The following test passes while I actually would like to see it fail. The order is important for my use case. However I thought groovy is always using linked lists so ordering should be testable.
def "test foo"() {
given:
def a = [a: 1, c: 3, b: 2]
when:
def b = [a: 1, b: 2, c: 3]
then:
a == b
}
If you want to test order of keys in those two LinkedHashMap instances you can do the following:
def "test foo"() {
given:
def a = [a: 1, c: 3, b: 2]
when:
def b = [a: 1, b: 2, c: 3]
then: "make sure maps are equal"
a == b
and: "make sure entries are defined in the same order"
a.collect { it.key } == b.collect { it.key }
}
LinkedHashMap does not override equals method (it uses the one defined in AbstractMap class, the same one used by e.g. HashMap) and it only defines order of the iteration (order in which entries are added to the map).
Both assertions can be simplified to a single:
def "test foo"() {
given:
def a = [a: 1, c: 3, b: 2]
when:
def b = [a: 1, b: 2, c: 3]
then: "compare ordered list of map entries"
a.collect { it } == b.collect { it }
}
You could use toMapString() in your comparison
a.toMapString() == b.toMapString()
toMapString converts the map into a String which means the order will impact the comparison
'[a:1, c:3, b:2]' == '[a:1, b:2, c:3]'
will return false.
Related
di = [{ "k": [1, 3, 5], "k1": { "k10" : 4, "k11": [4, 7, 9], "k12" : { "k120" : { "k121" : "v121" }}, "k14" : 6}}, {"k22": { "k221" : "v122"}}]
key_list = list()
val_list = list()
for i in di:
key_list.extend(i.keys())
val_list.extend(i.values())
for i in val_list:
if "dict" in str(type(i)):
key_list.extend(i.keys())
val_list.remove(i)
val_list.extend(i.values())
for i in val_list:
if "dict" in str(type(i)):
key_list.extend(i.keys())
val_list.remove(i)
val_list.extend(i.values())
print("Key list: ",key_list)
print("Vlaue list: ", val_list)
"""This is giving my answer but i need a optimised way and also for N nested dictionary how to get all keys and values, I need get all keys in a list and all values in a list."""
About optimization: If there is a reason of existing hierarchy there is not "optimised way" to disrespect the hierarchy, otherwise a real optimization needs to include a change not to construct the hierarchy that would be disrespected later.
I wrote the following, is not optimized but decently efficient and concise, it does something in the direction of what you imprecisely asked: (I let you refine what you intend to be returned, I noticed that you silently excluded to return those value whose type is dictionary, I didn't make such discrimination). To help to understand the flexibility of the code I made some examples as cases controlled by the behavioral variable bkv, this behavioral variable can be squeezed when you know what you want to obtain.
dictype=type({})
listype=type([])
def through(ref,bkv):
""" bkv in binary 01=1 yield key
10=2 yield value
11=3 yield (key,value) """
if type(ref)==listype:
for x in ref:
yield from through(x,bkv)
elif type(ref)==dictype:
for kv in ref.items():
if bkv==1:
yield kv[0]
elif bkv==2:
yield kv[1]
elif bkv==3:
yield kv
else:
throw(TypeError, "Not implemented")
yield from through(kv[1],bkv)
di = [
{ "k": [1, 3, 5],
"k1": { "k10" : 4,
"k11": [4, 7, 9],
"k12" : { "k120" : { "k121" : "v121" }},
"k14" : 6}},
{"k22": { "k221" : "v122"}}
]
list_keys=[x for x in through(di,1)]
list_values=[x for x in through(di,2)]
list_keyval=[x for x in through(di,3)]
print("Key list: ",list_keys)
print("Values list (all, not excluding dict values): ",list_values)
print("all (Key,Value) pairs:", list_keyval)
Here is the code using collect
def lst = [1,2,3,4];
def newlst = [];
newlst = lst.collect {element -> return element * element}
println(newlst);
Here is the code using findResults
def lst2 = [1,2,3,4];
def newlst2 = [];
newlst2 = lst2.findResults {element -> return element * element}
println(newlst2);
Both seem to return [1, 4, 9, 16] so what is the difference? Thanks!
Basically the difference is how they deal with null values
collect when sees null will collect it, while findResults won't pick it.
In other words, the size of resulting collection is the same as the size of input when using collect.
Of course you could filter out the results but its an additional step
Here is a link to the example I've found in the internet
Example:
def list = [1, 2, 3, 4]
println list.collect { it % 2 ? it : null}
// [1, null, 3, null]
println list.findResults { it % 2 ? it : null}
// [1,3]
When we need to check if returned list is empty then findResults seem more useful. Thanks to Mark for the answer.
def list = [1, 2, 3, 4]
def l1 = list.collect { it % 100 == 0 ? it : null}
def l2 = list.findResults { it % 100 == 0 ? it : null}
if(l1){
println("not null/empty " + l1)
}
if(l2){
println("not null/empty " + l2)
}
I have a map in Groovy:
['keyOfInterest' : 1, 'otherKey': 2]
There is a list containing a number of these maps. I want to know if a map exists in the list with keyOfInterest of a certain value.
If the data types were simple objects, I could use indexOf(), but I don't know how to do this with a more complicated type. E.g. (taken from the docs)
assert ['a', 'b', 'c', 'd', 'c'].indexOf('z') == -1 // 'z' is not in the list
I'd like to do something like:
def mapA = ['keyOfInterest' : 1, 'otherKey': 2]
def mapB = ['keyOfInterest' : 3, 'otherKey': 2]
def searchMap = ['keyOfInterest' : 1, 'otherKey': 5]
def list = [mapA, mapB]
assert list.indexOf(searchMap) == 0 // keyOfInterest == 1 for both mapA and searchMap
Is there a way to do this with more complicated objects, such as a map, easily?
While #dmahapatro is correct, and you can use find() to find the map in the list of maps that has the matching index... that's not what you asked for. So I'll show how you can get either the index of that entry in the list, or just whether a map with matching keyOfInterest exists.
def mapA = ['keyOfInterest' : 1, 'otherKey': 2]
def mapB = ['keyOfInterest' : 3, 'otherKey': 2]
def searchMap = ['keyOfInterest':1, 'otherKey': 55 ]
def list = [mapA, mapB]
// findIndexOf() returns the first index of the map that matches in the list, or -1 if none match
assert list.findIndexOf { it.keyOfInterest == searchMap.keyOfInterest } == 0
assert list.findIndexOf { it.keyOfInterest == 33 } == -1
// any() returns a boolean OR of all the closure results for each entry in the list.
assert list.any { it.keyOfInterest == searchMap.keyOfInterest } == true
assert list.any { it.keyOfInterest == 33 } == false
Note that there is no performance penalty for using one over the other as they all stop as soon as one match is found. find() gives you the most information, but if you're actually looking for the index or a boolean result, these others can also be used.
Simplest implementation would be to use find(). It returns null when criteria is not met in the supplied closure.
def mapA = ['keyOfInterest' : 1, 'otherKey': 2]
def mapB = ['keyOfInterest' : 3, 'otherKey': 2]
def list = [mapA, mapB]
assert list.find { it.keyOfInterest == 1 } == ['keyOfInterest':1, 'otherKey':2]
assert !list.find { it.keyOfInterest == 7 }
I have a map,
def map= [name:[Vin], email:[vin#gmail.com], phone:[9988888888], jobTitle:[SE]]
i want get the total number of values that a key holds
for ex,
key name can have many values like [name:[Vin,Hus,Rock] how to do it programatically?
def count = map.name.size() //gives wrong answer
You can use the following code to get a list of size for all key.
def map= [name:['Vin',''], email:['vin#gmail.com'], phone:['9988888888'], jobTitle:['SE']]
map.collect{it.value.size()}
Output:
[2, 1, 1, 1]
I think map.name.size() should work fine too in groovy.
def map= [name :['Vin', 'abc', 'xyz'],
email:['vin#gmail.com'],
phone:[9988888888],
jobTitle:['SE']]
//Spread operator to get size of each value
assert map.values()*.size == [3, 1, 1, 1]
//Implicit spread
assert map.values().size == [3, 1, 1, 1]
//use size() to get the size of the values collection
assert map.values().size() == 4
//Values
assert map.values() as List == [['Vin', 'abc', 'xyz'],
['vin#gmail.com'], [9988888888], ['SE']]
I'm trying to figure out how to remove an item from a list in groovy from within a loop.
static main(args) {
def list1 = [1, 2, 3, 4]
for(num in list1){
if(num == 2)
list1.remove(num)
}
println(list1)
}
list = [1, 2, 3, 4]
newList = list.findAll { it != 2 }
Should give you all but the 2
Of course you may have a reason for requiring the loop?
If you want to remove the item with index 2, you can do
list = [1,2,3,4]
list.remove(2)
assert list == [1,2,4]
// or with a loop
list = [1,2,3,4]
i = list.iterator()
2.times {
i.next()
}
i.remove()
assert list == [1,2,4]
If you want to remove the (first) item with value 2, you can do
list = [1,2,3,4]
list.remove(list.indexOf(2))
assert list == [1,3,4]
// or with a loop
list = [1,2,3,4]
i = list.iterator()
while (i.hasNext()) {
if (i.next() == 2) {
i.remove()
break
}
}
assert list == [1,3,4]
As you state in your comment that you do not specifically require a loop .... If you are happy to modify your original list you can use removeAll:
// Remove all negative numbers
list = [1, 2, -4, 8]
list.removeAll { it < 0 }
I think you can do:
list - 2;
or...
list.remove(2)
There's no loop required.
If you want to use a loop I guess you could look at using the iterator to actually remove the item.
import java.util.Iterator;
static main(args) { def list1 = [1, 2, 3, 4]
Iterator i = list1.iterator();
while (i.hasNext()) {
n = i.next();
if (n == 2) i.remove();
}
println(list1)
}
but I don't see why you'd want to do it that way.