I have 9 screen coordinates, each representing one of 9 positions. From the top right, I want that position to start as the 1st position, and the following clockwise coordinates to represent the 2nd, 3rd, 4th and so on, up until the 9th, which would be the top left coordinate.
Would anybody here be able to come up with some sort of mathematical means of determining which of the 9 coordinates is in which position? They're all relative to each other, and will always be THAT relative to each other.
Example coordinates could be:
(x,y)
X Y
663 382
543 454
303 454
183 382
418 459
543 209
303 209
653 259
183 259
Plotting into an image something like:
Would anybody have any ideas? I simply want some form of programmatic way of listing these coordinates in clockwise order...
Find the center of the "circle," i.e., the average X and average Y
Shift the X and Y values so all are relative to the new center.
Convert to polar coordinates and sort by angle.
Assuming I understand you correctly, I would just do something like
newPosition = (originalPosition + # of rotations) mod 8
Where I am assuming that the first position is 0, and that you can only make the things jump clockwise by integer increments (hence the # of rotations)
Related
I am trying to create a list of the 99th and 1st percentiles. Rather than a single percentile for today. I wanted percentiles for 500 days each using the prior 500 days. The functions I was using for this are the following
swin:{[f;w;s] f each { 1_x,y }\[w#0;s]}
percentile:{[x;y] y (100 xrank y:asc y) bin x}
swin[percentile[99;];500;List].
The issue I come across is that the 99th percentile calculates perfectly, but the 1st percentile makes the entire list = 0. a bit lost as to why it would do that. suggestions appreciated!
What's causing the zeros is two-fold:
What behaviour do you want for the earliest 500 days when there isn't 500 days of history to work with? On day 1 there's only 1 datapoint, on day 2 only 2 etc. Only on the 500th day is there 500 days of actual data to work with. By default that swin function fills the gaps with some seed value
You're using zero as that seed value, aka w#0
For example a 5 day lookback on each date looks something like:
q)swin[::;5;1 2 3 4 5]
0 0 0 0 1
0 0 0 1 2
0 0 1 2 3
0 1 2 3 4
1 2 3 4 5
You have zeros until you have data, so naturally the 1st percentile will pick up the zeros for the first roughly 500 dates.
So then you can decide to seed with a different value, or else possibly exclude zeros from your percentile function:
q)List:1000?1000
q)percentile:{[x;y] y (100 xrank y:asc y except 0) bin x}
q)swin[percentile[1;];500;List]
908 360 360 257 257 257 90 90 90 90 90 90 90 90...
If zeros are a legitimate value in your list and can't be excluded then maybe seed the swin with some other value that you know won't be in the list (negatives? infinity? null?) and then exclude that seed from the percentile function.
EDIT: A final alternative is to use a different sliding window function which doesn't fill gaps with a seed value, e.g.
q)swin2:{[f;w;s] f each(),/:{neg[x]sublist y,z}[w]\[s]}
q)swin2[::;5;1 2 3 4 5]
,1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
q)percentile:{[x;y] y (100 xrank y:asc y) bin x}
q)swin2[percentile[99;];500;List]
908 908 908 908 908 908 908 908 908 908 908 959 959..
q)swin2[percentile[1;];500;List]
908 360 360 257 257 257 90 90 90 90 90 90 90 90 90..
I am sure this is a easy formula but 1 am struggling, I have the following:
On tab 1 I want to enter a colour multiple times into column A using a drop down option, for example and I want to pull the how many information from a table on another sheet, so when I do my formula using xlookup (=XLOOKUP(A2,Sheet2!A2:A7,Sheet2!B2:B7)) it works for the top 4 options but not the rest. Can someone help? I ahve also tried the IF formula etc but with no success.
A B
Colours How Many
Black 17
Yellow 765
Purple 65
Orange 43
Red #N/A
Green #N/A
Purple #N/A
Orange #N/A
Sheet 2 table:
Colours How Many
Red 34
Black 17
Green 32
Yellow 765
Purple 65
Orange 43
I hope this make sense.
Thanks in advance
Wayne
I figured it out
=VLOOKUP(A2,Sheet2!$A$1:$B$7, 2, FALSE)
I want to measure the height and width of each individual pole in pixel.
But because the poles are not always stand straight, but i need the height of pole from the horizontal ground. Can anyone guide me how to handle this?
Note: I might need to get the angle it has slanted later on. Not sure I can ask so many question in here. But greatly appreciate if someone can help.
The image sample i have is at below link:
This should give you a good idea how to do it:
#!/usr/local/bin/python3
import cv2
# Open image in greyscale mode
img = cv2.imread('poles.png',cv2.IMREAD_GRAYSCALE)
# Threshold image to pure black and white AND INVERT because findContours looks for WHITE objects on black background
_, thresh = cv2.threshold(img,127,255,cv2.THRESH_BINARY_INV)
# Find contours
_, contours, _ = cv2.findContours(thresh,cv2.RETR_LIST,cv2.CHAIN_APPROX_SIMPLE)
# Print the contours
for c in contours:
x,y,w,h = cv2.boundingRect(c)
print(x,y,w,h)
The output is this, where each line corresponds to one vertical bar in your image:
841 334 134 154 <--- bar 6 is 154 pixels tall
190 148 93 340 <--- bar 2 is 340 pixels tall
502 79 93 409 <--- bar 4 is 409 pixels tall
633 55 169 433 <--- bar 5 is 433 pixels tall
1009 48 93 440 <--- bar 7 is 490 pixels tall
348 48 93 440 <--- bar 3 is 440 pixels tall
46 46 93 442 <--- bar 1 is 442 pixels tall (leftmost bar)
The first column is the distance from the left edge of the image and the last column is the height of the bar in pixels.
As you seem unsure about whether you want to do this in Python or C++, you may prefer not write any code at all - in which case you can simply use ImageMagick which is included in most Linux distros and is available for macOS and Windows.
Basically, you use "Connected Component" analysis by typing this into the Terminal:
convert poles.png -colorspace gray -threshold 50% \
-define connected-components:verbose=true \
-connected-components 8 null:
Output
Objects (id: bounding-box centroid area mean-color):
0: 1270x488+0+0 697.8,216.0 372566 srgb(255,255,255)
1: 93x442+46+46 92.0,266.5 41106 srgb(0,0,0)
2: 93x440+348+48 394.0,267.5 40920 srgb(0,0,0)
3: 93x440+1009+48 1055.0,267.5 40920 srgb(0,0,0)
4: 169x433+633+55 717.3,271.0 40269 srgb(0,0,0)
5: 93x409+502+79 548.0,283.0 38037 srgb(0,0,0)
6: 93x340+190+148 236.0,317.5 31620 srgb(0,0,0)
7: 134x154+841+334 907.4,410.5 14322 srgb(0,0,0)
That gives you a header line which tells you what all the fields are, then a line for each of the blobs it found in the image. Disregard the first one because that is the white background - you can see that from the last field which is rgb(255,255,255).
So, if we look at the last line, it is a blob that is 134 pixels wide and 154 pixels tall, starting at x=841 and y=334 from the top-left corner, i.e. it corresponds to the first contour that OpenCV found.
Not quite a homework problem, but it may as well be:
You have a long list of positive integer values stored in column A. These are packets in unit U.
A Drum can fit up to 500 U, but you cannot break up packets.
How many drums are required for any given list of values in column A?
This does not have to be the most efficient answer, processing in row order is absolutely fine.
I Think you should be able to solve this with a formula, but the closest I got was
=CEILING(SUM(A1:A1000)/500;1)
Of course, this breaks up packets.
Additionally, this problem requires me to be able to find the room left in each drum used, but emphasis for this question should remain on just the number required.
This cannot be done with a single simple formula. Each drum and packet needs to be counted. However contrary to my comment, for this particular problem a spreadsheet works well, and there is no need for a macro.
First, set B2 to 500 for use in other formulas. If column A is not yet filled, use the formula =RANDBETWEEN(1,B$2) to add some values.
Column C is the main formula that determines how full each drum is. Set C2 to =A2. C3 is =IF(C2+A3>B$2,A3,C2+A3). Fill C3 down to fill the remaining rows.
For column D, use =IF(C2+A3>B$2,B$2-C2,""). However the last row of column D is shorter: =B$2-C21 and change 21 to whatever the last row is.
Finally in column E we find the answer, which is simply =COUNT(D2:D21).
Packets Drum Size How Full Room left in each drum used Number of filled drums
------- --------- -------- --------------------------- ----------------------
206 500 206 294 13
309 309
68 377
84 461 39
305 305 195
387 387 113
118 118
8 126 374
479 479 21
492 492 8
120 120
291 411 89
262 262
108 370 130
440 440 60
88 88
100 188
102 290 210
478 478 22
87 87 413
For OpenOffice Calc, use semicolons ; instead of commas , in formulas.
I've encountered an interesting situation while calculating the inter-quartile range. Assuming we have a dataframe such as:
import pandas as pd
index=pd.date_range('2014 01 01',periods=10,freq='D')
data=pd.np.random.randint(0,100,(10,5))
data = pd.DataFrame(index=index,data=data)
data
Out[90]:
0 1 2 3 4
2014-01-01 33 31 82 3 26
2014-01-02 46 59 0 34 48
2014-01-03 71 2 56 67 54
2014-01-04 90 18 71 12 2
2014-01-05 71 53 5 56 65
2014-01-06 42 78 34 54 40
2014-01-07 80 5 76 12 90
2014-01-08 60 90 84 55 78
2014-01-09 33 11 66 90 8
2014-01-10 40 8 35 36 98
# test for q1 values (this works)
data.quantile(0.25)
Out[111]:
0 40.50
1 8.75
2 34.25
3 17.50
4 29.50
# break it by inserting row of nans
data.iloc[-1] = pd.np.NaN
data.quantile(0.25)
Out[115]:
0 42
1 11
2 34
3 12
4 26
The first quartile can be calculated by taking the median of values in the dataframe that fall below the overall median, so we can see what data.quantile(0.25) should have yielded. e.g.
med = data.median()
q1 = data[data<med].median()
q1
Out[119]:
0 37.5
1 8.0
2 19.5
3 12.0
4 17.0
It seems that quantile is failing to provide an appropriate representation of q1 etc. since it is not doing a good job of handling the NaN values (i.e. it works without NaNs, but not with NaNs).
I thought this may not be a "NaN" issue, rather it might be quantile failing to handle even-numbered data sets (i.e. where the median must be calculated as the mean of the two central numbers). However, after testing with dataframes with both even and odd-numbers of rows I saw that quantile handled these situations properly. The problem seems to arise only when NaN values are present in the dataframe.
I would like to use quntile to calculate the rolling q1/q3 values in my dataframe, however, this will not work with NaN's present. Can anyone provide a solution to this issue?
Internally, quantile uses numpy.percentile over the non-null values. When you change the last row of data to NaNs you're essentially left with an array array([ 33., 46., 71., 90., 71., 42., 80., 60., 33.]) in the first column
Calculating np.percentile(array([ 33., 46., 71., 90., 71., 42., 80., 60., 33.]) gives 42.
From the docstring:
Given a vector V of length N, the qth percentile of V is the qth ranked
value in a sorted copy of V. A weighted average of the two nearest
neighbors is used if the normalized ranking does not match q exactly.
The same as the median if q=50, the same as the minimum if q=0
and the same as the maximum if q=100.