I use the following command to export a jpg file:
gdaldem color-relief C:\input.tif C:\colors.txt C:\output.jpg
The file colors.txt:
-0.99 255 38 0
-0.5 255 81 0
-0.3 255 119 0
-0.2 255 153 0
-0.1 255 187 0
-0 255 221 0
0.1 255 255 0
0.2 206 224 0
0.3 164 196 0
0.4 122 171 0
0.5 85 145 0
0.6 48 120 0
1 0 97 0
The problem is that the file looks different when I display it in the standard Windows software to display images or when I show it in arcmap.
This is how it looks good and is shown in arcmap:
ARCMAP
This is how it is displayed in the Windows program:
WINDOWS
Does anyone have any idea how to display this in Windows software as it appears in arcmap software?
Thanks
Related
The PDF content below renders with the correct vertical positions, but how?
1 0 0 -1 0 792 cm
q
.75 0 0 .75 72 192.75 cm
BT
/F4 14.666667 Tf
1 0 0 -1 0 .80265617 Tm
0 -13.2773438 Td <0030> Tj
12.2087708 0 Td <0024> Tj
8.6870575 0 Td <003C> Tj
9.7756042 0 Td <0032> Tj
11.4001007 0 Td <0035> Tj
ET
Q
q
.75 0 0 .75 72 222.75 cm
BT
/F4 14.666667 Tf
1 0 0 -1 4.0719757 .80265617 Tm
0 -13.2773438 Td <002C> Tj
4.0719757 0 Td <0003> Tj
4.0719757 0 Td <0057> Tj
4.0719757 0 Td <004B> Tj
8.1511078 0 Td <004C> Tj
3.2561493 0 Td <0051> Tj
8.1511078 0 Td <004E> Tj
ET
Q
Renders correctly:
MAJOR
I think
However I can't understand how the y positions are calculated to do this (x is fine). The Text Rendering Matrix (TRM) is given by Text Matrix (TM) multiplied by Current Transformation Matrix (CTM) PDF1.7 Reference section 9.4.4. CTM is the identity matrix multiplied by each "cm" operation.
So for the first snippet,
CTM = [1 0 0 -1 0 792] x [0.75 0 0 0.75 72 192.75] = [0.75 0 0 -0.75 72 786.75]
TRM is TM x CTM:
TRM = [1 0 0 -1 0 0.8026] x [0.75 0 0 -0.75 72 786.75] = [0.75 0 0 0.75 72 786.1]
So, ignoring small details, the text will be rendered around y = 786 (actually 776 I reckon)
For the second snippet,
CTM = [1 0 0 -1 0 792] x [0.75 0 0 0.75 72 222.75] = [0.75 0 0 -0.75 72 816.75]
TRM = [1 0 0 -1 4.072 0.802] x [0.75 0 0 -0.75 72 816.75] = [0.75 0 0 0.75 75.05 816.4]
Again, ignoring small details, the text will be rendered around y = 816 (actually 806 I reckon)
But the y origin is the bottom of the page, and 816 is greater than 786. So how come the second snippet of text renders correctly below the first? I'm clearly missing something in the calculations, but I can't see what. Any ideas?
The error in your calculations is that you apply the cm matrix by multiplication from the right side. You instead have to apply it from the left side.
I.e. for the first snippet you have
CTM = [0.75 0 0 0.75 72 192.75] × [1 0 0 -1 0 792] = [0.75 0 0 -0.75 72 599.25]
and for the second snippet
CTM = [0.75 0 0 0.75 72 222.75] × [1 0 0 -1 0 792] = [0.75 0 0 -0.75 72 569.25]
With these current transformation matrices the rendered result is to be expected.
If you wonder how you should have known that you need to multiply from the left side...
This result is true in general for PDF: when a sequence of transformations is carried out, the matrix representing the combined transformation (M′) is calculated by premultiplying the matrix representing the additional transformation (MT) with the one representing all previously existing transformations (M):
𝑀′ = 𝑀𝑇 × 𝑀
(ISO 32000-2 section 8.3.4 "Transformation matrices")
Without going deep into matrices (not my forte, there is a slight error in my initial maths so images new corrected) you are working downwards from top left based on an inverted start point of 0 792 cm (Top Left corner)
The start of that snippet is above MAJOR 72 192.75 cm
Without outher transformations the text would be "UpsideDown" with M facing towards the bottom then the second 1 0 0 -1 mirrors it back upright and 0.8 "raises" it towards bottom so baseline is 193.5 ish from topleft at which point you "add" 0 -13.2773438 Td so the baseline is now about 205 from top left
Likewise, the origin for the second row is 72 222.75 cm down from above datum.
In both cases you placed their mirrored baseline even lower at 0 -13.2773438 Td thus both lines will be lower than shown above. In part due to the matrix inversions.
so here the second baseline is now at about 72 234 cm down from top left as subject to similar maths is roughly 222.75+.802+13.277 down but scale can also have effect.
Generally its best to use real time viewer of alterations (however this is not the best way just an example that by playing with rounded values I can see the effects).
I am attempting to use the pretty-print confusion-matrix library to create a confusion matrix.
When I run pp_matrix(df, cmap=cmap), I get the following error message:
Traceback (most recent call last):
File "/Users/name/folder/subfolder/subsubfolder/prepositions.py", line 27, in <module>
pp_matrix(df, cmap=cmap)
File "/opt/anaconda3/lib/python3.8/site-packages/pretty_confusion_matrix/pretty_confusion_matrix.py", line 222, in pp_matrix
txt_res = configcell_text_and_colors(
File "/opt/anaconda3/lib/python3.8/site-packages/pretty_confusion_matrix/pretty_confusion_matrix.py", line 59, in configcell_text_and_colors
tot_rig = array_df[col][col]
IndexError: index 37 is out of bounds for axis 0 with size 37
The first few lines of my DateFrame look like this:
in auf mit zu ... an-entlang auf-entlang ohne außerhalb
into 318 8 10 9 ... 0 0 0 0
in 4325 727 681 62 ... 0 0 0 0
on 253 3197 215 46 ... 0 0 0 0
at 206 280 54 9 ... 0 0 0 0
with 384 397 2097 24 ... 0 0 0 0
by 31 12 23 0 ... 0 0 0 0
in-front-of 15 15 25 0 ... 0 0 0 0
The total size is 49 rows x 36 columns.
I think the issue has to do with zero-indexing in Python, but to be honest, I'm not even sure how to go about debugging this.
Any suggestions would be greatly appreciated. Thank you in advance.
Hi I've tried to find similar, answered questions before sending this.
Im importing a csv into Jupyter, and it seems to import as a single column, with 800 rows and 1 column.
import pandas as pd
df = pd.read_csv ('pokemon_data.csv')
df.head()
where as the same method on another file, works exactly as expected.
Using this data that was in the comments:
#,Name,Type 1,Type 2,HP,Attack,Defense,Sp. Atk,Sp. Def,Speed,Generation,Legendary
1,Bulbasaur,Grass,Poison,45,49,49,65,65,45,1,FALSE
2,Ivysaur,Grass,Poison,60,62,63,80,80,60,1,FALSE
3,Venusaur,Grass,Poison,80,82,83,100,100,80,1,FALSE
3,Mega Venusaur,Grass,Poison,80,100,123,122,120,80,1,FALSE
4,Charmander,Fire,,39,52,43,60,50,65,1,FALSE
5,Charmeleon,Fire,,58,64,58,80,65,80,1,FALSE
And using
import pandas as pd
df = pd.read_csv('pokemon.csv',sep=',')
print(df.head())
Yields the following results
# Name Type 1 Type 2 HP Attack Defense Sp. Atk Sp. Def Speed Generation Legendary
0 1 Bulbasaur Grass Poison 45 49 49 65 65 45 1 FALSE
1 2 Ivysaur Grass Poison 60 62 63 80 80 60 1 FALSE
2 3 Venusaur Grass Poison 80 82 83 100 100 80 1 FALSE
3 3 Mega Venusaur Grass Poison 80 100 123 122 120 80 1 FALSE
4 4 Charmander Fire NaN 39 52 43 60 50 65 1 FALSE
I have the following df and I need to find all values which are equal or nearest to 660 in column value.
In detail this means I must iterate somehow through the column value to find all these 660 or nearest values. The values in column value are in a range from 1 to (the end varies) and when the end is reached the values start again from 1. Finally, I must select the entire row of all other columns where value == 660 (or neareest). I have more or less a 'helper' column helper which has the same value during a value range of 1 to end. It could be helpfull to get the result (column help is always 0 or 1). Here is the df example:
helper value
0 1
.
.
.
0 647
0 649
0 652
0 654
0 656
0 659
0 661
0 663
0 665
0 667
0 669
0 672
0 674
0 676
0 678
0 681
.
.
.
0 1000
1 1
.
.
.
1 647
1 649
1 652
1 654
1 656
1 659
1 661
1 663
1 665
1 667
1 669
1 672
1 674
1 676
1 678
1 681
.
.
1 1500
0 1
.
.
.
0 645
0 647
0 650
0 652
0 654
0 656
0 658
0 661
0 663
0 665
0 667
0 669
0 672
0 674
0 676
0 679
.
.
.
0 980
Thanks for any help or hints!
There is not enough info for a correct answer. What is the size of the dataframe? Also what do you mean by nearest? The solution to "where value == 660" can be done by applying a mask to the pandas series.
Something like df = df[(df['value'] == 660)] should do.
While if you have a range of numbers you can use boolean operators for the mask:
mask = df.value > 650 & df.value < 670
df = df[mask]
Let me know if this is what you are looking for
I need to create spline programmatically. I've made something like:
0
SECTION
2
HEADER
9
$ACADVER
1
AC1006
0
ENDSEC
0
SECTION
2
TABLES
0
TABLE
2
LAYER
0
LAYER
2
shape
70
64
62
250
6
CONTINUOUS
0
LAYER
2
holes
70
64
62
250
6
CONTINUOUS
0
ENDTAB
0
ENDSEC
0
SECTION
2
ENTITIES
0
SPLINE
8
shape
100
AcDbSpline
210
0
220
0
230
1
70
4
71
3
72
11
73
4
74
4
42
0.0000001
43
0.0000001
44
0.0000000001
40
0
40
0
40
0
40
0
40
1
40
1
40
1
40
2
40
2
40
2
40
2
10
0
20
0
30
0
10
100
20
50
30
0
10
40
20
40
30
0
10
15
20
23
30
0
11
0
21
0
31
0
11
200
21
200
31
0
11
80
21
80
31
0
11
432
21
234
31
0
0
ENDSEC
0
EOF
When I'm trying to open it in Autodesk TrueView, I'm getting an error:
Undefined group code 210 for object on line 54.
Invalid or incomplete DXF input -- drawing discarded.
Where is the error? When I'm copying just SPLINE section to the DXF generated by AI everything works fine. So I think I need to add something in the header section or something.
This file is DXF version AC1006 which is older than DXF R12. The SPLINE entity
requires at least DXF version AC1012 DXF R13/R14. But with DXF version AC1012
the tag structure of DXF files is changed (OBJECTS and CLASSES sections, SubClassMarkers ...), so just editing the DXF version
does not work.
See also: http://ezdxf.readthedocs.io/en/latest/dxfinternals/filestructure.html#minimal-dxf-content
Also the SPLINE entity seems to be invalid, it has no handle (5) and no owner
tag (330), and the whole AcDbEntity subclass is missing.
Your spline is of degree 3 with 11 knots (0, 0,0,0,1,1,1,2,2,2,2) and 4 control points ( (0,0), (100,50),(40,40),(15,23) ). This might be the problem culprit. You should either have 4 control points and 8 knots or 7 control points and 11 knots.
You may need to assign a handle to the SPLINE, since you're specifying $ACADVER = AC1018 = AutoCAD 2004 where item handles are required.
Try adding a 5-code pair right before the layer designation, like so, where AAAA is a unique hex-encoded handle:
0
SPLINE
5 <-- add these two lines
AAAA <--
8
shape
100
AcDbSpline