I iterate thru items of a dictionary "var_dict".
Then as I iterate in a for loop, I need to update the dictionary.
I understand that is not possible and that triggers the runtime error I experienced.
My question is, do I need to create a different dictionary to store data? As is now, I am trying to use same dictionary with different keys.
I know the problem is related to iteration thru the key and values of a dictionary and attempt to change it. I want to know if the best option in this case if to create a separate dictionary.
for k, v in var_dict.items():
match = str(match)
match = match.strip("[]")
match = match.strip("''")
result = [index for index, value in enumerate(v) if match in value]
result = str(result)
result = result.strip("[]")
result = result.strip("'")
#====> IF I print(var_dict), at this point I have no error *********
if result == "0":
#It means a match between interface on RP PSE2 model was found; Interface position is on PSE2 architecture
print (f'PSE-2 Line cards:{v} Interfaces on PSE2:{entry} Interface PortID:{port_id}')
port_id = int(port_id)
print(port_id)
if port_id >= 19:
#print(f'interface:{entry} portID={port_id} CPU_POS={port_cpu_pos} REPLICATION=YES')
if_info = [entry,'PSE2=YES',port_id,port_cpu_pos,'REPLICATION=YES']
var_dict['IF_PSE2'].append(if_info)
#===> *** This is the point that if i attempt to print var_dict, I get the Error during olist(): dictionary changed size during iteration
else:
#print(f'interface:{entry},portID={port_id} CPU_POS={port_cpu_pos} REPLICATION=NO')
if_info = [entry,'PSE2=YES',port_id,port_cpu_pos,'REPLICATION=NO']
var_dict['IF_PSE2'].append(if_info)
else:
#it means the interface is on single PSE. No replication is applicable. Just check threshold between incoming and outgoing rate.
if_info = [entry,'PSE2=NO',int(port_id),port_cpu_pos,'REPLICATION=NO']
var_dict['IF_PSE1'].append(if_info)
I did a shallow copy and that allowed me to iterate a dictionary copy and make modifications to the original dictionary. Problem solved. Thanks.
(...)
temp_var_dict = var_dict.copy()
for k, v in temp_var_dict.items():
(...)
Related
Swiftui dictionaries have the feature that the value returned by using key access is always of type "optional". For example, a dictionary that has type String keys and type String values is tricky to access because each returned value is of type optional.
An obvious need is to assign x=myDictionary[key] where you are trying to get the String of the dictionary "value" into the String variable x.
Well this is tricky because the String value is always returned as an Optional String, usually identified as type String?.
So how is it possible to convert the String?-type value returned by the dictionary access into a plain String-type that can be assigned to a plain String-type variable?
I guess the problem is that there is no way to know for sure that there exists a dictionary value for the key. The key used to access the dictionary could be anything so somehow you have to deal with that.
As described in #jnpdx answer to this SO question (How do you assign a String?-type object to a String-type variable?), there are at least three ways to convert a String? to a String:
import SwiftUI
var x: Double? = 6.0
var a = 2.0
if x != nil {
a = x!
}
if let b = x {
a = x!
}
a = x ?? 0.0
Two key concepts:
Check the optional to see if it is nil
if the optional is not equal to nil, then go ahead
In the first method above, "if x != nil" explicitly checks to make sure x is not nil be fore the closure is executed.
In the second method above, "if let a = b" will execute the closure as long as b is not equal to nil.
In the third method above, the "nil-coalescing" operator ?? is employed. If x=nil, then the default value after ?? is assigned to a.
The above code will run in a playground.
Besides the three methods above, there is at least one other method using "guard let" but I am uncertain of the syntax.
I believe that the three above methods also apply to variables other than String? and String.
It seems Both merge and compute Map methods are created to reduce if("~key exists here~") when put.
My problem is: add to map a [key, value] pair when I know nothing: neither key existing in map nor it exist but has value nor value == null nor key == null.
words.forEach(word ->
map.compute(word, (w, prev) -> prev != null ? prev + 1 : 1)
);
words.forEach(word ->
map.merge(word, 1, (prev, one) -> prev + one)
);
Is the only difference 1 is moved from Bifunction to parameter?
What is better to use? Does any of merge, compute suggests key/val are existing?
And what is essential difference in use case of them?
The documentation of Map#compute(K, BiFunction) says:
Attempts to compute a mapping for the specified key and its current mapped value (or null if there is no current mapping). For example, to either create or append a String msg to a value mapping:
map.compute(key, (k, v) -> (v == null) ? msg : v.concat(msg))
(Method merge() is often simpler to use for such purposes.)
If the remapping function returns null, the mapping is removed (or remains absent if initially absent). If the remapping function itself throws an (unchecked) exception, the exception is rethrown, and the current mapping is left unchanged.
The remapping function should not modify this map during computation.
And the documentation of Map#merge(K, V, BiFunction) says:
If the specified key is not already associated with a value or is associated with null, associates it with the given non-null value. Otherwise, replaces the associated value with the results of the given remapping function, or removes if the result is null. This method may be of use when combining multiple mapped values for a key. For example, to either create or append a String msg to a value mapping:
map.merge(key, msg, String::concat)
If the remapping function returns null, the mapping is removed. If the remapping function itself throws an (unchecked) exception, the exception is rethrown, and the current mapping is left unchanged.
The remapping function should not modify this map during computation.
The important differences are:
For compute(K, BiFunction<? super K, ? super V, ? extends V>):
The BiFunction is always invoked.
The BiFunction accepts the given key and the current value, if any, as arguments and returns a new value.
Meant for taking the key and current value (if any), performing an arbitrary computation, and returning the result. The computation may be a reduction operation (i.e. merge) but it doesn't have to be.
For merge(K, V, BiFunction<? super V, ? super V, ? extends V>):
The BiFunction is invoked only if the given key is already associated with a non-null value.
The BiFunction accepts the current value and the given value as arguments and returns a new value. Unlike with compute, the BiFunction is not given the key.
Meant for taking two values and reducing them into a single value.
If the mapping function, as in your case, only depends on the current mapped value, then you can use both. But I would prefer:
compute if you can guarantee that a value for the given key exists. In this case the extra value parameter taken by the merge method is not needed.
merge if it is possible that no value for the given key exists. In this case merge has the advantage that null does NOT have to be handled by the mapping function.
I am reading the source code of map in go1.10.3.It seemed there exist corresponding method about operation such as:
makemap(t *maptype, hint int, h *hmap) *hmap ==> m = make(map[xx]yy)
mapaccess1(t *maptype, h *hmap, key unsafe.Pointer)==> m['key']
but I cant find the correspond method for the operation which add key/value as below:
m['xx']='yy'
there exist a method called mapassign which has some similarity with this
operation.
mapassign(t *maptype, h *hmap, key unsafe.Pointer) unsafe.Pointer
this will add a new key to the map, but as we can see, the input arguments has no value. And another question is when it has already this key, it maybe update this key.
if !alg.equal(key, k) {
continue
}
// already have a mapping for key. Update it.
if t.needkeyupdate {//why??
typedmemmove(t.key, k, key)
}
since the two key is equal, why should update it?
summary:
1. the relation between add key/value operation and method mapassign?
2. why it maybe need to update the key since the insert key and the key which has already exist is equal in mapassign method?
In the operation m[k] = v, the caller copies the value v to the address returned by mapassign.
The comments in the function needkeyupdate explain why some types need key updates: floating point & complex -0 and 0 are equal, but different values; string might have smaller backing store.
I'm messing around a bit with F# and I'm not quite sure if I'm doing this correctly. In C# this could be done with an IDictionary or something similar.
type School() =
member val Roster = Map.empty with get, set
member this.add(grade: int, studentName: string) =
match this.Roster.ContainsKey(grade) with
| true -> // Can I do something like this.Roster.[grade].Insert([studentName])?
| false -> this.Roster <- this.Roster.Add(grade, [studentName])
Is there a way to insert into the map if it contains a specified key or am I just using the wrong collection in this case?
The F# Map type is a mapping from keys to values just like ordinary .NET Dictionary, except that it is immutable.
If I understand your aim correctly, you're trying to keep a list of students for each grade. The type in that case is a map from integers to lists of names, i.e. Map<int, string list>.
The Add operation on the map actually either adds or replaces an element, so I think that's the operation you want in the false case. In the true case, you need to get the current list, append the new student and then replace the existing record. One way to do this is to write something like:
type School() =
member val Roster = Map.empty with get, set
member this.Add(grade: int, studentName: string) =
// Try to get the current list of students for a given 'grade'
let studentsOpt = this.Roster.TryFind(grade)
// If the result was 'None', then use empty list as the default
let students = defaultArg studentsOpt []
// Create a new list with the new student at the front
let newStudents = studentName::students
// Create & save map with new/replaced mapping for 'grade'
this.Roster <- this.Roster.Add(grade, newStudents)
This is not thread-safe (because calling Add concurrently might not update the map properly). However, you can access school.Roster at any time, iterate over it (or share references to it) safely, because it is an immutable structure. However, if you do not care about that, then using standard Dictionary would be perfectly fine too - depends on your actual use case.
For a homework assignment, we've been instructed to complete a task without introducing any "side-effects". I've looked up "side-effects" on Wikipedia, and though I get that in theory it means "modifies a state or has an observable interaction with calling functions", I'm having trouble figuring out specifics.
For example, would creating a value that holds a non-compile time result be introducing side effects?
Say I had (might not be syntactically perfect):
val myList = (someFunction x y);;
if List.exists ((=) 7) myList then true else false;;
Would this introduce side-effects? I guess maybe I'm confused on what "modifies a state" means in the definition of side-effects.
No; a side-effect refers to e.g. mutating a ref cell with the assignment operator :=, or other things where the value referred to by a name changes over time. In this case, myList is an immutable value that never changes during the program, thus it is effect-free.
See also
http://en.wikipedia.org/wiki/Referential_transparency_(computer_science)
A good way to think about it is "have I changed anything which any later code (including running this same function again later) could ever possibly see other than the value I'm returning?" If so, that's a side effect. If not, then you can know that there isn't one.
So, something like:
let inc_nosf v = v+1
has no side effects because it just returns a new value which is one more than an integer v. So if you run the following code in the ocaml toplevel, you get the corresponding results:
# let x = 5;;
val x : int = 5
# inc_nosf x;;
- : int = 6
# x;;
- : int = 5
As you can see, the value of x didn't change. So, since we didn't save the return value, then nothing really got incremented. Our function itself only modifies the return value, not x itself. So to save it into x, we'd have to do:
# let x = inc_nosf x;;
val x : int = 6
# x;;
- : int = 6
Since the inc_nosf function has no side effects (that is, it only communicates with the outside world using its return value, not by making any other changes).
But something like:
let inc_sf r = r := !r+1
has side effects because it changes the value stored in the reference represented by r. So if you run similar code in the top level, you get this, instead:
# let y = ref 5;;
val y : int ref = {contents = 5}
# inc_sf y;;
- : unit = ()
# y;;
- : int ref = {contents = 6}
So, in this case, even though we still don't save the return value, it got incremented anyway. That means there must have been changes to something other than the return value. In this case, that change was the assignment using := which changed the stored value of the ref.
As a good rule of thumb, in Ocaml, if you avoid using refs, records, classes, strings, arrays, and hash tables, then you will avoid any risk of side effects. Although you can safely use string literals as long as you avoid modifying the string in place using functions like String.set or String.fill. Basically, any function which can modify a data type in place will cause a side effect.