I am using arguments from command line which is in the form of an Array and I would like to convert that into a Map.
So for example when I run my code with scala abc.scala A 10 B 20 C 30 I want to have a Map(A->10, B->20, C->30). Also I can use only val so I cannot reassign it because it is immutable. I am using the following piece of code unsuccessfully:
val names = args.filter(x => for(i <- 0 to args.length-1) i%2==0)
val numbers = args.partition(args(i) => i%2==1)
names.zip(numbers).toMap
You want grouped:
args.grouped(2).map { case Array(n,v) => (n,v) }.toMap
You're using for in completely the wrong way. That runs a new iteration every element of your args. If you want an index that you can work off of, try args.zipWithIndex (which pairs an index with each element). Alternatively, look at args.grouped(2).
Related
I have two lists and I want to return a result in the following way:
the result should contain elements that are in list one and list two
output should be same order as per first list
Input :
val first = listOf(1, 2, 3, 4, 5,7,9,15,11)
val second = listOf(2, 15 , 4,3, 11)
Output:
val output = listOf(2,3,4,15,11)
Please help me to learn how to get common values in both lists in order of list first in Kotlin.
You can do
val output = first.filter { second.contains(it) }
What you are looking for is the intersection of the two lists:
val output = first.intersect(second)
As pointed out by #Ivo the result is a Set which can be turned into a list with output.toList(). However, since the result is a set, it contains no duplicates, e.g. if first is listOf(1,2,3,1,2,3) and second is listOf(2,4,2,4), the result will be equal to setOf(2).
If this is not acceptable, the solution of #Ivo should be used instead.
I am trying to generate the 1000 random key-val map pairs from the given(statically defined) 2 map key-val pairs in scala and also later i would want to break the key and value pairs and store them into separate variables
Whatever i have tried:
object methodTest extends App{
val testMap = Map("3875835"->"ABCDE","316067107"->"EFGHI")
def getRandomElement(seq: Map[String,String]): Map[String,String] = {
seq(Random.nextInt(seq.length))
}
var outList = List.empty[Map[String,String]]
for(i<-0 to 1000){
outList+=getRandomElement(testMap)
}
print(outList)
}
The Output should generate 1000 Key-val pairs of map like i am showing below
[3875835,ABCDE]
[316067107,EFGHI]
[3875835,ABCDE]
[316067107,EFGHI]
[316067107,EFGHI]
............
............
............. upto 1000 random key-val pairs
Please help me figure out where i am going wrong and let me know How to resolve it, if any issue regarding the requirement, please feel free to comment for it
You can transform your seed map testMap into sequence of key/value tuples using .toSeq and then generate key/value pairs by iterating over the list of numbers from 0 until 1000, associating each number to a random choice between first or second element of the seed:
import scala.util.Random
val testMap = Map("3875835" -> "ABCDE", "316067107" -> "EFGHI")
val seed = testMap.toSeq
val keyValuesList = (0 until 1000).map(index => seed(Random.nextInt(seed.size)))
Note: 0 until 1000 will return all the numbers from 0 to 1000 excluded, so 1000 numbers. If you use 0 to 1000 you will get all the numbers from 0 to 1000 included, so 1001 numbers.
If you want to print the resulting list, you can use .foreach method with println function as argument:
keyValuesList.foreach(println)
And you will get:
(3875835,ABCDE)
(316067107,EFGHI)
(3875835,ABCDE)
(3875835,ABCDE)
(316067107,EFGHI)
(3875835,ABCDE)
...
if you want to keep only the keys, you can iterate on the list using .map method, taking only the first element of the tuple by using ._1 method, that retrieve the first element of a tuple:
val keys = keyValuesList.map(keyValuePair => keyValuePair._1)
And if you want only a list containing all second elements of each pair:
val values = keyValuesList.map(keyValuePair => keyValuePair._2)
Assume I have the following matrix:
X = np.array([[1,2,3], [4,5,6], [7,8,9], [70,80,90], [45,43,68], [112,87,245]])
I want to draw a batch of 2 random rows at each time loop, and send it to a function. For instance, a batch in iteration i can be batch = [[4,5,6], [70,80,90]]
I do the following:
X = np.array([[1,2,3], [4,5,6], [7,8,9], [70,80,90], [45,43,68], [112,87,245]])
def caclulate_batch(batch):
pass
for i in range(X.shape[0]/2):
batch = np.array([])
for _ in range(2):
r = random.randint(0, 5)
batch = np.append(batch, X[r])
caclulate_batch(batch)
There are two problems here: (1) It returns appended array (2) The random number can be repeated which can choose the same row many times. How can modify the code to fit my requirement.
r = np.random.randint(0, len(x), 2) should get you the indices. That lets you use fancy indexing to get the subset: batch = x[r, :].
If you want to accumulate arrays along a new dimension, as your loop does, use np.stack or np.block instead of np.append.
(1) You can use numpy.stack instead of append. EDIT: But this function would be called when you have all your batch in a list like:
list = ([1,2], [3,4])
numpy.stack(list)
# gives [[1,2],
# [3,4]]
(2) You can shuffle X array, loop through the results and extract two by two. Look at numpy.random.shuffle
It would look like that:
S = np.random.shuffle(X)
for i in range(S.shape[0]/2):
batch = S[i*2:i*2+1]
caclulate_batch(batch)
I used the split() function to convert string to a list time = time.split() and this is how my output looks like :
[u'1472120400.107']
[u'1472120399.999']
[u'1472120399.334']
[u'1472120397.633']
[u'1472120397.261']
[u'1472120394.328']
[u'1472120393.762']
[u'1472120393.737']
Then I tried accessing the contents of the list using print time[1] which gives the index out of range error (cause only a single value is stored in one list). I checked questions posted by other people and used print len(time). This is the output for that:
1
[u'1472120400.107']
1
[u'1472120399.999']
1
[u'1472120399.334']
1
[u'1472120397.633']
1
[u'1472120397.261']
1
[u'1472120394.328']
1
[u'1472120393.762']
1
[u'1472120393.737']
I do this entire thing inside a for loop because I get logs dynamically and have to extract out just the time.
This is part of my code:
line_collect = lines.collect() #spark function
for line in line_collect :
a = re.search(rx1,line)
time = a.group()
time = time.split()
#print time[1] #index out of range error which is why I wrote another for below
for k in time :
time1 = time[k]#trying to put those individual list values into one variable but get type error
print len(time1)
I get the following error :
time1 = time[k]
TypeError: list indices must be integers, not unicode
Can someone tell me how to read each of those single list values into just one list so I can access each of them using a single index[value]. I'm new to python.
My required output:
time =['1472120400.107','1472120399.999','1472120399.334','1472120397.633','1472120397.261','1472120394.328','1472120393.762','1472120393.737']
so that i can use time[1] to give 1472120399.999 as result.
Update: I misunderstood what you wanted. You have the correct output already and it's a string. The reason you have a u before the string is because it's a unicode string that has 16 bits. u is a python flag to distinguish it from a normal string. Printing it to the screen will give you the correct string. Use it normally as you would any other string.
time = [u'1472120400.107'] # One element just to show
for k in time:
print(k)
Looping over a list using a for loop will give you one value at a time, not the index itself. Consider using enumerate:
for k, value in enumerate(time):
time1 = value # Or time1 = time[k]
print(time1)
Or just getting the value itself:
for k in time:
time1 = k
print(time1)
--
Also, Python is zero based language, so to get the first element out of a list you probably want to use time[0].
Thanks for your help. I finally got the code right:
newlst = []
for line in line_collect :
a = re.search(rx1,line)
time = a.group()
newlst.append(float(time))
print newlst
This will put the whole list values into one list.
Output:
[1472120400.107, 1472120399.999, 1472120399.334, 1472120397.633,
1472120397.261, 1472120394.328, 1472120393.762, 1472120393.737]
I am trying to create a sequence of similar dictionaries to further store them in a tuple. I tried two approaches, using and not using a for loop
Without for loop
dic0 = {'modo': lambda x: x[0]}
dic1 = {'modo': lambda x: x[1]}
lst = []
lst.append(dic0)
lst.append(dic1)
tup = tuple(lst)
dic0 = tup[0]
dic1 = tup[1]
f0 = dic0['modo']
f1 = dic1['modo']
x = np.array([0,1])
print (f0(x) , f1(x)) # 0 , 1
With a for loop
lst = []
for j in range(0,2):
dic = {}
dic = {'modo': lambda x: x[j]}
lst.insert(j,dic)
tup = tuple(lst)
dic0 = tup[0]
dic1 = tup[1]
f0 = dic0['modo']
f1 = dic1['modo']
x = np.array([0,1])
print (f0(x) , f1(x)) # 1 , 1
I really don't understand why I am getting different results. It seems that the last dictionary I insert overwrite the previous ones, but I don't know why (the append method does not work neither).
Any help would be really welcomed
This is happening due to how scoping works in this case. Try putting j = 0 above the final print statement and you'll see what happens.
Also, you might try
from operator import itemgetter
lst = [{'modo': itemgetter(j)} for j in range(2)]
You have accidentally created what is know as a closure. The lambda functions in your second (loop-based) example include a reference to a variable j. That variable is actually the loop variable used to iterate your loop. So the lambda call actually produces code with a reference to "some variable named 'j' that I didn't define, but it's around here somewhere."
This is called "closing over" or "enclosing" the variable j, because even when the loop is finished, there will be this lambda function you wrote that references the variable j. And so it will never get garbage-collected until you release the references to the lambda function(s).
You get the same value (1, 1) printed because j stops iterating over the range(0,2) with j=1, and nothing changes that. So when your lambda functions ask for x[j], they're asking for the present value of j, then getting the present value of x[j]. In both functions, the present value of j is 1.
You could work around this by creating a make_lambda function that takes an index number as a parameter. Or you could do what #DavisYoshida suggested, and use someone else's code to create the appropriate closure for you.