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Assume I have two arrays, both of them are sorted, for example:
A: [1, 4, 5, 8, 10, 24]
B: [3, 6, 9, 29, 50, 65]
And then I merge these two array into one array and keep original relative order of both two array
C: [1, 4, 3, 5, 6, 9, 8, 29, 10, 24, 50, 65]
Is there any way to split C into two sorted array in O(n) time?
note: not necessarily into the original A and B
Greedily assign your integers to list 1 if they can go there. If they can't, assign them to list 2.
Here's some Ruby code to play around with this idea. It randomly splits the integers from 0 to n-1 into two sorted lists, then randomly merges them, then applies the greedy approach.
def f(n)
split1 = []
split2 = []
0.upto(n-1) do |i|
if rand < 0.5
split1.append(i)
else
split2.append(i)
end
end
puts "input 1: #{split1.to_s}"
puts "input 2: #{split2.to_s}"
merged = []
split1.reverse!
split2.reverse!
while split1.length > 0 && split2.length > 0
if rand < 0.5
merged.append(split1.pop)
else
merged.append(split2.pop)
end
end
merged += split1.reverse
merged += split2.reverse
puts "merged: #{merged.to_s}"
merged.reverse!
greedy1 = [merged.pop]
greedy2 = []
while merged.length > 0
if merged[-1] >= greedy1[-1]
greedy1.append(merged.pop)
else
greedy2.append(merged.pop)
end
end
puts "greedy1: #{greedy1.to_s}"
puts "greedy2: #{greedy2.to_s}"
end
Here's sample output:
> f(20)
input 1: [2, 3, 4, 5, 8, 9, 10, 18, 19]
input 2: [0, 1, 6, 7, 11, 12, 13, 14, 15, 16, 17]
merged: [2, 0, 1, 6, 3, 4, 5, 8, 9, 7, 10, 11, 18, 12, 13, 19, 14, 15, 16, 17]
greedy1: [2, 6, 8, 9, 10, 11, 18, 19]
greedy2: [0, 1, 3, 4, 5, 7, 12, 13, 14, 15, 16, 17]
> f(20)
input 1: [1, 3, 5, 6, 8, 9, 10, 11, 13, 15]
input 2: [0, 2, 4, 7, 12, 14, 16, 17, 18, 19]
merged: [0, 2, 4, 7, 12, 14, 16, 1, 3, 5, 6, 8, 17, 9, 18, 10, 19, 11, 13, 15]
greedy1: [0, 2, 4, 7, 12, 14, 16, 17, 18, 19]
greedy2: [1, 3, 5, 6, 8, 9, 10, 11, 13, 15]
> f(20)
input 1: [0, 1, 2, 6, 7, 9, 11, 14, 15, 18]
input 2: [3, 4, 5, 8, 10, 12, 13, 16, 17, 19]
merged: [3, 4, 5, 8, 10, 12, 0, 13, 16, 17, 1, 19, 2, 6, 7, 9, 11, 14, 15, 18]
greedy1: [3, 4, 5, 8, 10, 12, 13, 16, 17, 19]
greedy2: [0, 1, 2, 6, 7, 9, 11, 14, 15, 18]
Let's take your example.
[1, 4, 3, 5, 6, 9, 8, 29, 10, 24, 50, 65]
In time O(n) you can work out the minimum of the tail.
[1, 3, 3, 5, 6, 8, 8, 10, 10, 24, 50, 65]
And now the one stream is all cases where it is the minimum, and the other is the cases where it isn't.
[1, 3, 5, 6, 8, 10, 24, 50, 65]
[ 4, 9, 29, ]
This is all doable in time O(n).
We can go further and now split into 3 streams based on which values in the first stream could have gone in the last without changing it being increasing.
[ 3, 5, 6, 8, 10, 24, ]
[1, 5, 6, 8, 50, 65]
[ 4, 9, 29, ]
And now we can start enumerating the 2^6 = 64 different ways of splitting the original stream back into 2 increasing streams.
I have an array of products with 234 items.
I need to create another array with a pagination (every 10 items)
example:
[
[1,2,3,4,5,6,7,8,9,10],
[1,2,3,4,5,6,7,8,9,10],
[1,2,3,4,5,6,7,8,9,10],
...
]
How can I solve this?
I've tried in_groups_of but I don't have success.
You're looking for each_slice
Whenever you have an array problem, check the Enumerable. in_groups_of is a Rails method and uses each_slice under the hood.
Just use Enumerable#each_slice
[*1..34].each_slice(10).to_a
# =>
# [
# [1, 2, 3, 4, 5, 6, 7, 8, 9, 10],
# [11, 12, 13, 14, 15, 16, 17, 18, 19, 20],
# [21, 22, 23, 24, 25, 26, 27, 28, 29, 30],
# [31, 32, 33, 34]
# ]
Based on the documentation provided here, https://github.com/huggingface/transformers/blob/v4.21.3/src/transformers/modeling_outputs.py#L101, how can i read all the outputs, last_hidden_state (), pooler_output and hidden_state. in my sample code below, i get the outputs
from transformers import BertModel, BertConfig
config = BertConfig.from_pretrained("xxx", output_hidden_states=True)
model = BertModel.from_pretrained("xxx", config=config)
outputs = model(inputs)
when i print one of the output (sample below) . i looked through the documentation to see if i can use some functions of this class to just get the last_hidden_state values , but i'm not sure of the type here.
the value for the last_hidden_state =
tensor([[...
is it some class or tuple or array .
how can i get the values or array of values such as
[0, 1, 2, 3 , ...]
BaseModelOutputWithPoolingAndNoAttention(
last_hidden_state=tensor([
[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,
11, 12, 13, 14, 15, 16, 17,
18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29],
[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14,
15, 16, 17,
18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29]
...
hidden_states= ...
The BaseModelOutputWithPoolingAndCrossAttentions you retrieve is class that inherits from OrderedDict (code) that holds pytorch tensors. You can access the keys of the OrderedDict like properties of a class and, in case you do not want to work with Tensors, you can them to python lists or numpy. Please have a look at the example below:
from transformers import BertTokenizer, BertModel
t = BertTokenizer.from_pretrained("bert-base-cased")
m = BertModel.from_pretrained("bert-base-cased")
i = t("This is a test", return_tensors="pt")
o = m(**i, output_hidden_states=True)
print(o.keys())
print(type(o.last_hidden_state))
print(o.last_hidden_state.tolist())
print(o.last_hidden_state.detach().numpy())
Output:
odict_keys(['last_hidden_state', 'pooler_output', 'hidden_states'])
<class 'transformers.modeling_outputs.BaseModelOutputWithPoolingAndCrossAttentions'>
<class 'torch.Tensor'>
[[[0.36328405141830444, 0.018902940675616264, 0.1893523931503296, ..., 0.09052444249391556, 1.4617693424224854, 0.0774402841925621]]]
[[[ 0.36328405 0.01890294 0.1893524 ... -0.0259465 0.38701165
0.19099694]
[ 0.30656984 -0.25377586 0.76075834 ... 0.2055152 0.29494798
0.4561815 ]
[ 0.32563183 0.02308523 0.665546 ... 0.34597045 -0.0644953
0.5391255 ]
[ 0.3346715 -0.02526359 0.12209094 ... 0.50101244 0.36993945
0.3237842 ]
[ 0.18683438 0.03102166 0.25582778 ... 0.5166369 -0.1238729
0.4419385 ]
[ 0.81130844 0.4746894 -0.03862225 ... 0.09052444 1.4617693
0.07744028]]]
I got this question in an interview and got almost all the way to the answer but got stuck on the last part. If I want to get the multiplication table for 5, for instance, I want to get the output to be formatted like so:
1, 2, 3, 4, 5
2, 4, 6, 8, 10
3, 6, 9, 12, 15
4, 8, 12, 16, 20
5, 10, 15, 20, 25
My answer to this is:
def make_table(n)
s = ""
1.upto(n).each do |i|
1.upto(n).each do |j|
s += (i*j).to_s
end
s += "\n"
end
p s
end
But the output for make_table(5) is:
"12345\n246810\n3691215\n48121620\n510152025\n"
I've tried variations with array but I'm getting similar output.
What am I missing or how should I think about the last part of the problem?
You can use map and join to get a String in one line :
n = 5
puts (1..n).map { |x| (1..n).map { |y| x * y }.join(', ') }.join("\n")
It iterates over rows (x=1, x=2, ...). For each row, it iterates over cells (y=1, y=2, ...) and calculates x*y. It joins every cells in a row with ,, and joins every rows in the table with a newline :
1, 2, 3, 4, 5
2, 4, 6, 8, 10
3, 6, 9, 12, 15
4, 8, 12, 16, 20
5, 10, 15, 20, 25
If you want to keep the commas aligned, you can use rjust :
puts (1..n).map { |x| (1..n).map { |y| (x * y).to_s.rjust(3) }.join(',') }.join("\n")
It outputs :
1, 2, 3, 4, 5
2, 4, 6, 8, 10
3, 6, 9, 12, 15
4, 8, 12, 16, 20
5, 10, 15, 20, 25
You could even go fancy and calculate the width of n**2 before aligning commas :
n = 11
width = Math.log10(n**2).ceil + 1
puts (1..n).map { |x| (1..n).map { |y| (x * y).to_s.rjust(width) }.join(',') }.join("\n")
It outputs :
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22
3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33
4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44
5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55
6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66
7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77
8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88
9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99
10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110
11, 22, 33, 44, 55, 66, 77, 88, 99, 110, 121
Without spaces between the figures, the result is indeed unreadable. Have a look at the % operator, which formats strings and numbers. Instead of
s += (i*j).to_s
you could write
s += '%3d' % (i*j)
If you really want to get the output formatted in the way you explained in your posting (which I don't find that much readable), you could do a
s += "#{i*j}, "
This leaves you with two extra characters at the end of the line, which you have to remove. An alternative would be to use an array. Instead of the inner loop, you would have then something like
s += 1.upto(n).to_a.map {|j| i*j}.join(', ') + "\n"
You don't need to construct a string if you're only interested in printing the table and not returning the table(as a string).
(1..n).each do |a|
(1..n-1).each { |b| print "#{a * b}, " }
puts a * n
end
This is how I'd do it.
require 'matrix'
n = 5
puts Matrix.build(n) { |i,j| (i+1)*(j+1) }.to_a.map { |row| row.join(', ') }
1, 2, 3, 4, 5
2, 4, 6, 8, 10
3, 6, 9, 12, 15
4, 8, 12, 16, 20
5, 10, 15, 20, 25
See Matrix::build.
You can make it much shorter but here's my version.
range = Array(1..12)
range.each do |element|
range.map { |item| print "#{element * item} " } && puts
end
I am trying to find all the possible differences between the elements of one list.
For example:
x=[1,4,10,17,20,35].
I would like to have as an answer an array:
y=[3, 9, 16, 19, 34, 3, 6, 13, 16, 31, 9, 6, 7, 10, 25, 16, 13, 10, 3, 18, 19, 16, 10, 3, 15, 34, 31, 25, 18, 15]
corresponding to
[1-4, 1-10, 1-17, 1-20, 1-35, 4-1, 4-10, 4-17, ....]
I have tried to do that with diff, but I only get the difference of two consecutive numbers. and I do not really know how to compute it in a loop.
Can you please help?
A Python 1-liner:
>>> [abs(a - b) for i,a in enumerate(x) for j,b in enumerate(x) if i != j]
[3, 9, 16, 19, 34, 3, 6, 13, 16, 31, 9, 6, 7, 10, 25, 16, 13, 7, 3, 18, 19, 16, 10, 3, 15, 34, 31, 25, 18, 15]
A solution written in python
elements = [1,4,10,17,20,35]
differences = []
for i , element in enumerate(elements):
for j, element2 in enumerate(elements):
if i != j:
differences.append( abs(element - element2) )
Writing in Java, it is as simple as this:
List<Integer> diff = new ArrayList<Integer>();
for(int i=0; i<list.size(); i++) {
for(int j=0; j<list.size(); j++) {
if(i != j)
diff.add(Math.abs(list.get(i) - list.get(j)));
}
}