How to have multiple OR operators in a single if statement? I tried this way but there's an error of:
Incompatible data types in expression or assignment.
I looked at documentation of length function and it is LENGTH ( {string | raw-expression | blob-field } [ , type ] )?
Here's the code:
DEFINE VARIABLE cMonth AS CHARACTER.
DEFINE VARIABLE cDay AS CHARACTER.
DEFINE VARIABLE cYear AS CHARACTER.
UPDATE cDateFromUser.
cDay = (SUBSTRING(cDateFromUser,1,2)).
cMonth = (SUBSTRING(cDateFromUser,3,2)).
cYear = (SUBSTRING(cDateFromUser,5,4)).
IF (LENGTH(cDay <> 2)) OR (LENGTH(cMonth <> 2)) OR (LENGTH(cYear <> 4)) THEN DO:
/*Code*/
END.
ELSE DO:
/*Code*/
END.
The syntax is not right here. Use the following IF:
IF (LENGTH(cDay) <> 2) OR (LENGTH(cMonth) <> 2) OR (LENGTH(cYear) <> 4) THEN DO:
On a simple grammar I am in the bad situation that one of my ParseActions is not called.
For me this is strange as parseActions of a base symbol ("logic_oper") and a derived symbol ("cmd_line") are called correctly. Just "pa_logic_cmd" is not called. You can see this on the output which is included at the end of the code.
As there is no exception on parsing the input string, I am assuming that the grammar is (basically) correct.
import io, sys
import pyparsing as pp
def diag(msg, t):
print("%s: %s" % (msg , str(t)) )
def pa_logic_oper(t): diag('logic_oper', t)
def pa_operand(t): diag('operand', t)
def pa_ident(t): diag('ident', t)
def pa_logic_cmd(t): diag('>>>>>> logic_cmd', t)
def pa_cmd_line(t): diag('cmd_line', t)
def make_grammar():
semi = pp.Literal(';')
ident = pp.Word(pp.alphas, pp.alphanums).setParseAction(pa_ident)
operand = (ident).setParseAction(pa_operand)
op_and = pp.Keyword('A')
op_or = pp.Keyword('O')
logic_oper = (( op_and | op_or) + pp.Optional(operand))
logic_oper.setParseAction(pa_logic_oper)
logic_cmd = logic_oper + pp.Suppress(semi)
logic_cmd.setParseAction(pa_logic_cmd)
cmd_line = (logic_cmd)
cmd_line.setParseAction(pa_cmd_line)
grammar = pp.OneOrMore(cmd_line) + pp.StringEnd()
return grammar
if __name__ == "__main__":
inp_str = '''
A param1;
O param2;
A ;
'''
grammar = make_grammar()
print( "pp-version:" + pp.__version__)
parse_res = grammar.parseString( inp_str )
'''USAGE/Output: python test_4.py
pp-version:2.0.3
operand: ['param1']
logic_oper: ['A', 'param1']
cmd_line: ['A', 'param1']
operand: ['param2']
logic_oper: ['O', 'param2']
cmd_line: ['O', 'param2']
logic_oper: ['A']
cmd_line: ['A']
'''
Can anybody give me a hint on this parseAction problem?
Thanks,
The problem is here:
cmd_line = (logic_cmd)
cmd_line.setParseAction(pa_cmd_line)
The first line assigns cmd_line to be the same expression as logic_cmd. You can verify by adding this line:
print("???", cmd_line is logic_cmd)
Then the second line calls setParseAction, which overwrites the parse action of logic_cmd, so the pa_logic_cmd will never get called.
Remove the second line, since you are already testing the calling of the parse action with pa_logic_cmd. You could change to using the addParseAction method instead, but to my mind that is an invalid test (adding 2 parse actions to the same pyparsing expression object).
Or, change the definition of cmd_line to:
cmd_line = pp.Group(logic_cmd)
Now you will have wrapped logic_cmd inside another expression, and you can then independently set and test the running of parse actions on the two different expressions.
I'm having trouble combining expression and string interpolation. I want to do something like this:
a = [:foo, :bar]
b = [:printfoo, :printbar]
for (str, fn) in zip(a,b)
#eval begin
"""
```
$fn()
```
Prints $str.
"""
$fn() = println("$str")
end
end
My desired output are two functions, one which might look, as typed, like
"""
```
printfoo()
```
Prints foo.
"""
printfoo() = println("foo")
Instead, I get expressions that look like this (when I use quote rather than #eval begin):
```\n$(fn)()\n```\nPrints $(str).\n" # /home/..., line 12:
printfoo() = begin # /home/..., line 12:
println("$(str)")
end
I don't understand why using the same $x syntax, interpolation is performed as I want only outside of quoted ("...") regions. I've tried some combinations of $($x) etc. but am missing some of the subtleties:
"$str" -> "$(str)"
"$($str)" -> "$(foo)"
What would map to "foo"?
Thanks.
The answer seems to be $(string(str)).
The following version worked for me, but it took some tweaking to arrive at:
a = [:foo, :bar]
b = [:printfoo, :printbar]
for (str, fn) in zip(a,b)
#eval begin
$fn() = println($(string(str)))
#doc $("```\n$fn()\n```\nPrints $str.\n") $fn
end
end
One problem with the OP code was when the function bodies interpolated the variable during execution - this would cause printfoo to annoyingly print bar.
This question already has answers here:
What is the difference between __str__ and __repr__?
(28 answers)
Closed 2 years ago.
I really don't understand where are __str__ and __repr__ used in Python. I mean, I get that __str__ returns the string representation of an object. But why would I need that? In what use case scenario? Also, I read about the usage of __repr__
But what I don't understand is, where would I use them?
__repr__
Called by the repr() built-in function and by string conversions (reverse quotes) to compute the "official" string representation of an object. If at all possible, this should look like a valid Python expression that could be used to recreate an object with the same value (given an appropriate environment).
__str__
Called by the str() built-in function and by the print statement to compute the "informal" string representation of an object.
Use __str__ if you have a class, and you'll want an informative/informal output, whenever you use this object as part of string. E.g. you can define __str__ methods for Django models, which then gets rendered in the Django administration interface. Instead of something like <Model object> you'll get like first and last name of a person, the name and date of an event, etc.
__repr__ and __str__ are similar, in fact sometimes equal (Example from BaseSet class in sets.py from the standard library):
def __repr__(self):
"""Return string representation of a set.
This looks like 'Set([<list of elements>])'.
"""
return self._repr()
# __str__ is the same as __repr__
__str__ = __repr__
The one place where you use them both a lot is in an interactive session. If you print an object then its __str__ method will get called, whereas if you just use an object by itself then its __repr__ is shown:
>>> from decimal import Decimal
>>> a = Decimal(1.25)
>>> print(a)
1.25 <---- this is from __str__
>>> a
Decimal('1.25') <---- this is from __repr__
The __str__ is intended to be as human-readable as possible, whereas the __repr__ should aim to be something that could be used to recreate the object, although it often won't be exactly how it was created, as in this case.
It's also not unusual for both __str__ and __repr__ to return the same value (certainly for built-in types).
Building up and on the previous answers and showing some more examples. If used properly, the difference between str and repr is clear. In short repr should return a string that can be copy-pasted to rebuilt the exact state of the object, whereas str is useful for logging and observing debugging results. Here are some examples to see the different outputs for some known libraries.
Datetime
print repr(datetime.now()) #datetime.datetime(2017, 12, 12, 18, 49, 27, 134411)
print str(datetime.now()) #2017-12-12 18:49:27.134452
The str is good to print into a log file, where as repr can be re-purposed if you want to run it directly or dump it as commands into a file.
x = datetime.datetime(2017, 12, 12, 18, 49, 27, 134411)
Numpy
print repr(np.array([1,2,3,4,5])) #array([1, 2, 3, 4, 5])
print str(np.array([1,2,3,4,5])) #[1 2 3 4 5]
in Numpy the repr is again directly consumable.
Custom Vector3 example
class Vector3(object):
def __init__(self, args):
self.x = args[0]
self.y = args[1]
self.z = args[2]
def __str__(self):
return "x: {0}, y: {1}, z: {2}".format(self.x, self.y, self.z)
def __repr__(self):
return "Vector3([{0},{1},{2}])".format(self.x, self.y, self.z)
In this example, repr returns again a string that can be directly consumed/executed, whereas str is more useful as a debug output.
v = Vector3([1,2,3])
print str(v) #x: 1, y: 2, z: 3
print repr(v) #Vector3([1,2,3])
One thing to keep in mind, if str isn't defined but repr, str will automatically call repr. So, it's always good to at least define repr
Grasshopper, when in doubt go to the mountain and read the Ancient Texts. In them you will find that __repr__() should:
If at all possible, this should look like a valid Python expression that could be used to recreate an object with the same value.
Lets have a class without __str__ function.
class Employee:
def __init__(self, first, last, pay):
self.first = first
self.last = last
self.pay = pay
emp1 = Employee('Ivan', 'Smith', 90000)
print(emp1)
When we print this instance of the class, emp1, this is what we get:
<__main__.Employee object at 0x7ff6fc0a0e48>
This is not very helpful, and certainly this is not what we want printed if we are using it to display (like in html)
So now, the same class, but with __str__ function:
class Employee:
def __init__(self, first, last, pay):
self.first = first
self.last = last
self.pay = pay
def __str__(self):
return(f"The employee {self.first} {self.last} earns {self.pay}.")
# you can edit this and use any attributes of the class
emp2 = Employee('John', 'Williams', 90000)
print(emp2)
Now instead of printing that there is an object, we get what we specified with return of __str__ function:
The employee John Williams earns 90000
str will be informal and readable format whereas repr will give official object representation.
class Complex:
# Constructor
def __init__(self, real, imag):
self.real = real
self.imag = imag
# "official" string representation of an object
def __repr__(self):
return 'Rational(%s, %s)' % (self.real, self.imag)
# "informal" string representation of an object (readable)
def __str__(self):
return '%s + i%s' % (self.real, self.imag)
t = Complex(10, 20)
print (t) # this is usual way we print the object
print (str(t)) # this is str representation of object
print (repr(t)) # this is repr representation of object
Answers :
Rational(10, 20) # usual representation
10 + i20 # str representation
Rational(10, 20) # repr representation
str and repr are both ways to represent. You can use them when you are writing a class.
class Fraction:
def __init__(self, n, d):
self.n = n
self.d = d
def __repr__(self):
return "{}/{}".format(self.n, self.d)
for example when I print a instance of it, it returns things.
print(Fraction(1, 2))
results in
1/2
while
class Fraction:
def __init__(self, n, d):
self.n = n
self.d = d
def __str__(self):
return "{}/{}".format(self.n, self.d)
print(Fraction(1, 2))
also results in
1/2
But what if you write both of them, which one does python use?
class Fraction:
def __init__(self, n, d):
self.n = n
self.d = d
def __str__(self):
return "str"
def __repr__(self):
return "repr"
print(Fraction(None, None))
This results in
str
So python actually uses the str method not the repr method when both are written.
Suppose you have a class and wish to inspect an instance, you see the print doesn't give much useful information
class Person:
def __init__(self, name, age):
self.name = name
self.age = age
p1 = Person("John", 36)
print(p1) # <__main__.Animal object at 0x7f9060250410>
Now see a class with a str, it shows the instance information and with repr you even don't need the print. Nice no?
class Animal:
def __init__(self, color, age, breed):
self.color = color
self.age = age
self.breed = breed
def __str__(self):
return f"{self.color} {self.breed} of age {self.age}"
def __repr__(self):
return f"repr : {self.color} {self.breed} of age {self.age}"
a1 = Animal("Red", 36, "Dog")
a1 # repr : Red Dog of age 36
print(a1) # Red Dog of age 36
I'm trying to figure out how to have a quote block, when evaluated, return a symbol. See the example below.
function func(symbol::Symbol)
quote
z = $symbol
symbol
end
end
a = 1
eval(func(:a)) #this returns :symbol. I would like it to return :a
z
The symbol your function returned where the symbol function, due to the last symbol in your qoute did not have $ in front. The second problem is you would like to return the symbol it self, which requires you make a quote inside the quote similar to this question
Julia: How do I create a macro that returns its argument?
function func(s::Symbol)
quote
z = $s
$(Expr(:quote, s)) # This creates an expresion inside the quote
end
end
a = 1
eval(func(:a)) #this returns :a
z