For the tfidf result matrix, I wanted to get the top tfidf values. I saw how one could set max features amount for the tfidf vectorizer, but that is for the words with the top tf count. I want to still get the high values for the tfidf, which could include words with low tf. One idea I looked up is doing something like tf_idf_matrix.sum(axis=0), which would sum up the columns. This works in my code, but because of 113k columns, print wont show them all. If I could use something like argsort() to access the top K column sum values, that would be helpful.
This question stems off my original question which is here.
The reason is that I want to know which words are the ones I should look at closer, and not necessarily the ones that have the highest frequency. I would also like to know about the "anomalies" that is, words that might not appear in all or many documents/posts but could have a high tfidf in a one or fewer documents. In case there are other approaches I should consider, I wanted to explain this.
Thanks
Related
I have a dataset of n observations (nx1 vector) and would like to create a subset of this data, whose mean is known in advance, by selecting at random only n/3 observations (or within some constraint, ie where the mean of the data subset is within a range about the known mean).
Can someone please help me with the code do this in matlab?
Note, I don't want to use the rand function to create random data as I already have my data collected.
For example on a smaller scale: If I had the following dataset of 12 observations:
data = [8;7;4;6;9;6;4;7;3;2;1;1];
but then wanted to randomly select a subset of this data containing only 4 observations with a mean of 4 (or with a mean between 3.5-4.5 for example):
Then the answer might be datasubset=[7;3;2;4] but the answer could also be datasubset=[6;4;2;4] or datasubset=[6;4;3;4].
It doesn't matter if there are several possible solutions, I just need one of them, though I'd like to know the alternative solutions also.
I am trying to create a graph with two lines, with two filters from the same dimension.
I have a dimension which has 20+ values. I'd like one line to show data based on just one of the selected values and the other line to show a line excluding that same value.
I've tried the following:
-Creating a duplicate/copy dimension and filtering the original one with the first, and the copy with the 2nd. When I do this, the graphic disappears.
-Creating a calculated field that tries to split the measures up. This isn't letting me track the count.
I want this on the same axis; the best I've been able to do is create two sheets, one with the first filter and one with the 2nd, and stack them in a dashboard.
My end user wants the lines in the same visual, otherwise I'd be happy with the dashboard approach. Right now, though, I'd also like to know how to do this.
It is a little hard to tell exactly what you want to achieve, but the problem with filtering is common.
The principle that is important is that Tableau will filter the whole dataset by row. So duplicating the dimension you want to filter won't help as the filter on the original dimension will also filter the corresponding rows in the second dimension. Any solution has to be clever enough to work around this issue.
One solution is to build two new dimensions that use a calculation rather than a filter to create the new result. Let's say you have a dimension, [size] that has a range of numbers from 1 to 10 and you want to compare the total number of rows including and excluding the number 5. You could create a new field using a formula like if [size] <> 5 then 1 else 0 end
Summing the new field will give a count of the number of rows that don't contain a 5 and this can be compared directly to a rowcount of the original [size] field which will give the number including the value 5.
This basic principle can be extended to much more complex logic. The essential point is to realise that filters act on every row in your data and can't, by themselves, show comparisons with alternative filter choices on a single visualisation.
Depending on the nature of your problem there may be other solutions worth looking at including sets and groups but you would need to provide more specific details for users here to tell you whether they would be useful.
We can make a a set out of the values of the dimension and then place it in the required shelf. So, you will have your dimension which will plot accordingly and set which will have data as per the requirement because with filter you can't have that independence of showing data everytime you want.
I'm trying to make a decent implementation of the hungarian algorithm however I'm stuck at how to find the minimum number of lines that cover all the zeros in an array
also I need to know these lines to make some computations later
here is the explanation:
http://www.ams.jhu.edu/~castello/362/Handouts/hungarian.pdf
in step 3 it says
Use as few lines as possible to cover all the zeros in the matrix. There is no easy rule to do this - basically trial and error.
what does trial and error mean in terms of computation? If I have for example an 2d array of 5 rows and 5 columns then
The first row can cover all the zeros, the first and second, the first row and first column, etc etc too many combinations
isn't there something more efficient than this?
thanks in advance
You need to implement bipartite matching algorithm here. You have two partitions in the graph- the vertices in one of them represent the rows and the vertices in the other one represent the columns in the table. There is an edge between rowi and columnj iff there is a 1 in cell (i,j). Then you create maximum bipartite matching. After the last iteration of the bipartite matching algorithm you need to figure out which vertices are connected via a incremental path with the source for your matching. An incremental path is path using only edges with positive capacity. You should have picture like:
row_1 col_1
/ \
/ - row_2 col_2 -\
source - .... some_edges \ sink
\ /
\ - row_n col_n /
.... /
col_m
After you find the maximum bipartite matching, find which rows and columns are reachable via positive-capacity edges from sink. Now the minimum number of rows and columns you need to scratch is found using the following algorithm - you cross out all the rows that are not reachable from the source and all the columns that are reachable and this is your optimal solution.
Hope this answer helps you.
I'm not quite sure why they told you to do trial and error. The Hungarian algorithm, however, does not take exponential time. Take a look at wikipedia, which walks you through an example of how to figure out the minimum number of lines (look at Step 3):
http://en.wikipedia.org/wiki/Hungarian_algorithm#Matrix_interpretation
The article also includes links to implementations, and some online course notes which give more complete (although also more technical) descriptions of the Hungarian algorithm than the one you're using.
Trial and error means O((n+m)!) complexity.
At most you will only need to pick min(n,m) lines, as selecting all rows or columns will cover all 0s.
I would implement a dynamic programming algorithm to solve this step for large problems.
I read that it's possible to make quicksort run at O(nlogn)
the algorithm says on each step choose the median as a pivot
but, suppose we have this array:
10 8 39 2 9 20
which value will be the median?
In math if I remember correct the median is (39+2)/2 = 41/2 = 20.5
I don't have a 20.5 in my array though
thanks in advance
You can choose either of them; if you consider the input as a limit, it does not matter as it scales up.
We're talking about the exact wording of the description of an algorithm here, and I don't have the text you're referring to. But I think in context by "median" they probably meant, not the mathematical median of the values in the list, but rather the middle point in the list, i.e. the median INDEX, which in this cade would be 3 or 4. As coffNjava says, you can take either one.
The median is actually found by sorting the array first, so in your example, the median is found by arranging the numbers as 2 8 9 10 20 39 and the median would be the mean of the two middle elements, (9+10)/2 = 9.5, which doesn't help you at all. Using the median is sort of an ideal situation, but would work if the array were at least already partially sorted, I think.
With an even numbered array, you can't find an exact pivot point, so I believe you can use either of the middle numbers. It'll throw off the efficiency a bit, but not substantially unless you always ended up sorting even arrays.
Finding the median of an unsorted set of numbers can be done in O(N) time, but it's not really necessary to find the true median for the purposes of quicksort's pivot. You just need to find a pivot that's reasonable.
As the Wikipedia entry for quicksort says:
In very early versions of quicksort, the leftmost element of the partition would often be chosen as the pivot element. Unfortunately, this causes worst-case behavior on already sorted arrays, which is a rather common use-case. The problem was easily solved by choosing either a random index for the pivot, choosing the middle index of the partition or (especially for longer partitions) choosing the median of the first, middle and last element of the partition for the pivot (as recommended by R. Sedgewick).
Finding the median of three values is much easier than finding it for the whole collection of values, and for collections that have an even number of elements, it doesn't really matter which of the two 'middle' elements you choose as the potential pivot.
I have several datasets i.e. matrices that have a 2 columns, one with a matlab date number and a second one with a double value. Here an example set of one of them
>> S20_EavesN0x2DEAir(1:20,:)
ans =
1.0e+05 *
7.345016409722222 0.000189375000000
7.345016618055555 0.000181875000000
7.345016833333333 0.000177500000000
7.345017041666667 0.000172500000000
7.345017256944445 0.000168750000000
7.345017465277778 0.000166875000000
7.345017680555555 0.000164375000000
7.345017888888889 0.000162500000000
7.345018104166667 0.000161250000000
7.345018312500001 0.000160625000000
7.345018527777778 0.000158750000000
7.345018736111110 0.000160000000000
7.345018951388888 0.000159375000000
7.345019159722222 0.000159375000000
7.345019375000000 0.000160625000000
7.345019583333333 0.000161875000000
7.345019798611111 0.000162500000000
7.345020006944444 0.000161875000000
7.345020222222222 0.000160625000000
7.345020430555556 0.000160000000000
Now that I have those different sensor values, I need to get them together into a matrix, so that I could perform clustering, neural net and so on, the only problem is, that the sensor data was taken with slightly different timings or timestamps and there is nothing I can do about that from a data collection point of view.
My first thought was interpolation to make one sensor data set fit another one, but that seems like a messy approach and I was thinking maybe I am missing something, a toolbox or function that would enable me to do this quicker without me fiddling around. To even complicate things more, the number of sensors grew over time, therefore I am looking at different start dates as well.
Someone a good idea on how to go about this? Thanks
I think your first thought about interpolation was the correct one, at least if you plan to use NNs. Another option would be to use approaches which are designed to deal with missing data, like http://en.wikipedia.org/wiki/Dempster%E2%80%93Shafer_theory for example.
It's hard to give an answer for the clustering part, because I have no idea what you're looking for in the data.
For the neural network, beside interpolating there are at least two other methods that come to mind:
training separate networks for each matrix
feeding them all together to the same network, with a flag specifying which matrix the data is coming from, i.e. something like: input (timestamp, flag_m1, flag_m2, ..., flag_mN) => target (value) where the flag_m* columns are mutually exclusive boolean values - i.e. flag_mK is 1 iff the line comes from matrix K, 0 otherwise.
These are the only things I can safely say with the amount of information you provided.