I have a file "a_test.dat" with two data blocks that I can select via the corresponding index.
# first
x1 y1
3 1
6 2
9 8
# second
x2 y2
4 5
8 2
2 7
Now I want to connect the data points of both indices with an arrow.
set arrow from (x1,y1) to (x2,y2).
I can plot both blocks with one plot statement. But I cannot get the points to set the arrows.
plot "a_test.dat" index "first" u 1:2, "" index "second" u 1:2
From version 5.2 you can use gnuplot arrays:
stats "a_test.dat" nooutput
array xx[STATS_records]
array yy[STATS_records]
# save all data into two arrays
i = 1
fnset(x,y) = (xx[i]=x, yy[i]=y, i=i+1)
# parse data ignoring output
set table $dummy
plot "" using (fnset($1,$2)) with table
unset table
# x2,y2 data starts at midpoint in array
numi = int((i-1)/2)
plot for [i=1:numi] $dummy using (xx[i]):(yy[i]):(xx[numi+i]-xx[i]):(yy[numi+i]-yy[i]) with vectors
Use stats to count the number of lines in the file, so that the array can
be large enough. Create an array xx and another yy to hold the data.
Use plot ... with table to read the file again, calling your function
fnset() for each data line with the x and y column values. The function
saves them at the current index i, which it increments. It was
initialised to 1.
For 3+3 data lines, i ends up at 7, so we set numi to (i-1)/2 i.e. 3.
Use plot for ... vectors to draw the arrows. Each arrow needs 4 data
items from the array. Note that the second x,y must be a relative delta,
not an absolute position.
I have a ~4.4M dataframe with purchase orders. I'm interested in a column that indicates the presence of certain items in that purchase order. It is structured like this:
df['item_arr'].head()
1 [a1, a2, a5]
2 [b1, b2, c3...
3 [b3]
4 [a2]
There are 4k different items, and in each row there is always at least one. I have generated another 4.4M x 4k dataframe df_te_sub with a sparse structure indicating the same array in terms of booleans, i.e.
c = df_te_sub.columns[[10, 20, 30]]
df_te_sub[c].head()
>>a10 b8 c1
0 0 0 0
1 0 0 0
2 0 0 0
3 0 0 0
4 0 0 0
The name of the columns is not important, although it is in alphabetical order, for what is worth.
Given a subset g of items, I am trying to extract the orders (rows) for two different cases:
At least one of the items is present in the row
The items is present in the row are all a subset of g
The first rule I found it the best way was to do:
c = df_te_sub[g].sparse.to_coo()
rows = pd.unique(c.row)
The second rule presented a challenge. I tried different things but they are all slow:
# using set
s = set(g)
df['item_arr'].apply(s.issuperset)
# using the "non selected items"
c = df_te_sub[df_te_sub.columns[~df_te_sub.columns.isin(g)]].sparse.to_coo()
x = np.ones(len(df_te_sub), dtype='bool')
x[c.row] = False
# mix
s = set(g)
c = df_te_sub[g].sparse.to_coo()
rows = pd.unique(c.row)
df['item_arr'].iloc[rows].apply(s.issuperset)
Any ideas to improve performance? I need to do this for several subsets.
The output can be given either in rows (e.g. [0, 2, 3]) or as a boolean mask (e.g. True False True True ....), as both will work to slice the order dataframe.
I feel like you're overthinking this. If you have a boolean array of membership you've already done 90% of the work.
from scipy.sparse import csc_matrix
# Turn your sparse data into a sparse array
arr = csc_matrix(df_te_sub.sparse.to_coo())
# Get the number of items per row
row_len = arr.sum(axis=1).A.flatten()
# Get the column indices for each item and slice your array
arr_col_idx = [df.columns.get_loc(g_val) for g_val in g]
# Sum the number of items in g in the slice per row
arr_g = arr[:, arr_col_idx].sum(axis=1).A.flatten()
# Find all the rows with at least one thing in g
arr_one_g = arr_g > 0
# Find all the things in the rows which are subsets of G
# This assumes row_len is always greater than 0, if it isnt add a test for that
arr_subset_g = (row_len - arr_g) == 0
arr_one_g and arr_subset_g are 1d boolean arrays that should index for the things you want.
I have a model that predicts 10 words for a particular course in order of likelihood, and I'd like the first 5 words of those words that appear in the course's description.
This is the format of the data:
course_name course_title course_description predicted_word_10 predicted_word_9 predicted_word_8 predicted_word_7 predicted_word_6 predicted_word_5 predicted_word_4 predicted_word_3 predicted_word_2 predicted_word_1
Xmath 32 Precalculus Polynomial and rational functions, exponential... directed scholars approach build african different visual cultures placed global
Xphilos 2 Morality Introduction to ethical and political philosop... make presentation weekly european ways general range questions liberal speakers
My idea is for each row to start iterating from predicted_word_1 until I get the first 5 that are in the description. I'd like to save those words in the order they appear into additional columns description_word_1 ... description_word_5. (If there are <5 predicted words in the description I plan to return NAN in the corresponding columns).
To clarify with an example: if the course_description of a course is 'Polynomial and rational functions, exponential and logarithmic functions, trigonometry and trigonometric functions. Complex numbers, fundamental theorem of algebra, mathematical induction, binomial theorem, series, and sequences. ' and its first few predicted words are irrelevantword1, induction, exponential, logarithmic, irrelevantword2, polynomial, algebra...
I would want to return induction, exponential, logarithmic, polynomial, algebra for that in that order and do the same for the rest of the courses.
My attempt was to define an apply function that will take in a row and iterate from the first predicted word until it finds the first 5 that are in the description, but the part I am unable to figure out is how to create these additional columns that have the correct words for each course. This code will currently only keep the words for one course for all the rows.
def find_top_description_words(row):
print(row['course_title'])
description_words_index=1
for i in range(num_words_per_course):
description = row.loc['course_description']
word_i = row.loc['predicted_word_' + str(i+1)]
if (word_i in description) & (description_words_index <=5) :
print(description_words_index)
row['description_word_' + str(description_words_index)] = word_i
description_words_index += 1
df.apply(find_top_description_words,axis=1)
The end goal of this data manipulation is to keep the top 10 predicted words from the model and the top 5 predicted words in the description so the dataframe would look like:
course_name course_title course_description top_description_word_1 ... top_description_word_5 predicted_word_1 ... predicted_word_10
Any pointers would be appreciated. Thank you!
If I understand correctly:
Create new DataFrame with just 100 predicted words:
pred_words_lists = df.apply(lambda x: list(x[3:].dropna())[::-1], axis = 1)
Please note that, there are lists in each row with predicted words. The order is nice, I mean the first, not empty, predicted word is on the first place, the second on the second place and so on.
Now let's create a new DataFrame:
pred_words_df = pd.DataFrame(pred_words_lists.tolist())
pred_words_df.columns = df.columns[:2:-1]
And The final DataFrame:
final_df = df[['course_name', 'course_title', 'course_description']].join(pred_words_df.iloc[:,0:11])
Hope this works.
EDIT
def common_elements(xx, yy):
temp = pd.Series(range(0, len(xx)), index= xx)
return list(df.reindex(yy).sort_values()[0:10].dropna().index)
pred_words_lists = df.apply(lambda x: common_elements(x[2].replace(',','').split(), list(x[3:].dropna())), axis = 1)
Does it satisfy your requirements?
Adapted solution (OP):
def get_sorted_descriptions_words(course_description, predicted_words, k):
description_words = course_description.replace(',','').split()
predicted_words_list = list(predicted_words)
predicted_words = pd.Series(range(0, len(predicted_words_list)), index=predicted_words_list)
predicted_words = predicted_words[~predicted_words.index.duplicated()]
ordered_description = predicted_words.reindex(description_words).dropna().sort_values()
ordered_description_list = pd.Series(ordered_description.index).unique()[:k]
return ordered_description_list
df.apply(lambda x: get_sorted_descriptions_words(x['course_description'], x.filter(regex=r'predicted_word_.*'), k), axis=1)
I'm trying to create 20 unique cards with numbers, but I struggle a bit.. So basically I need to create 20 unique matrices 3x3 having numbers 1-10 in first column, numbers 11-20 in the second column and 21-30 in the third column.. Any ideas? I'd prefer to have it done in r, especially as I don't know Visual Basic. In excel I know how to generate the cards, but not sure how to ensure they are unique..
It seems to be quite precise and straightforward to me. Anyway, i needed to create 20 matrices that would look like :
[,1] [,2] [,3]
[1,] 5 17 23
[2,] 8 18 22
[3,] 3 16 24
Each of the matrices should be unique and each of the columns should consist of three unique numbers ( the 1st column - numbers 1-10, the 2nd column 11-20, the 3rd column - 21-30).
Generating random numbers is easy, though how to make sure that generated cards are unique?Please have a look at the post that i voted for as an answer - as it gives you thorough explanation how to achieve it.
(N.B. : I misread "rows" instead of "columns", so the following code and explanation will deal with matrices with random numbers 1-10 on 1st row, 11-20 on 2nd row etc., instead of columns, but it's exactly the same just transposed)
This code should guarantee uniqueness and good randomness :
library(gtools)
# helper function
getKthPermWithRep <- function(k,n,r){
k <- k - 1
if(n^r< k){
stop('k is greater than possibile permutations')
}
v <- rep.int(0,r)
index <- length(v)
while ( k != 0 )
{
remainder<- k %% n
k <- k %/% n
v[index] <- remainder
index <- index - 1
}
return(v+1)
}
# get all possible permutations of 10 elements taken 3 at a time
# (singlerowperms = 720)
allperms <- permutations(10,3)
singlerowperms <- nrow(allperms)
# get 20 random and unique bingo cards
cards <- lapply(sample.int(singlerowperms^3,20),FUN=function(k){
perm2use <- getKthPermWithRep(k,singlerowperms,3)
m <- allperms[perm2use,]
m[2,] <- m[2,] + 10
m[3,] <- m[3,] + 20
return(m)
# if you want transpose the result just do:
# return(t(m))
})
Explanation
(disclaimer tl;dr)
To guarantee both randomness and uniqueness, one safe approach is generating all the possibile bingo cards and then choose randomly among them without replacements.
To generate all the possible cards, we should :
generate all the possibilities for each row of 3 elements
get the cartesian product of them
Step (1) can be easily obtained using function permutations of package gtools (see the object allPerms in the code). Note that we just need the permutations for the first row (i.e. 3 elements taken from 1-10) since the permutations of the other rows can be easily obtained from the first by adding 10 and 20 respectively.
Step (2) is also easy to get in R, but let's first consider how many possibilities will be generated. Step (1) returned 720 cases for each row, so, in the end we will have 720*720*720 = 720^3 = 373248000 possible bingo cards!
Generate all of them is not practical since the occupied memory would be huge, thus we need to find a way to get 20 random elements in this big range of possibilities without actually keeping them in memory.
The solution comes from the function getKthPermWithRep, which, given an index k, it returns the k-th permutation with repetition of r elements taken from 1:n (note that in this case permutation with repetition corresponds to the cartesian product).
e.g.
# all permutations with repetition of 2 elements in 1:3 are
permutations(n = 3, r = 2,repeats.allowed = TRUE)
# [,1] [,2]
# [1,] 1 1
# [2,] 1 2
# [3,] 1 3
# [4,] 2 1
# [5,] 2 2
# [6,] 2 3
# [7,] 3 1
# [8,] 3 2
# [9,] 3 3
# using the getKthPermWithRep you can get directly the k-th permutation you want :
getKthPermWithRep(k=4,n=3,r=2)
# [1] 2 1
getKthPermWithRep(k=8,n=3,r=2)
# [1] 3 2
Hence now we just choose 20 random indexes in the range 1:720^3 (using sample.int function), then for each of them we get the corresponding permutation of 3 numbers taken from 1:720 using function getKthPermWithRep.
Finally these triplets of numbers, can be converted to actual card rows by using them as indexes to subset allPerms and get our final matrix (after, of course, adding +10 and +20 to the 2nd and 3rd row).
Bonus
Explanation of getKthPermWithRep
If you look at the example above (permutations with repetition of 2 elements in 1:3), and subtract 1 to all number of the results you get this :
> permutations(n = 3, r = 2,repeats.allowed = T) - 1
[,1] [,2]
[1,] 0 0
[2,] 0 1
[3,] 0 2
[4,] 1 0
[5,] 1 1
[6,] 1 2
[7,] 2 0
[8,] 2 1
[9,] 2 2
If you consider each number of each row as a number digit, you can notice that those rows (00, 01, 02...) are all the numbers from 0 to 8, represented in base 3 (yes, 3 as n). So, when you ask the k-th permutation with repetition of r elements in 1:n, you are also asking to translate k-1 into base n and return the digits increased by 1.
Therefore, given the algorithm to change any number from base 10 to base n :
changeBase <- function(num,base){
v <- NULL
while ( num != 0 )
{
remainder = num %% base # assume K > 1
num = num %/% base # integer division
v <- c(remainder,v)
}
if(is.null(v)){
return(0)
}
return(v)
}
you can easily obtain getKthPermWithRep function.
One 3x3 matrix with the desired value range can be generated with the following code:
mat <- matrix(c(sample(1:10,3), sample(11:20,3), sample(21:30, 3)), nrow=3)
Furthermore, you can use a for loop to generate a list of 20 unique matrices as follows:
for (i in 1:20) {
mat[[i]] <- list(matrix(c(sample(1:10,3), sample(11:20,3), sample(21:30,3)), nrow=3))
print(mat[[i]])
}
Well OK I may fall on my face here but I propose a checksum (using Excel).
This is a unique signature for each bingo card which will remain invariate if the order of numbers within any column is changed without changing the actual numbers. The formula is
=SUM(10^MOD(A2:A4,10)+2*10^MOD(B2:B4,10)+4*10^MOD(C2:C4,10))
where the bingo numbers for the first card are in A2:C4.
The idea is to generate a 10-digit number for each column, then multiply each by a constant and add them to get the signature.
So here I have generated two random bingo cards using a standard formula from here plus two which are deliberately made to be just permutations of each other.
Then I check if any of the signatures are duplicates using the formula
=MAX(COUNTIF(D5:D20,D5:D20))
which shouldn't given an answer more than 1.
In the unlikely event that there were duplicates, then you would just press F9 and generate some new cards.
All formulae are array formulae and must be entered with CtrlShiftEnter
Here is an inelegant way to do this. Generate all possible combinations and then sample without replacement. These are permutations, combinations: order does matter in bingo
library(dplyr)
library(tidyr)
library(magrittr)
generate_samples = function(n) {
first = data_frame(first = (n-9):n)
first %>%
merge(first %>% rename(second = first)) %>%
merge(first %>% rename(third = first)) %>%
sample_n(20)
}
suffix = function(df, suffix)
df %>%
setNames(names(.) %>%
paste0(suffix))
generate_samples(10) %>% suffix(10) %>%
bind_cols(generate_samples(20) %>% suffix(20)) %>%
bind_cols(generate_samples(30) %>% suffix(30)) %>%
rowwise %>%
do(matrix = t(.) %>% matrix(3)) %>%
use_series(matrix)
I'm very confused on the apply function for pandas. I have a big dataframe where one column is a column of strings. I'm then using a function to count part-of-speech occurrences. I'm just not sure the way of setting up my apply statement or my function.
def noun_count(row):
x = tagger(df['string'][row].split())
# array flattening and filtering out all but nouns, then summing them
return num
So basically I have a function similar to the above where I use a POS tagger on a column that outputs a single number (number of nouns). I may possibly rewrite it to output multiple numbers for different parts of speech, but I can't wrap my head around apply.
I'm pretty sure I don't really have either part arranged correctly. For instance, I can run noun_count[row] and get the correct value for any index but I can't figure out how to make it work with apply how I have it set up. Basically I don't know how to pass the row value to the function within the apply statement.
df['num_nouns'] = df.apply(noun_count(??),1)
Sorry this question is all over the place. So what can I do to get a simple result like
string num_nouns
0 'cat' 1
1 'two cats' 1
EDIT:
So I've managed to get something working by using list comprehension (someone posted an answer, but they've deleted it).
df['string'].apply(lambda row: noun_count(row),1)
which required an adjustment to my function:
def tagger_nouns(x):
list_of_lists = st.tag(x.split())
flat = [y for z in list_of_lists for y in z]
Parts_of_speech = [row[1] for row in flattened]
c = Counter(Parts_of_speech)
nouns = c['NN']+c['NNS']+c['NNP']+c['NNPS']
return nouns
I'm using the Stanford tagger, but I have a big problem with computation time, and I'm using the left 3 words model. I'm noticing that it's calling the .jar file again and again (java keeps opening and closing in the task manager) and maybe that's unavoidable, but it's really taking far too long to run. Any way I can speed it up?
I don't know what 'tagger' is but here's a simple example with a word count that ought to work more or less the same way:
f = lambda x: len(x.split())
df['num_words'] = df['string'].apply(f)
string num_words
0 'cat' 1
1 'two cats' 2