I have an array. I sorted it, so I have sorted array and indeces of sorted elements in the initial array.
Fo example, from [4 5 4 4 4 4 5 4] I got [4 4 4 4 4 4 5 5] and [1 3 4 5 6 8 2 7].
How to place recieved indeces in a cell array, so that in one cell will be indeces of equal elements? For my example, it will be: {1 3 4 5 6 8}, {2 7}.
I'm searching for non-loop way to solve it.
Use accumarray:
x = [4 5 4 4 4 4 5 4]; %// data
[~, ~, jj] = unique(x);
result = accumarray(jj(:), 1:numel(x), [], #(v) {v(:).'});
Or, if you need each set of indices sorted:
result = accumarray(jj(:), 1:numel(x), [], #(v) {sort(v(:)).'});
Related
Consider an array as the following:
[1,4,5,9]
The goal is to duplicate each value and append it next to it's copy in Julia, hence:
[1,1,4,4,5,5,9,9]
Is there a neat "julian" way of doing this avoiding a for- loop?
You can use the repeat function:
julia> v = [1, 4, 5, 9]
4-element Vector{Int64}:
1
4
5
9
julia> repeat(v; inner=2)
8-element Vector{Int64}:
1
1
4
4
5
5
9
9
Lets say I have the following array:
board = np.random.randint(1, 9, size=(2, 5))
How do I remove duplicates from each element in the array
e.g.
[[6 1 2 8 4]
[8 3 2 3 6]]
So here there are two 3s, and I want one of those to be deleted, how can I perform such an action?
Given your example, it seems that you don't want repetition relatively to rows. You may be interested in numpy.random.choice and try something like this:
import numpy as np
nb_lines = 2
nb_columns = 5
min_value = 1
max_value = 9
range_value = max_value-min_value
# The number of columns should be <= than the integer range to have a solution
assert(range_value+1 >= nb_columns)
board = min_value + np.array([
np.random.choice(np.arange(range_value+1), nb_columns, replace=False)
for l in range(nb_lines)
])
print(board)
Output:
% python3 script.py
[[7 4 6 3 1]
[2 8 6 4 3]]
This question already has answers here:
Convert pandas dataframe to NumPy array
(15 answers)
Closed 4 years ago.
I am using a pandas data frame to clean and process data. However, I need to then convert it into a numpy ndarray in order to use exploit matrix multiplication. I turn the data frame into a list of lists with the following:
x = df.tolist()
This returns the following structure:
[[1, 2], [3, 4], [5, 6], [7, 8] ...]
I then convert it into a numpy array like this:
x = np.array(x)
However, the following print:
print(type(x))
print(type(x[0]))
gives this result:
'numpy.ndarray'
'numpy.float64'
However, I need them both to be numpy arrays. If it's not from a pandas data frame and I just convert a hard-coded list of lists then they are both ndarrays. How do I get the list, and the lists in that list to be ndarrays when that list has been made from a data frame? Many thanks for reading, this has had me stumped for hours.
I think you need values:
df = pd.DataFrame({'C':[7,8,9,4,2,3],
'D':[1,3,5,7,1,0]})
print (df)
C D
0 7 1
1 8 3
2 9 5
3 4 7
4 2 1
5 3 0
x = df.values
print (x)
[[7 1]
[8 3]
[9 5]
[4 7]
[2 1]
[3 0]]
And then select by indexing:
print (x[:,0])
[7 8 9 4 2 3]
print (x[:,1])
[1 3 5 7 1 0]
print (type(x[:,0]))
<class 'numpy.ndarray'>
Also is possible transpose array:
x = df.values.T
print (x)
[[7 8 9 4 2 3]
[1 3 5 7 1 0]]
print (x[0])
[7 8 9 4 2 3]
print (x[1])
[1 3 5 7 1 0]
How about as_matrix:
x = df.as_matrix()
You may want to try df.get_values(), and eventually np.reshape it.
I have a cell array that looks like this:
Column1 Column2
[1 2 3 4] [2 5 6 9]
[1 3 4] [3 4 7 8]
[2 3 4] [1 3 7 9]
[1 2 4] [1 4 6 8]
There are a few more columns that have similar styles of data. I need to create a way to make a graph of each column (separate graphs for each column of the array), that plots each point as a number from each double as the x-coordinate, and the row as the y-coordinate. It should look something like this:
(Row)
1 x x x x
2 x x x
3 x x x
4 x x x
1 2 3 4
X is just a point on the graph.
Does this make enough sense? I feel like I'm making 0 progress in explaining what I want. If anyone doesn't understand this, feel free to ask questions and I'll answer them as best I can.
Something like this?
cin = { {[1 2 3 4] , [1 3 4], [2 3 4], [1 2 4]}, {[1 2 3 8] , [1 3 4], [2 3 4], [1 2 4]} };
for k=1:numel(cin)
col_k = cin{k};
figure(); %// 1 figure per column
for y=1:numel(col_k)
plot(col_k{y}, y);
hold on;
end
end
I have n different length cell vectors, call it c{i}, i=1,2,...,n.
I wanna find those c{i} which equal with others, for example:
c{1}=[1 2 3 4 5 6]; c{2}=[1 3 5 7]; c{3}=[2 4 6 8];
c{4}=[1 4 6]; c{5}=[3 7]; c{6}=[2 4 6 8]; c{7}=[1 3 5 7];
I hope I can find [2 4 6 8] and [1 3 5 7] with a simple way instead of using two loops.
Thanks!
You can do it with unique. You need to convert vectors to strings, because unique works with cell arrays of strings but not with cell arrays of numeric vectors. After unique, you can count how many strings (vector) are repeated with histc, and them some indexing lets you retrieve the corresponding vectors:
strcell = cellfun(#(e) num2str(e), c, 'uniformoutput', 0); %// convert to strings
[~, ii, jj] = unique(strcell); %// apply unique. Second and third outputs needed
ind = find(histc(jj,min(jj)-.5:max(jj)+.2)>1); %// which appear more than once
result = c(ii(ind)); %// indexing to obtain corresponding original vectors