i have created a script that use user inputs to greet them by hello but it prints the alphabets not the complete name in one sentence. How?
val = input()
for i in val:
print("Hello", i)
but this prints
Hello p
Hello r
Hello i
Hello n
Hello c
Hello e
The function input() takes in an input from the command line. For example, if you input prince into the command line, now the variable val has a value of "prince".
With the for loop, you are using the for-each notation. Strings are also a type of iterator--in fact, strings are simply arrays of characters. Think of it like a regular list, but instead of having a list like [1, 2, 3, 4], you have the list ['p', 'r', 'i', 'n', 'c', 'e']. So each iteration of the for loop only prints the character it is currently iterating on.
You could simplify your code by avoiding the for loop and only using the code print("Hello", val).
However, if you only want to practice with for loops, you can use the code below. Try to understand how and why you can simplify it!
val = input() //stores the user input into val
name = "" //creates an empty string called name
for s in val: //iterates through each character in val
name += s //adds that character to name
//when the for loop ends, the user input is stored in name
print("Hello", name) //prints "Hello" and the name
Related
New to Julia, trying to simply ask the user to choose 5 numbers and put it into an array and print the array. My output only says pick 5 numbers with "nothing" followed underneath. I cant seem to figure out why it wont read my inputs.
function ask()
lst = []
i = 0
println("pick 5 numbers to add to a list")
while i < 5
choice = readline
choice = push!(lst, choice);
i += 1
end
end
println(ask())
You were assigning function reference to list elements rather than calling the function.
This should be:
function ask()
lst = String[]
i = 0
println("pick 5 numbers to add to a list")
while i < 5
choice = readline()
choice = push!(lst, choice);
i += 1
end
lst
end
If you want numbers rather than Strings the last line could be parse.(Int, lst) or you could add this conversion near readline
Note that if you do not plan to introduce some error checking etc. this all code could be simply written as:
println("pick 5 numbers to add to a list")
lst = [parse(Int, readline()) for _ in 1:5]
I am new to Elixir language and I am having some issues while writing a piece of code.
What I am given is a 2D array like
list1 = [
[1 ,2,3,4,"nil"],
[6,7,8,9,10,],
[11,"nil",13,"nil",15],
[16,17,"nil",19,20] ]
Now, what I've to do is to get all the elements that have values between 10 and 20, so what I'm doing is:
final_list = []
Enum.each(list1, fn row ->
Enum.each(row, &(if (&1 >= 10 and &1 <= 99) do final_list = final_list ++ &1 end))
end
)
Doing this, I'm expecting that I'll get my list of numbers in final_list but I'm getting blank final list with a warning like:
warning: variable "final_list" is unused (there is a variable with the same name in the context, use the pin operator (^) to match on it or prefix this variable with underscore if it is not meant to be used)
iex:5
:ok
and upon printing final_list, it is not updated.
When I try to check whether my code is working properly or not, using IO.puts as:
iex(5)> Enum.each(list1, fn row -> ...(5)> Enum.each(row, &(if (&1 >= 10 and &1 <= 99) do IO.puts(final_list ++ &1) end))
...(5)> end
...(5)> )
The Output is:
10
11
13
15
16
17
19
20
:ok
What could I possibly be doing wrong here? Shouldn't it add the elements to the final_list?
If this is wrong ( probably it is), what should be the possible solution to this?
Any kind of help will be appreciated.
As mentioned in Adam's comments, this is a FAQ and the important thing is the message "warning: variable "final_list" is unused (there is a variable with the same name in the context, use the pin operator (^) to match on it or prefix this variable with underscore if it is not meant to be used)" This message actually indicates a very serious problem.
It tells you that the assignment "final_list = final_list ++ &1" is useless since it just creates a local variable, hiding the external one. Elixir variables are not mutable so you need to reorganize seriously your code.
The simplest way is
final_list =
for sublist <- list1,
n <- sublist,
is_number(n),
n in 10..20,
do: n
Note that every time you write final_list = ..., you actually declare a new variable with the same name, so the final_list you declared inside your anonymous function is not the final_list outside the anonymous function.
I'm studying Standard ML and one of the exercices I have to do is to write a function called opPairs that receives a list of tuples of type int, and returns a list with the sum of each pair.
Example:
input: opPairs [(1, 2), (3, 4)]
output: val it = [3, 7]
These were my attempts, which are not compiling:
ATTEMPT 1
type T0 = int * int;
fun opPairs ((h:TO)::t) = let val aux =(#1 h + #2 h) in
aux::(opPairs(t))
end;
The error message is:
Error: unbound type constructor: TO
Error: operator and operand don't agree [type mismatch]
operator domain: {1:'Y; 'Z}
operand: [E]
in expression:
(fn {1=1,...} => 1) h
ATTEMPT 2
fun opPairs2 l = map (fn x => #1 x + #2 x ) l;
The error message is: Error: unresolved flex record (need to know the names of ALL the fields
in this context)
type: {1:[+ ty], 2:[+ ty]; 'Z}
The first attempt has a typo: type T0 is defined, where 0 is zero, but then type TO is referenced in the pattern, where O is the letter O. This gets rid of the "operand and operator do not agree" error, but there is a further problem. The pattern ((h:T0)::t) does not match an empty list, so there is a "match nonexhaustive" warning with the corrected type identifier. This manifests as an exception when the function is used, because the code needs to match an empty list when it reaches the end of the input.
The second attempt needs to use a type for the tuples. This is because the tuple accessor #n needs to know the type of the tuple it accesses. To fix this problem, provide the type of the tuple argument to the anonymous function:
fun opPairs2 l = map (fn x:T0 => #1 x + #2 x) l;
But, really it is bad practice to use #1, #2, etc. to access tuple fields; use pattern matching instead. Here is a cleaner approach, more like the first attempt, but taking full advantage of pattern matching:
fun opPairs nil = nil
| opPairs ((a, b)::cs) = (a + b)::(opPairs cs);
Here, opPairs returns an empty list when the input is an empty list, otherwise pattern matching provides the field values a and b to be added and consed recursively onto the output. When the last tuple is reached, cs is the empty list, and opPairs cs is then also the empty list: the individual tuple sums are then consed onto this empty list to create the output list.
To extend on exnihilo's answer, once you have achieved familiarity with the type of solution that uses explicit recursion and pattern matching (opPairs ((a, b)::cs) = ...), you can begin to generalise the solution using list combinators:
val opPairs = map op+
How can I access the name property of a ProtoField after I declare it?
For example, something along the lines of:
myproto = Proto("myproto", "My Proto")
myproto.fields.foo = ProtoField.int8("myproto.foo", "Foo", base.DEC)
print(myproto.fields.foo.name)
Where I get the output:
Foo
An alternate method that's a bit more terse:
local fieldString = tostring(field)
local i, j = string.find(fieldString, ": .* myproto")
print(string.sub(fieldString, i + 2, j - (1 + string.len("myproto")))
EDIT: Or an even simpler solution that works for any protocol:
local fieldString = tostring(field)
local i, j = string.find(fieldString, ": .* ")
print(string.sub(fieldString, i + 2, j - 1))
Of course the 2nd method only works as long as there are no spaces in the field name. Since that's not necessarily always going to be the case, the 1st method is more robust. Here is the 1st method wrapped up in a function that ought to be usable by any dissector:
-- The field is the field whose name you want to print.
-- The proto is the name of the relevant protocol
function printFieldName(field, protoStr)
local fieldString = tostring(field)
local i, j = string.find(fieldString, ": .* " .. protoStr)
print(string.sub(fieldString, i + 2, j - (1 + string.len(protoStr)))
end
... and here it is in use:
printFieldName(myproto.fields.foo, "myproto")
printFieldName(someproto.fields.bar, "someproto")
Ok, this is janky, and certainly not the 'right' way to do it, but it seems to work.
I discovered this after looking at the output of
print(tostring(myproto.fields.foo))
This seems to spit out the value of each of the members of ProtoField, but I couldn't figure out the correct way to access them. So, instead, I decided to parse the string. This function will return 'Foo', but could be adapted to return the other fields as well.
function getname(field)
--First, convert the field into a string
--this is going to result in a long string with
--a bunch of info we dont need
local fieldString= tostring(field)
-- fieldString looks like:
-- ProtoField(188403): Foo myproto.foo base.DEC 0000000000000000 00000000 (null)
--Split the string on '.' characters
a,b=fieldString:match"([^.]*).(.*)"
--Split the first half of the previous result (a) on ':' characters
a,b=a:match"([^.]*):(.*)"
--At this point, b will equal " Foo myproto"
--and we want to strip out that abreviation "abvr" part
--Count the number of times spaces occur in the string
local spaceCount = select(2, string.gsub(b, " ", ""))
--Declare a counter
local counter = 0
--Declare the name we are going to return
local constructedName = ''
--Step though each word in (b) separated by spaces
for word in b:gmatch("%w+") do
--If we hav reached the last space, go ahead and return
if counter == spaceCount-1 then
return constructedName
end
--Add the current word to our name
constructedName = constructedName .. word .. " "
--Increment counter
counter = counter+1
end
end
Say I have a string.
"poop"
I want to change "poop" to "peep".
In fact, I also want all of the o's in poop to change to e's for any word I put in.
Here's my attempt to do the above.
def getword():
x = (input("Please enter a word."))
return x
def main():
y = getword()
for i in range (len(y)):
if y[i] == "o":
y = y[:i] + "e"
print (y)
main()
As you can see, when you run it, it doesn't amount to what I want. Here is my expected output.
Enter a word.
>>> brother
brether
Something like this. I need to do it using slicing. I just don't know how.
Please keep your answer simple, since I'm somewhat new to Python. Thanks!
This uses slicing (but keep in mind that slicing is not the best way to do it):
def f(s):
for x in range(len(s)):
if s[x] == 'o':
s = s[:x]+'e'+s[x+1:]
return s
Strings in python are non-mutable, which means that you can't just swap out letters in a string, you would need to create a whole new string and concatenate letters on one-by-one
def getword():
x = (input("Please enter a word."))
return x
def main():
y = getword()
output = ''
for i in range(len(y)):
if y[i] == "o":
output = output + 'e'
else:
output = output + y[i]
print(output)
main()
I'll help you this once, but you should know that stack overflow is not a homework help site. You should be figuring these things out on your own to get the full educational experience.
EDIT
Using slicing, I suppose you could do:
def getword():
x = (input("Please enter a word."))
return x
def main():
y = getword()
output = '' # String variable to hold the output string. Starts empty
slice_start = 0 # Keeps track of what we have already added to the output. Starts at 0
for i in range(len(y) - 1): # Scan through all but the last character
if y[i] == "o": # If character is 'o'
output = output + y[slice_start:i] + 'e' # then add all the previous characters to the output string, and an e character to replace the o
slice_start = i + 1 # Increment the index to start the slice at to be the letter immediately after the 'o'
output = output + y[slice_start:-1] # Add the rest of the characters to output string from the last occurrence of an 'o' to the end of the string
if y[-1] == 'o': # We still haven't checked the last character, so check if its an 'o'
output = output + 'e' # If it is, add an 'e' instead to output
else:
output = output + y[-1] # Otherwise just add the character as-is
print(output)
main()
Comments should explain what is going on. I'm not sure if this is the most efficient or best way to do it (which really shouldn't matter, since slicing is a terribly inefficient way to do this anyways), just the first thing I hacked together that uses slicing.
EDIT Yeah... Ourous's solution is much more elegant
Can slicing even be used in this situation??
The only probable solution I think would work, as MirekE stated, is y.replace("o","e").