I'm trying to script something in blender3D using python.
I've got a bunch of objects in my scene and want to translate them using a the numerical part of their objectname.
First of all i collect objects from the scene by matching a part of their name.
root_obj = [obj for obj in scene.objects if fnmatch.fnmatchcase(obj.name, "*_Root")]
This gives me a list with:[bpy.data.objects['01_Root'],bpy.data.objects['02_Root'],bpy.data.objects['03_Root'],bpy.data.objects['00_Root']]
My goal is to move these objects 15x their corresponding part of the name. So '00_Root' doesnt have to move, but '01_Root' has to move 15 blender units and '02_Root' 30 blender units.
How do i exctract the numberpart of the names and use them as translation values.
I'm a pretty newb with python so i would appreciate all the help i can get.
A string is a list of characters, each character can be accessed by index starting with 0, get the first character with name[0], the second with name[1]. As with any list you can use slicing to get a portion of the list. If the value is always the first two characters you can get the value with name[:2] you can them turn that into an integer with int() or a float with float(). Combined that becomes,
val = int(name[:2])
You then have a number you can calculate the new location with.
obj.location.x = val * 15
If the number of digits in the name might vary you can use split() to break the string on a specific separating character. This returns a list of items between the specified character, so if you want the first item to turn into an integer.
name = '02_item'
val = int(name.split('_')[0])
Using split also allows multiple values in a name.
name = '2_12_item'
val1 = int(name.split('_')[0])
val2 = int(name.split('_')[1])
Related
I will state the obvious that I am a beginner. I should also mention that I have been coding in Zybooks, which affects things. My textbook hasn't helped me much
I tried sub_lyric= rhyme_lyric[ : ]
Zybooks should be able to input an index number can get only that part of the sentence but my book doesnt explain how to do that. If it throws a [4:7] then it would output cow. Hopefully I have exolained everything well.
You need to set there:
sub_lyric = rhyme_lyric[start_index:end_index]
The string is as a sequence of characters and you can use string slicing to extract any sub-text from the main one. As you have observed:
sub_lyric = rhyme_lyric[:]
will copy the entire content of rhyme_lyric to sub_lyric.
To select only a portion of the text, specify the start_index (strings start with index 0) to end_index (not included).
sub_lyric = rhyme_lyric[4:7]
will extract characters in rhyme_lyric from position 4 (included) to position 7 (not included) so the result will be cow.
You can check more on string slicing here: Python 3 introduction
I'm trying to implement in a different way what I can already do implementing some custom matlab functions. Let us suppose to have this string 'AAAAAAAAAAAaaaaaaaaaaaTTTTTTTTTTTTTTTTsssssssssssTTTTTTTTTT' I know to remove each lowercase sub strings with
regexprep(String, '[a-z]*', '')
But since I want to understand how to take indexes of these substrings and using them to check them and remove them maybe with a for loop I'm investigating about how to do it.
Regexp give the indexes :
[Start,End] = regexp(Seq,'[a-z]{1,}');
but i'm not succeeding in figuring out how to use them to check these sequences and eliminate them.
With the indexing approach you get several start and end indices (two in your example), so you need a loop to remove the corresponding sections from the string. You should remove them from last to first, otherwise indices that haven't been used yet will become invalid as you remove sections:
x = 'AAAAAAAAAAAaaaaaaaaaaaTTTTTTTTTTTTTTTTsssssssssssTTTTTTTTTT'; % input
y = x; % initiallize result
[Start, End] = regexp(x, '[a-z]{1,}');
for k = numel(Start):-1:1 % note: from last to first
y(Start(k):End(k)) = []; % remove section
end
What is the difference between string and character class in MATLAB?
a = 'AX'; % This is a character.
b = string(a) % This is a string.
The documentation suggests:
There are two ways to represent text in MATLAB®. You can store text in character arrays. A typical use is to store short pieces of text as character vectors. And starting in Release 2016b, you can also store multiple pieces of text in string arrays. String arrays provide a set of functions for working with text as data.
This is how the two representations differ:
Element access. To represent char vectors of different length, one had to use cell arrays, e.g. ch = {'a', 'ab', 'abc'}. With strings, they can be created in actual arrays: str = [string('a'), string('ab'), string('abc')].
However, to index characters in a string array directly, the curly bracket notation has to be used:
str{3}(2) % == 'b'
Memory use. Chars use exactly two bytes per character. strings have overhead:
a = 'abc'
b = string('abc')
whos a b
returns
Name Size Bytes Class Attributes
a 1x3 6 char
b 1x1 132 string
The best place to start for understanding the difference is the documentation. The key difference, as stated there:
A character array is a sequence of characters, just as a numeric array is a sequence of numbers. A typical use is to store short pieces of text as character vectors, such as c = 'Hello World';.
A string array is a container for pieces of text. String arrays provide a set of functions for working with text as data. To convert text to string arrays, use the string function.
Here are a few more key points about their differences:
They are different classes (i.e. types): char versus string. As such they will have different sets of methods defined for each. Think about what sort of operations you want to do on your text, then choose the one that best supports those.
Since a string is a container class, be mindful of how its size differs from an equivalent character array representation. Using your example:
>> a = 'AX'; % This is a character.
>> b = string(a) % This is a string.
>> whos
Name Size Bytes Class Attributes
a 1x2 4 char
b 1x1 134 string
Notice that the string container lists its size as 1x1 (and takes up more bytes in memory) while the character array is, as its name implies, a 1x2 array of characters.
They can't always be used interchangeably, and you may need to convert between the two for certain operations. For example, string objects can't be used as dynamic field names for structure indexing:
>> s = struct('a', 1);
>> name = string('a');
>> s.(name)
Argument to dynamic structure reference must evaluate to a valid field name.
>> s.(char(name))
ans =
1
Strings do have a bit of overhead, but still increase by 2 bytes per character. After every 8 characters it increases the size of the variable. The red line is y=2x+127.
figure is created using:
v=[];N=100;
for ct = 1:N
s=char(randi([0 255],[1,ct]));
s=string(s);
a=whos('s');v(ct)=a.bytes;
end
figure(1);clf
plot(v)
xlabel('# characters')
ylabel('# bytes')
p=polyfit(1:N,v,1);
hold on
plot([0,N],[127,2*N+127],'r')
hold off
One important practical thing to note is, that strings and chars behave differently when interacting with square brackets. This can be especially confusing when coming from python. consider following example:
>>['asdf' '123']
ans =
'asdf123'
>> ["asdf" "123"]
ans =
1×2 string array
"asdf" "123"
Extracting a specific word and a number of tokens on each side of it from each string in a column in SAS EG ?
For example,
row1: the sun is nice
row2: the sun looks great
row3: the sun left me
Is there a code that would produce the following result column (2 words where sun is the first):
SUN IS
SUN LOOKS
SUN LEFT
and possibly a second column with COUNT in case of duplicate matches.
So if there was 20 SUN LOOKS then it they would be grouped and have a count of 20.
Thanks
I think you can use functions findw() and scan() to do want you want. Both of those functions operate on the concept of word boundaries. findw() returns the position of the word in the string. Once you know the position, you can use scan() in a loop to get the next word or words following it.
Here is a simple example to show you the concept. It is by no means a finished or polished solution, but intended you point you in the right direction. The input data set (text) contains the sentences you provided in your question with slight modifications. The data step finds the word "sun" in the sentence and creates a variable named fragment that contains 3 words ("sun" + the next 2 words).
data text2;
set text;
length fragment $15;
word = 'sun'; * search term;
fragment_len = 3; * number of words in target output;
word_pos = findw(sentence, word, ' ', 'e');
if word_pos then do;
do i = 0 to fragmen_len-1;
fragment = catx(' ', fragment, scan(sentence, word_pos+i));
end;
end;
run;
Here is a partial print of the output data set.
You can use a combination of the INDEX, SUBSTR and SCAN functions to achieve this functionality.
INDEX - takes two arguments and returns the position at which a given substring appears in a string. You might use:
INDEX(str,'sun')
SUBSTR - simply returns a substring of the provided string, taking a second numeric argument referring to the starting position of the substring. Combine this with your INDEX function:
SUBSTR(str,INDEX(str,'sun'))
This returns the substring of str from the point where the word 'sun' first appears.
SCAN - returns the 'words' from a string, taking the string as the first argument, followed by a number referring to the 'word'. There is also a third argument that specifies the delimiter, but this defaults to space, so you wouldn't need it in your example.
To pick out the word after 'sun' you might do this:
SCAN(SUBSTR(str,INDEX(str,'sun')),2)
Now all that's left to do is build a new string containing the words of interest. That can be achieved with concatenation operators. To see how to concatenate two strings, run this illustrative example:
data _NULL_;
a = 'Hello';
b = 'World';
c = a||' - '||b;
put c;
run;
The log should contain this line:
Hello - World
As a result of displaying the value of the c variable using the put statement. There are a number of functions that can be used to concatenate strings, look in the documentation at CAT,CATX,CATS for some examples.
Hopefully there is enough here to help you.
My data looks like this
-3442.77 -16749.64 893.08 -3442.77 -16749.64 1487.35 -3231.45 -16622.36 902.29
.....
159*2539.87 10*0.00 162*2539.87 10*0.00
which means I start with either 7 or 8 reals per line and then (towards the end) have 159 values of 2539.87 followed by 10 values of 0 followed by 162 of 2539.87 etc. This seems to be a space-saving method as previous versions of this file format were regular 6 reals per line.
I am already reading the data into a string because of not knowing whether there are 7 or 8 numbers per line. I can therefore easily spot lines that contain *. But what then? I suppose I have to identify the location of each * and then identify the integer number before and real value after before assigning to an array. Am I missing anything?
Read the line. Split it into tokens delimited by whitespace(s). Replace the * in tokens that have it with space. Then read from the string one or two values, depending on wheather there was an asterisk or not. Sample code follows:
REAL, DIMENSION(big) :: data
CHARACTER(LEN=40) :: token
INTEGER :: iptr, count, idx
REAL :: val
iptr = 1
DO WHILE (there_are_tokens_left)
... ! Get the next token into "token"
idx = INDEX(token, "*")
IF (idx == 0) THEN
READ(token, *) val
count = 1
ELSE
! Replace "*" with space and read two values from the string
token(idx:idx) = " "
READ(token, *) count, val
END IF
data(iptr:iptr+count-1) = val ! Add "val" "count" times to the list of values
iptr = iptr + count
END DO
Here I have arbitrarily set the length of the token to be 40 characters. Adjust it according to what you expect to find in your input files.
BTW, for the sake of completeness, this method of compressing something by replacing repeating values with value/repetition-count pairs is called run-length encoding (RLE).
Your input data may have been written in a form suitable for list directed input (where the format specification in the READ statement is simply ''*''). List directed input supports the r*c form that you see, where r is a repeat count and c is the constant to be repeated.
If the total number of input items is known in advance (perhaps it is fixed for that program, perhaps it is defined by earlier entries in the file) then reading the file is as simple as:
REAL :: data(size_of_data)
READ (unit, *) data
For example, for the last line shown in your example on its own ''size_of_data'' would need to be 341, from 159+10+162+10.
With list directed input the data can span across multiple records (multiple lines) - you don't need to know how many items are on each line in advance - just how many appear in the next "block" of data.
List directed input has a few other "features" like this, which is why it is generally not a good idea to use it to parse "arbitrary" input that hasn't been written with it in mind - use an explicit format specification instead (which may require creating the format specification on the fly to match the width of the input field if that is not know ahead of time).
If you don't know (or cannot calculate) the number of items in advance of the READ statement then you will need to do the parsing of the line yourself.