python: Name Error:name 'data_x' is not defined - abaqus

I am doing my project on incremental deep drawing using ABAQUS.
I am trying to import a text file of loop program into abaqus script so that there is no need of entering amplitude values manually.
But I am getting an error when trying to import the data using the following code
f = open('data_x', 'r')
values=f.read()
values=f.readline()
Error:
data_x is not defined

Error NameError: name 'data_x' is not defined points that you are using data_x as a name in your code, not as a string (with quotes).
This means that in your code, you probably have something like
f = open(data_x)
Python is trying to figure out which value is associated with data_x, which is a Python name, not a string. Since it's not defined before getting to that line, you are getting an error.
If you want to store the name of a file and then open a file, write
data_x = 'data_x.txt'
f = open(data_x)
You could also directly write
f = open('data_x.txt')
Whichever solution you adopt, make sure that a correct path to the file is passed to the function open, so that it could find the file.

Related

Using io.tmpfile() with shell command, ran via io.popen, in Lua?

I'm using Lua in Scite on Windows, but hopefully this is a general Lua question.
Let's say I want to write a temporary string content to a temporary file in Lua - which I want to be eventually read by another program, - and I tried using io.tmpfile():
mytmpfile = assert( io.tmpfile() )
mytmpfile:write( MYTMPTEXT )
mytmpfile:seek("set", 0) -- back to start
print("mytmpfile" .. mytmpfile .. "<<<")
mytmpfile:close()
I like io.tmpfile() because it is noted in https://www.lua.org/pil/21.3.html :
The tmpfile function returns a handle for a temporary file, open in read/write mode. That file is automatically removed (deleted) when your program ends.
However, when I try to print mytmpfile, I get:
C:\Users\ME/sciteLuaFunctions.lua:956: attempt to concatenate a FILE* value (global 'mytmpfile')
>Lua: error occurred while processing command
I got the explanation for that here Re: path for io.tmpfile() ?:
how do I get the path used to generate the temp file created by io.tmpfile()
You can't. The whole point of tmpfile is to give you a file handle without
giving you the file name to avoid race conditions.
And indeed, on some OSes, the file has no name.
So, it will not be possible for me to use the filename of the tmpfile in a command line that should be ran by the OS, as in:
f = io.popen("python myprog.py " .. mytmpfile)
So my questions are:
Would it be somehow possible to specify this tmpfile file handle as the input argument for the externally ran program/script, say in io.popen - instead of using the (non-existing) tmpfile filename?
If above is not possible, what is the next best option (in terms of not having to maintain it, i.e. not having to remember to delete the file) for opening a temporary file in Lua?
You can get a temp filename with os.tmpname.
local n = os.tmpname()
local f = io.open(n, 'w+b')
f:write(....)
f:close()
os.remove(n)
If your purpose is sending some data to a python script, you can also use 'w' mode in popen.
--lua
local f = io.popen(prog, 'w')
f:write(....)
#python
import sys
data = sys.stdin.readline()

Why the lua function io.write() did not work. It only display the results on the terminal, rather than writing to a file

I am learning the Lua IO library. I'm having trouble with io.write(). In Programming Design in Lua, there is a piece of code that iterates through the file line by line and precedes each line with a serial number.
This is the file I`m working on:
test file: "iotest.txt"
This is my code
io.input("iotest.txt")
-- io.output("iotest.txt")
local count = 0
for line in io.lines() do
count=count+1
io.write(string.format("%6d ",count), line, "\n")
end
This is the result of the terminal display, but this result cannot be written to the file, whether I add IO. Output (" iotest.txt ") or not.
the results in terminal
This is the result of file, we can see there is no change
The result after code running
Just add io.flush() after your write operations to save the data to the file.
io.input("iotest.txt")
io.output("iotestout.txt")
local count = 0
for line in io.lines() do
count=count+1
io.write(string.format("%6d ",count), line, "\n")
end
io.flush()
io.close()
Refer to Lua 5.4 Reference Manual : 6.8 - Input and Output Facilities
io.flush() will save any written data to the output file which you set with io.output
See koyaanisqatsi's answer for the optional use of file handles. This becomes especially useful if you're working on multiple files at a time and gives you more control on how to interact with the file.
That said you should also have different files for input and output. You'll agree that it doesn't make sense to read and write from and to the same file alternatingly.
For writing to a file you need a file handle.
This handle comes from: io.open()
See: https://www.lua.org/manual/5.4/manual.html#6.8
A file handle has methods that acts on self.
Thats the function after the : at file handle.
So io.write() puts out on stdout and file:write() in a file.
Example function that can dump a defined function to a file...
fdump=function(func,path)
assert(type(func)=="function")
assert(type(path)=="string")
-- Get the file handle (file)
local file,err = io.open(path, "wb")
assert(file, err)
local chunk = string.dump(func,true)
file:write(chunk)
file:flush()
file:close()
return 'DONE'
end
Here are the methods, taken from io.stdin
close = function: 0x566032b0
seek = function: 0x566045f0
flush = function: 0x56603d10
setvbuf = function: 0x56604240
write = function: 0x56603e70
lines = function: 0x566040c0
read = function: 0x56603c90
This makes it able to use it directly like...
( Lua console: lua -i )
> do io.stdout:write('Input: ') local result=io.stdin:read() return result end
Input: d
d
You are trying to open the same file for reading and writing at the same time. You cannot do that.
There are two possible solutions:
Read from file X, iterate through it and write the result to another file Y.
Read the complete file X into memory, close file X, then delete file X, open the same filename for writing and write to it while iterating through the original file (in memory).
Otherwise, your approach is correct although file operations in Lua are more often done using io.open() and file handles instead of io.write() and io.read().

Command line args in F# fsx

I run my .fsx file like
>fsi A.fsx
In this file I read csv with CsvProvider that has to have path to csv data.
type Data = CsvProvider<"my_data.txt", ";", Schema
I need to pass file name as command line argument and it is possible
>fsi A.fsx my_data.txt
I can read it like
let originalPath = fsi.CommandLineArgs.ElementAt(1)
Problem is, that file name used in CsvProvider constructor needs to be constant and command line argument is not. How I can initialize CsvProvider from command line argument?
The value inside the angle brackes <"my_data.txt"...> specifies an example format file and is checked at compile time, hence the need for it to be a constant string. Assuming your .fsx script merely wants to load a different CSV file of the same general format, you would use
let contents = Data.Load(originalPath)

Writing a file to a specific path in ruby taking the file name from excel

I am having some trouble writing a file to a specific path taking the file name from excel. Here is the code which I am using
out_file = File.new (#temp_path/ "#{obj_info[3].to_s}","w")
"#{obj_info[3].to_s}" = sample.txt
The value sample.txt comes from Excel during run time
#temp_path = "C:/Users/Somefolder/"
The error displayed is:
NoMethodError: undefined method `[]' for nil:NilClass
However, if the code is:
out_file = File.new ("#{obj_info[3].to_s}","w")
it successfully creates a file called sample.txt in the default directory. However, I want it to be stored in a specific directory and the file name needs to be passed from Excel.
Any help is much appreciated.
I believe your problem is because there a space between / and "
#temp_path/ "#{obj_info[3].to_s}
and I guess you want to build a path.
My advice is that you use File.join
f_path = File.join(#temp_path,obj_info[3].to_s)
out_file = File.new (f_path,"w")
Let me know if that solved the problem
You have 2 problems:
obj_info is nil so you make an error reading the value from excel, the error you get indicates it is on an array, in the code you published the only thing that's an array is what you read from excel.
Print the contents with p obj_info right before your code to check.
#temp_path and {obj_info[3].to_s} need to be concatenated to make a path.
You can do that with File.join like Mauricio suggests or like this
out_file = File.new ("#{#temp_path}/#{obj_info[3]}","w")
You can drop the to_s in this case.
It would be better if you publish the whole of your script that is meaningful.

How to source a syntax file from another syntax file in SPSS?

In R there is the source function where you can source an R script from another R script.
I want to be able to do the same in SPSS.
How can I source an SPSS syntax file from another SPSS syntax file?
Updated Following #AndyW's comments.
There is the INSERT and INCLUDE commands. INSERT is newer and more versatile than INCLUDE.
See documentation on INSERT here.
The following is the basic syntax template:
INSERT FILE='file specification'
[SYNTAX = {INTERACTIVE*}]
{BATCH }
[ERROR = {CONTINUE*}]
{STOP }
[CD = {NO*}]
{YES}
[ENCODING = 'encoding specification']
Thus, the following command can be placed in an SPSS syntax file
INSERT FILE='foo.sps'.
and it would import foo.sps syntax file.
By default, syntax must follow the rules of interactive mode, and the code wont stop on an error.
To avoid having to specify the full path to the file, the working directory can be specified as an argument in the INSERT statement or with a separate CD command.
E.g.,
CD '/user/jimbo/long/path/to/project'
Another option is to use FILE HANDLE.
For more information see the SPSS Syntax Reference (available here as a large PDF file).

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