When I compile the following Metapost file:
beginfig(1);
def f(expr n) =
if n=0: 0
else: 1
fi;
enddef;
show f(0)+1;
endfig;
end
I expect to get the output 1 (since f(0) is equal to 0, f(0)+1 should be 1!). However, Metapost complains about an Isolated expression.
When I put the expression in parentheses: show (f(0)+1), things get even stranger. The error message becomes : Missing ')' has been inserted. (The first quote should be a backquote, but I couldn't figure out how to escape it). Where on earth was there a mismatched parenthesis??
Thanks for your help!
The def command just expands a name into its definition, so you get literally:
show if 0=0: 0 else: 1;+1;
The semicolon in the middle is what's wrong, so let us remove it:
beginfig(1);
def f(expr n) =
if n=0: 0
else: 1
fi
enddef;
show f(0)+1;
endfig;
end
This produces the correct expansion:
show if 0=0: 0 else: 1+1;
And outputs 1 as expected.
On a side note, I'd recommend using begingroup...endgroup for heavier macro definitions, and at least parentheses for lighter ones: for example,
def f = 1 + 2 enddef;
show f * 2;
gives 1 + 2 * 2, which is not the same as what is probably expected:
def f = (1 + 2) enddef;
show f * 2;
which gives (1 + 2) * 2.
Related
How the game works is that there is a 3-digit number, and you have to guess it. If you guess a digit in the right spot, you get a strike, and if you guess a digit but in the wrong spot you get a ball. I've coded it like this.
x = random.randint(1, 9)
y = random.randint(1, 9)
z = random.randint(1, 9)
userguessunlisted = input('What number do you want to guess?')
numbertoguess = list[x, y, z]
userguess = list(userguessunlisted)
b = 0
s = 0
while 0 == 0:
if userguess[0] == numbertoguess[0]:
s = s + 1
if userguess[0] == numbertoguess[1]:
b = b + 1
if userguess[0] == numbertoguess[2]:
b = b + 1
if userguess[1] == numbertoguess[0]:
b = b + 1
if userguess[1] == numbertoguess[1]:
s = s + 1
if userguess[1] == numbertoguess[2]:
b = b + 1
if userguess[2] == numbertoguess[0]:
b = b + 1
if userguess[2] == numbertoguess[1]:
b = b + 1
if userguess[2] == numbertoguess[2]:
s = s + 1
print(s + "S", b + "B")
if s != 3:
b = 0
s = 0
else:
print('you win!')
break
When you said list[x, y, z] on line 5, you used square brackets, which python interprets to be a type annotation. For example, if I wanted to specify that a variable is a list of ints, I could say
my_list_of_ints: list[int] = [1, 2, 3]
I think what you meant to do is create a new list from x, y, and z. One way to do this is
numbertoguess = list([x, y, z])
which is probably what you meant to write. This is valid because the list function takes an iterable as its one and only argument.
However, the list portion is redundant; square brackets on the right-hand side of an assignment statement already means "create a list with this content," so instead you should simply say
numbertoguess = [x, y, z]
A few other notes:
input will return a string, but you are comparing that string to integers further down, so none of the comparisons will ever be true. What you want to say is something like the following:
while True:
try:
userguessunlisted = int(input('What number do you want to guess?'))
except:
continue
break
What this code does is attempts to parse the string returned from input into an int. If it fails to do so, which would happen if the user inputted something other than a valid integer, an exception would be thrown, and the except block would be entered. continue means go to the top of the loop, so the input line runs repeatedly until a valid int is entered. When that happens, the except block is skipped, so break runs, which means "exit the loop."
userguessunlisted is only ever going to contain 1 number as written, so userguess will be a list of length 1, and all of the comparisons using userguess[1] and userguess[2] will throw an IndexError. Try to figure out how to wrap the code from (1) in another loop to gather multiple guesses from the user. Hint: use a for loop with range.
It might also be that you meant for the user to input a 3-digit number all at once. In that case, you can use a list comprehension to grab each character from the input and parse it into a separate int. This is probably a bit complicated for a beginner, so I'll help you out:
[int(char) for char in input('What number do you want to guess?')]
print(s + "S", b + "B") will throw TypeError: unsupported operand type(s) for +: 'int' and 'str'. There are lots of ways to combine non-string types with strings, but the most modern way is using f-strings. For example, to combine s with "S", you can say f"{s}S".
When adding some amount to a variable, instead of saying e.g. b = b + 1, you can use the += operator to more concisely say b += 1.
It's idiomatic in python to use snake_case for variables and Pascal case for classes. So instead of writing e.g. numbertoguess, you should use number_to_guess. This makes your code more readable and familiar to other python programmers.
Happy coding!
I'm getting a SyntaxError for:
housing['Lot Area'].apply(lambda x: x + 50000 if x > 20000)
When I add else, the code runs fine:
housing['Lot Area'].apply(lambda x: x + 50000 if x > 20000 else x)
Does if only work in combination with else here? I'd like to increment x with 50000 only if x > 20000 -- otherwise I'd like x to remain unchanged. I find the else part a bit redundant here. Besides the first question before, is there any way to write this code without the else part?
Base on your description , even apply is not need here
housing.loc[housing['Lot Area']>20000,'Lot Area']+=50000
Comment from Alex :
if the if statement resolves to False for a value, then apply() doesn't return and just lets the value in the Series as it is
you're getting a SyntaxError because you are typing invalid syntax. the ternary operator must be used like
expression if bool else other_expression
I'm sorry for asking such a basic question here, but I'm getting a syntax error when trying to compile the following code,
let sum_of_squares_of_two_largest x y z =
let a :: b :: _ = List.sort (fun x y -> -(compare x y)) [x; y; z] in
a * a + b * b;
let rec factorial n =
if n = 0 then 1 else n * factorial (n - 1);
let e_term n = 1.0 /. float_of_int (factorial n);
let rec e_approximation n =
if n = 0 then (e_term 0) else (e_term n) +. (e_approximation (n - 1));
let rec is_even x = if x = 0 then true else is_odd (x - 1);
and is_odd x = not (is_even x);
let rec f_rec n =
if n < 3 then n else f_rec(n - 1) + 2 * f_rec(n - 2) + 3 * f_rec(n - 3);
The uninformative compiler tells me there is syntax error on line 19, which is the last line of the file.
File "source.ml", line 19, characters 0-0:
Error: Syntax error
That is, line 19 is a blank line, only with a new-line character.
I can "fix" this syntax error by adding ;; at the end of each function definition instead of the ;.
Am I missing a semicolon somewhere?
As has been pointed out in the comments, ; is not a statement terminator like in many other (Algol-inspired) languages, but a sequence operator. It takes two values, throws away the first (but warns if it is not unit) and returns the second. a; b is therefore roughly equivalent to let _ = a in b.
You say you would have expected the error to say something like ';' is missing the second operand., but the thing is: it doesn't. Not until it reaches the end of the file (at which point the error message certainly could have been more intelligent, but probably not very accurate anyway).
What follows each ; in your code looks like a perfectly valid expression that might yield a value. For example, if let rec factorial n = ...; had been let rec factorial n = ... in 2 The value of the expression would have been 2. The problem here, from the compiler's point of view, is that it runs out of file before the expression is finished.
As you've also found out, ;; actually is a statement terminator. Or perhaps a toplevel definition terminator is a better term (I don't know if it even has an official name). It's used in the REPL to terminate input, but isn't needed in normal OCaml code unless you mix toplevel definitions and expressions, which you never should.
;; can still be useful for beginners as a "bulkhead", however. If you put just one ;; in place of a ; in the middle of your file, you'll find the compiler now points the error location there. That's because ;; terminates the definition without the preceding expression being complete. So you now know there's an error before that. (Actually in the preceding expression, but since your entire file is one single expression, "before that" is really the best we can do).
I saw this pseudo-code on another stackoverflow question found here Split a string to a string of valid words using Dynamic Programming.
The problem is a dynamic programming question to see if an input string can be split into words from a dictionary.
The third line, means to set an array b of size [N+1] to all false values? I'm pretty sure about that. But what I am really not sure about is the fifth line. Is that a for-loop or what? I feel like pseudo-code saying 'for i in range' would only have 2 values. What is that line saying?
def try_to_split(doc):
N = len(doc)
b = [False] * (N + 1)
b[N] = True
for i in range(N - 1, -1, -1):
for word starting at position i:
if b[i + len(word)]:
b[i] = True
break
return b
It's confusing syntax, and I'm pretty sure there's a mistake. It should be:
for i in range(N - 1, 0, -1) //0, not -1
which I believe means
for i from (N - 1) downto 0 //-1 was the step, like i-- or i -= 1
This makes sense with the algorithm, as it simply starts at the end of the string, and solves each trailing substring until it gets to the beginning. If b[0] is true at the end, then the input string can be split into words from the dictionary. for word starting at position i just checks all words in the dictionary to see if they start at that position.
If one wants to be able to reconstruct a solution, they can change b to an int array, initialize to 0s, and change the if to this:
if b[i + len(word)] != 0
b[i] = i + len(word) //len(word) works too
break
My code below is meant to generate the square of natural numbers in order
(i.e sq(X). -> X=0; X=1; X=4; X=9; X=16; ...)
nat(0).
nat(X) :- nat(Y), Z is Y+1, X is Z*Z.
but the answer I am getting is:
1
0 ?- nat(X).
X = 0 ;
X = 1 ;
X = 4 ;
X = 25 ;
X = 676
Should be a quick fix, but I've spent longer on this than I'd like to say. Any help is greatly appreciated!
your nat/1 really seems to return a different sequence. Should be
nat(0).
nat(X) :- nat(Y), X is Y+1.
and then, a different predicate for square
sq(X) :- % call nat/1, square it...
please complete the code