011
4. Permute data by choosing indices 0, 7, 14, 21, 28, 35, 42, 49,
56, 3, 10..., i.e. incrementing 7 modulo 60. (Note: unpermute by
incrementing 43 mod 60).
The encoded sequence after permuting is
111100110001110101101000011110010110101100111111101000111000
5. The final step is to map back to bar lengths 0 to 7 using the
gray map (000,001,011,010,110,111,101,100). This gives the 20 bar
encoding. As noted before, add three bars: short one on each end
and a long one in the middle.
UPDATE: I've added a barcode (levels) decoder (assuming no errors) and an alternate encoder that follows the description above rather than the equivalent linear algebra method. Hopefully that is a bit more clear.
UPDATE 2: Got rid of most of the hard-coded arrays to illustrate how they are generated.
The linear algebra method defines the linear transformation (spotify_generator) and mask to map the 37 bit input into the 60 bit convolutionally encoded data. The mask is result of the 8-bit inverted CRC being convolutionally encoded. The spotify_generator is a 37x60 matrix that implements the product of generators for the CRC (a 37x45 matrix) and convolutional codes (a 45x60 matrix). You can create the generator matrix from an encoding function by applying the function to each row of an appropriate size generator matrix. For example, a CRC function that add 8 bits to each 37 bit data vector applied to each row of a 37x37 identity matrix.
import numpy as np
import crccheck
# Utils for conversion between int, array of binary
# and array of bytes (as ints)
def int_to_bin(num, length, endian):
if endian == 'l':
return [num >> i & 1 for i in range(0, length)]
elif endian == 'b':
return [num >> i & 1 for i in range(length-1, -1, -1)]
def bin_to_int(bin,length):
return int("".join([str(bin[i]) for i in range(length-1,-1,-1)]),2)
def bin_to_bytes(bin, length):
b = bin[0:length] + [0] * (-length % 8)
return [(b[i]<<7) + (b[i+1]<<6) + (b[i+2]<<5) + (b[i+3]<<4) +
(b[i+4]<<3) + (b[i+5]<<2) + (b[i+6]<<1) + b[i+7] for i in range(0,len(b),8)]
# Return the circular right shift of an array by 'n' positions
def shift_right(arr, n):
return arr[-n % len(arr):len(arr):] + arr[0:-n % len(arr)]
gray_code = [0,1,3,2,7,6,4,5]
gray_code_inv = [[0,0,0],[0,0,1],[0,1,1],[0,1,0],
[1,1,0],[1,1,1],[1,0,1],[1,0,0]]
# CRC using Rocksoft model:
# NOTE: this is not quite any of their predefined CRC's
# 8: number of check bits (degree of poly)
# 0x7: representation of poly without high term (x^8+x^2+x+1)
# 0x0: initial fill of register
# True: byte reverse data
# True: byte reverse check
# 0xff: Mask check (i.e. invert)
spotify_crc = crccheck.crc.Crc(8, 0x7, 0x0, True, True, 0xff)
def calc_spotify_crc(bin37):
bytes = bin_to_bytes(bin37, 37)
return int_to_bin(spotify_crc.calc(bytes), 8, 'b')
def check_spotify_crc(bin45):
data = bin_to_bytes(bin45,37)
return spotify_crc.calc(data) == bin_to_bytes(bin45[37:], 8)[0]
# Simple convolutional encoder
def encode_cc(dat):
gen1 = [1,0,1,1,0,1,1]
gen2 = [1,1,1,1,0,0,1]
punct = [1,1,0]
dat_pad = dat[-6:] + dat # 6 bits are needed to initialize
# register for tail-biting
stream1 = np.convolve(dat_pad, gen1, mode='valid') % 2
stream2 = np.convolve(dat_pad, gen2, mode='valid') % 2
enc = [val for pair in zip(stream1, stream2) for val in pair]
return [enc[i] for i in range(len(enc)) if punct[i % 3]]
# To create a generator matrix for a code, we encode each row
# of the identity matrix. Note that the CRC is not quite linear
# because of the check mask so we apply the lamda function to
# invert it. Given a 37 bit media reference we can encode by
# ref * spotify_generator + spotify_mask (mod 2)
_i37 = np.identity(37, dtype=bool)
crc_generator = [_i37[r].tolist() +
list(map(lambda x : 1-x, calc_spotify_crc(_i37[r].tolist())))
for r in range(37)]
spotify_generator = 1*np.array([encode_cc(crc_generator[r]) for r in range(37)], dtype=bool)
del _i37
spotify_mask = 1*np.array(encode_cc(37*[0] + 8*[1]), dtype=bool)
# The following matrix is used to "invert" the convolutional code.
# In particular, we choose a 45 vector basis for the columns of the
# generator matrix (by deleting those in positions equal to 2 mod 4)
# and then inverting the matrix. By selecting the corresponding 45
# elements of the convolutionally encoded vector and multiplying
# on the right by this matrix, we get back to the unencoded data,
# assuming there are no errors.
# Note: numpy does not invert binary matrices, i.e. GF(2), so we
# hard code the following 3 row vectors to generate the matrix.
conv_gen = [[0,1,0,1,1,1,1,0,1,1,0,0,0,1]+31*[0],
[1,0,1,0,1,0,1,0,0,0,1,1,1] + 32*[0],
[0,0,1,0,1,1,1,1,1,1,0,0,1] + 32*[0] ]
conv_generator_inv = 1*np.array([shift_right(conv_gen[(s-27) % 3],s) for s in range(27,72)], dtype=bool)
# Given an integer media reference, returns list of 20 barcode levels
def spotify_bar_code(ref):
bin37 = np.array([int_to_bin(ref, 37, 'l')], dtype=bool)
enc = (np.add(1*np.dot(bin37, spotify_generator), spotify_mask) % 2).flatten()
perm = [enc[7*i % 60] for i in range(60)]
return [gray_code[4*perm[i]+2*perm[i+1]+perm[i+2]] for i in range(0,len(perm),3)]
# Equivalent function but using CRC and CC encoders.
def spotify_bar_code2(ref):
bin37 = int_to_bin(ref, 37, 'l')
enc_crc = bin37 + calc_spotify_crc(bin37)
enc_cc = encode_cc(enc_crc)
perm = [enc_cc[7*i % 60] for i in range(60)]
return [gray_code[4*perm[i]+2*perm[i+1]+perm[i+2]] for i in range(0,len(perm),3)]
# Given 20 (clean) barcode levels, returns media reference
def spotify_bar_decode(levels):
level_bits = np.array([gray_code_inv[levels[i]] for i in range(20)], dtype=bool).flatten()
conv_bits = [level_bits[43*i % 60] for i in range(60)]
cols = [i for i in range(60) if i % 4 != 2] # columns to invert
conv_bits45 = np.array([conv_bits[c] for c in cols], dtype=bool)
bin45 = (1*np.dot(conv_bits45, conv_generator_inv) % 2).tolist()
if check_spotify_crc(bin45):
return bin_to_int(bin45, 37)
else:
print('Error in levels; Use real decoder!!!')
return -1
And example:
>>> levels = [5,7,4,1,4,6,6,0,2,4,3,4,6,7,5,5,6,0,5,0]
>>> spotify_bar_decode(levels)
57639171874
>>> spotify_barcode(57639171874)
[5, 7, 4, 1, 4, 6, 6, 0, 2, 4, 3, 4, 6, 7, 5, 5, 6, 0, 5, 0]
Using the MNIST tutorial of Tensorflow, I try to make a convolutional network for face recognition with the "Database of Faces".
The images size are 112x92, I use 3 more convolutional layer to reduce it to 6 x 5 as adviced here
I'm very new at convolutional network and most of my layer declaration is made by analogy to the Tensorflow MNIST tutorial, it may be a bit clumsy, so feel free to advice me on this.
x_image = tf.reshape(x, [-1, 112, 92, 1])
h_conv1 = tf.nn.relu(conv2d(x_image, W_conv1) + b_conv1)
h_pool1 = max_pool_2x2(h_conv1)
W_conv2 = weight_variable([5, 5, 32, 64])
b_conv2 = bias_variable([64])
h_conv2 = tf.nn.relu(conv2d(h_pool1, W_conv2) + b_conv2)
h_pool2 = max_pool_2x2(h_conv2)
W_conv3 = weight_variable([5, 5, 64, 128])
b_conv3 = bias_variable([128])
h_conv3 = tf.nn.relu(conv2d(h_pool2, W_conv3) + b_conv3)
h_pool3 = max_pool_2x2(h_conv3)
W_conv4 = weight_variable([5, 5, 128, 256])
b_conv4 = bias_variable([256])
h_conv4 = tf.nn.relu(conv2d(h_pool3, W_conv4) + b_conv4)
h_pool4 = max_pool_2x2(h_conv4)
W_conv5 = weight_variable([5, 5, 256, 512])
b_conv5 = bias_variable([512])
h_conv5 = tf.nn.relu(conv2d(h_pool4, W_conv5) + b_conv5)
h_pool5 = max_pool_2x2(h_conv5)
W_fc1 = weight_variable([6 * 5 * 512, 1024])
b_fc1 = bias_variable([1024])
h_pool5_flat = tf.reshape(h_pool5, [-1, 6 * 5 * 512])
h_fc1 = tf.nn.relu(tf.matmul(h_pool5_flat, W_fc1) + b_fc1)
keep_prob = tf.placeholder("float")
h_fc1_drop = tf.nn.dropout(h_fc1, keep_prob)
print orlfaces.train.num_classes # 40
W_fc2 = weight_variable([1024, orlfaces.train.num_classes])
b_fc2 = bias_variable([orlfaces.train.num_classes])
y_conv = tf.nn.softmax(tf.matmul(h_fc1_drop, W_fc2) + b_fc2)
My problem appear when the session run the "correct_prediction" op which is
tf.equal(tf.argmax(y_conv, 1), tf.argmax(y_, 1))
At least I think given the error message:
W tensorflow/core/common_runtime/executor.cc:1027] 0x19369d0 Compute status: Invalid argument: Incompatible shapes: [8] vs. [20]
[[Node: Equal = Equal[T=DT_INT64, _device="/job:localhost/replica:0/task:0/cpu:0"](ArgMax, ArgMax_1)]]
Traceback (most recent call last):
File "./convolutional.py", line 133, in <module>
train_accuracy = accuracy.eval(feed_dict = {x: batch[0], y_: batch[1], keep_prob: 1.0})
File "/usr/local/lib/python2.7/dist-packages/tensorflow/python/framework/ops.py", line 405, in eval
return _eval_using_default_session(self, feed_dict, self.graph, session)
File "/usr/local/lib/python2.7/dist-packages/tensorflow/python/framework/ops.py", line 2728, in _eval_using_default_session
return session.run(tensors, feed_dict)
File "/usr/local/lib/python2.7/dist-packages/tensorflow/python/client/session.py", line 345, in run
results = self._do_run(target_list, unique_fetch_targets, feed_dict_string)
File "/usr/local/lib/python2.7/dist-packages/tensorflow/python/client/session.py", line 419, in _do_run
e.code)
tensorflow.python.framework.errors.InvalidArgumentError: Incompatible shapes: [8] vs. [20]
[[Node: Equal = Equal[T=DT_INT64, _device="/job:localhost/replica:0/task:0/cpu:0"](ArgMax, ArgMax_1)]]
Caused by op u'Equal', defined at:
File "./convolutional.py", line 125, in <module>
correct_prediction = tf.equal(tf.argmax(y_conv, 1), tf.argmax(y_, 1))
File "/usr/local/lib/python2.7/dist-packages/tensorflow/python/ops/gen_math_ops.py", line 328, in equal
return _op_def_lib.apply_op("Equal", x=x, y=y, name=name)
File "/usr/local/lib/python2.7/dist-packages/tensorflow/python/ops/op_def_library.py", line 633, in apply_op
op_def=op_def)
File "/usr/local/lib/python2.7/dist-packages/tensorflow/python/framework/ops.py", line 1710, in create_op
original_op=self._default_original_op, op_def=op_def)
File "/usr/local/lib/python2.7/dist-packages/tensorflow/python/framework/ops.py", line 988, in __init__
self._traceback = _extract_stack()
It looks like the y_conv output a matrix of shape 8 x batch_size instead of number_of_class x batch_size
If I change the batch size from 20 to 10, the error message stay the same but instead [8] vs. [20] I get [4] vs. [10]. So from that I conclude that the problem may come from the y_conv declaration (last line of the code above).
The loss function, optimizer, training, etc declarations is the same as in the MNIST tutorial:
cross_entropy = -tf.reduce_sum(y_ * tf.log(y_conv))
train_step = tf.train.AdamOptimizer(1e-4).minimize(cross_entropy)
correct_prediction = tf.equal(tf.argmax(y_conv, 1), tf.argmax(y_, 1))
accuracy = tf.reduce_mean(tf.cast(correct_prediction, "float"))
sess.run((tf.initialize_all_variables()))
for i in xrange(1000):
batch = orlfaces.train.next_batch(20)
if i % 100 == 0:
train_accuracy = accuracy.eval(feed_dict = {x: batch[0], y_: batch[1], keep_prob: 1.0})
print "Step %d, training accuracy %g" % (i, train_accuracy)
train_step.run(feed_dict = {x: batch[0], y_: batch[1], keep_prob: 0.5})
print "Test accuracy %g" % accuracy.eval(feed_dict = {x: orlfaces.test.images, y_: orlfaces.test.labels, keep_prob: 1.0})
Thanks for reading, have a good day
Well, after a lot debugging, I found that my issue was due to a bad instantiation of the labels. Instead of creating arrays full of zeros and replace one value by one, I created them with random value! Stupid mistake. In case someone wondering what I did wrong there and how I fix it here is the change I made.
Anyway during all the debugging I made, to find this mistake, I found some useful information to debug this kind of problem:
For the cross entropy declaration, the tensorflow's MNIST tutorial use a formula that can lead to NaN value
This formula is
cross_entropy = -tf.reduce_sum(y_ * tf.log(y_conv))
Instead of this, I found two ways to declare it in a safer fashion:
cross_entropy = -tf.reduce_sum(y_ * tf.log(tf.clip_by_value(y_conv, 1e-10, 1.0)))
or also:
cross_entropy = tf.reduce_mean(tf.nn.softmax_cross_entropy_with_logits(logit, y_))
As mrry says. printing the shape of the tensors can help to detect shape anomaly.
To get the shape of a tensor just call his get_shape() method like this:
print "W shape:", W.get_shape()
user1111929 in this question use a debug print that help me assert where the problem come from.
I am trying to implement another pooling function for neural network with Theano, expect of already existing maxpool, for example average pool.
Using to this source, where average pooling is already implemented, my code looks like:
Random initialization just to test:
invals = numpy.random.RandomState(1).rand(3,2,5,5)
Definition of Theano scalars and functions:
pdim = T.scalar('pool dim', dtype='float32')
pool_inp = T.tensor4('pool input', dtype='float32')
pool_sum = TSN.images2neibs(pool_inp, (pdim, pdim))
pool_out = pool_sum.mean(axis=-1)
pool_fun = theano.function([pool_inp, pdim], pool_out, name = 'pool_fun', allow_input_downcast=True)
TSN is theano.sandbox.neighbours
And the call of the function:
pool_dim = 2
temp = pool_fun(invals, pool_dim)
temp.shape = (invals.shape[0], invals.shape[1], invals.shape[2]/pool_dim,
invals.shape[3]/pool_dim)
print ('invals[1,0,:,:]=\n', invals[1,0,:,:])
print ('output[1,0,:,:]=\n',temp[1,0,:,:])
And I am getting an error:
TypeError: neib_shape[0]=2, neib_step[0]=2 and ten4.shape[2]=5 not consistent
Apply node that caused the error: Images2Neibs{valid}(pool input, MakeVector.0, MakeVector.0)
Inputs shapes: [(3, 2, 5, 5), (2,), (2,)]
Inputs strides: [(200, 100, 20, 4), (4,), (4,)]
Inputs types: [TensorType(float32, 4D), TensorType(float32, vector), TensorType(float32, vector)]
Use the Theano flag 'exception_verbosity=high' for a debugprint of this apply node.
I don't really understand this error. Would be glad to have any suggestions how to correct this error or example of other pooling techniques, programmed in Theano.
Thanks!
Edit: with the ignoring the border, it works perfectly
pool_sum = TSN.images2neibs(pool_inp, (pdim, pdim), mode='ignore_borders')
invals[1,0,:,:]=
[[ 0.01936696 0.67883553 0.21162812 0.26554666 0.49157316]
[ 0.05336255 0.57411761 0.14672857 0.58930554 0.69975836]
[ 0.10233443 0.41405599 0.69440016 0.41417927 0.04995346]
[ 0.53589641 0.66379465 0.51488911 0.94459476 0.58655504]
[ 0.90340192 0.1374747 0.13927635 0.80739129 0.39767684]]
output[1,0,:,:]=
[[ 0.33142066 0.30330223]
[ 0.42902038 0.64201581]]
invals has shape (5, 5) in the last two dimensions, however you want to pool over (2, 2) subsets. This only works if you ignore the border (i.e. the last column and the last row of invals).
Let's say I roll a 6-sided die 60 times and I get 16, 5, 9, 7, 6, 15 roles for the numbers 1 through 6, respectively. The numbers 1 and 6 are showing up too much and there's only about a 1.8% chance of that being random. If I use Statistics::ChiSquare, it prints out:
There's a >1% chance, and a <5% chance, that this data is random.
So not only is it a bad interface (I can't get those numbers back directly), but the rounding error is significant.
What's worse, what if I'm rolling 2 six sided dice? The odds of getting any particular number are:
Sum Frequency Relative Frequency
2 1 1/36
3 2 2/36
4 3 3/36
5 4 4/36
6 5 5/36
7 6 6/36
8 5 5/36
9 4 4/36
10 3 3/36
11 2 2/36
12 1 1/36
Statistics::ChiSquare used to have a chisquare_nonuniform() function, but it was removed.
So the numbers are rounded poorly and I can't use it for a non-uniform distribution. Given a list of actual frequency and a list of expected frequency, what's the best way of calculating the chi-square test in Perl? The various modules I'm finding on the CPAN aren't helping me, so I'm guessing I missed something obvious.
Implementing this yourself is so simple that I wouldn't want to upload Yet Another Statistics Module just for this.
use Carp qw< croak >;
use List::Util qw< sum >;
use Statistics::Distributions qw< chisqrprob >;
sub chi_squared_test {
my %args = #_;
my $observed = delete $args{observed} // croak q(Argument "observed" required);
my $expected = delete $args{expected} // croak q(Argument "expected" required);
#$observed == #$expected or croak q(Input arrays must have same length);
my $chi_squared = sum map {
($observed->[$_] - $expected->[$_])**2 / $expected->[$_];
} 0 .. $#$observed;
my $degrees_of_freedom = #$observed - 1;
my $probability = chisqrprob($degrees_of_freedom, $chi_squared);
return $probability;
}
say chi_squared_test
observed => [16, 5, 9, 7, 6, 17],
expected => [(10) x 6];
Output: 0.018360