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There are two arrays, an array of images and an array of the corresponding labels. (e.g pictures of figures and it's values)
The occurrences in the labels are unevenly distributed.
What I want is to cut both arrays in such a way, that the labels are evenly distributed. E.g. every label occurs 2 times.
To test I've just created two 1D arrays and it was working:
labels = np.array([1, 2, 3, 3, 1, 2, 1, 3, 1, 3, 1,])
images = np.array(['A','B','C','C','A','B','A','C','A','C','A',])
x, y = zip(*sorted(zip(images, labels)))
label = list(set(y))
new_images = []
new_labels = []
amount = 2
for i in label:
start = y.index(i)
stop = start + amount
new_images = np.append(new_images, x[start: stop])
new_labels = np.append(new_labels, y[start: stop])
What I get/want is this:
new_labels: [ 1. 1. 2. 2. 3. 3.]
new_images: ['A' 'A' 'B' 'B' 'C' 'C']
(It is not necessary, that the arrays are sorted)
But when I tried it with the right data (images.shape = (35000, 32, 32, 3), labels.shape = (35000)) I've got an error:
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
This does not help me a lot:
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
I think that my solution is quite dirty anyhow. Is there a way to do it right?
Thank you very much in advance!
When your labels are equal, the sort function tries to sort on the second value of the tuples it has as input, since this is an array in the case of your real data, (instead of the 1D data), it cannot compare them and raises this error.
Let me explain it a bit more detailed:
x, y = zip(*sorted(zip(images, labels)))
First, you zip your images and labels. What this means, is that you create tuples with the corresponding elements of images and lables. The first element from images by the first element of labels, etc.
In case of your real data, each label is paired with an array with shape (32, 32, 3).
Second you sort all those tuples. This function tries first to sort on the first element of the tuple. However, when they are equal, it will try to sort on the second element of the tuples. Since they are arrays it cannot compare them en throws an error.
You can solve this by explicitly telling the sorted function to only sort on the first tuple element.
x, y = zip(*sorted(zip(images, labels), key=lambda x: x[0]))
If performance is required, using itemgetter will be faster.
from operator import itemgetter
x, y = zip(*sorted(zip(images, labels), key=itemgetter(0)))
I am fairly new to Go. I have coded in JavaScript where I could do this:
var x = [];
x[0] = 1;
This would work fine. But in Go, I am trying to implement the same thing with Go syntax. But that doesn't help. I need to have a array with unspecified index number.
I did this:
var x []string
x[0] = "name"
How do I accomplish that?
When you type:
var x []string
You create a slice, which is similar to an array in Javascript. But unlike Javascript, a slice has a set length and capacity. In this case, you get a nil slice which has the length and capacity of 0.
A few examples of how you can do it:
x := []string{"name"} // Creates a slice with length 1
y := make([]string, 10) // Creates a slice with length 10
y[0] = "name" // Set the first index to "name". The remaining 9 will be ""
var z []string // Create an empty nil slice
z = append(z, "name") // Appends "name" to the slice, creating a new slice if required
More indepth reading about slices:
Go slices usage and internals
In JavaScript arrays are dynamic in the sense that if you set the element of an array using an index which is greater than or equal to its length (current number of elements), the array will be automatically extended to have the required size to set the element (so the index you use will become the array's new length).
Arrays and slices in Go are not that dynamic. When setting elements of an array or slice, you use an index expression to designate the element you want to set. In Go you can only use index values that are in range, which means the index value must be 0 <= index < length.
In your code:
var x []string
x[0] = "name"
The first line declares a variable named x of type []string. This is a slice, and its value will be nil (the zero value of all slice types, because you did not provide an initialization value). It will have a length of 0, so the index value 0 is out of range as it is not less that the length.
If you know the length in advance, create your array or slice with that, e.g.:
var arr [3]string // An array with length of 3
var sli = make([]string, 3) // A slice with length of 3
After the above declarations, you can refer to (read or write) values at indicies 0, 1, and 2.
You may also use a composite literal to create and initialize the array or slice in one step, e.g.
var arr = [3]string{"one", "two", "three"} // Array
var sli = []string{"one", "two", "three"} // Slice
You can also use the builtin append() function to add a new element to the end of a slice. The append() function allocates a new, bigger array/slice under the hood if needed. You also need to assign the return value of append():
var x []string
x = append(x, "name")
If you want dynamic "arrays" similar to arrays of JavaScript, the map is a similar construct:
var x = map[int]string{}
x[0] = "name"
(But a map also needs initialization, in the above example I used a composite literal, but we could have also written var x = make(map[int]string).)
You may assign values to keys without having to declare the intent in prior. But know that maps are not slices or arrays, maps typically not hold values for contiguous ranges of index keys (but may do so), and maps do not maintain key or insertion order. See Why can't Go iterate maps in insertion order? for details.
Must read blog post about arrays and slices: Go Slices: usage and internals
Recommended questions / answers for a better understanding:
Why have arrays in Go?
How do I initialize an array without using a for loop in Go?
How do I find the size of the array in go
Keyed items in golang array initialization
Are golang slices pass by value?
Can you please use var x [length]string; (where length is size of the array you want) instead of var x []string; ?
In Go defining a variable like var x=[]int creates a slice of type integer. Slices are dynamic and when you want to add an integer to the slice, you have to append it like x = append(x, 1) (or x = append(x, 2, 3, 4) for multiple).
As srxf mentioned, have you done the Go tour? There is a page about slices.
I found out that the way to do it is through a dynamic array. Like this
type mytype struct {
a string
}
func main() {
a := []mytype{mytype{"name1"}}
a = append(a, mytype{"name 2"})
fmt.Println(a);
}
golang playground link: https://play.golang.org/p/owPHdQ6Y6e
I have a 1x10 structure array with plenty of fields and I would like to remove from the struct array the element with a specific value on one of the field variables.
I know the value im looking for and the field I should be looking for and I also know how to delete the element from the struct array once I find it. Question is how(if possible) to elegantly identify it without going through a brute force solution ie a for-loop that goes through elements of the struct array to compare with the value I m looking for.
Sample code: buyers as 1x10 struct array with fields:
id,n,Budget
and the variable to find in the id values like id_test = 12
You can use the fact that if you have an array of structs, and you use the dot referencing, this creates a comma-separated list. If you enclose this in [] it will attempt to create an array and if you enclose it in {} it will be coerced into a cell array.
a(1).value = 1;
a(2).value = 2;
a(3).value = 3;
% Into an array
[a.value]
% 1 2 3
% Into a cell array
{a.value}
% [1] [2] [3]
So to do your comparison, you can convert the field you care about into either an array of cell array to do the comparison. This comparison will then yield a logical array which you can use to index into the original structure.
For example
% Some example data
s = struct('id', {1, 2, 3}, 'n', {'a', 'b', 'c'}, 'Budget', {100, 200, 300});
% Remove all entries with id == 2
s = s([s.id] ~= 2);
% Remove entries that have an id of 2 or 3
s = s(~ismember([s.id], [2 3]));
% Find ones with an `n` of 'a' (uses a cell array since it's strings)
s = s(ismember({s.id}, 'a'));
On this page, it shows how to initialize an array, and if you scroll down a bit, under the section called "The Lists" it "explains" what lists are and how they're different from arrays.
Except it uses an example that's just exactly the same as declaring an array, and doesn't explain it whatsoever.
What is the difference?
Take a look at perldoc -q "list and an array". The biggest difference is that an array is a variable, but all of Perl's data types (scalar, array and hash) can provide a list, which is simply an ordered set of scalars.
Consider this code
use strict;
use warnings;
my $scalar = 'text';
my #array = (1, 2, 3);
my %hash = (key1 => 'val1', key2 => 'val2');
test();
test($scalar);
test(#array);
test(%hash);
sub test { printf "( %s )\n", join ', ', #_ }
which outputs this
( )
( text )
( 1, 2, 3 )
( key2, val2, key1, val1 )
A Perl subroutine takes a list as its parameters. In the first case the list is empty; in the second it has a single element ( $scalar); in the third the list is the same size as #array and contains ( $array[0], $array[1], $array[2], ...), and in the last it is twice as bug as the number of elements in %hash, and contains ( 'key1', $hash{key1}, 'key2', $hash{key2}, ...).
Clearly that list can be provided in several ways, including a mix of scalar variables, scalar constants, and the result of subroutine calls, such as
test($scalar, $array[1], $hash{key2}, 99, {aa => 1, bb => 2}, \*STDOUT, test2())
and I hope it is clear that such a list is very different from an array.
Would it help to think of arrays as list variables? There is rarely a problem distinguishing between scalar literals and scalar variables. For instance:
my $str = 'string';
my $num = 99;
it is clear that 'string' and 99 are literals while $str and $num are variables. And the distinction is the same here:
my #numbers = (1, 2, 3, 4);
my #strings = qw/ aa bb cc dd /;
where (1, 2, 3, 4) and qw/ aa bb cc dd / are list literals, while #numbers and #strings are variables.
Actually, this question is quite well answered in Perl's FAQ. Lists are (one of) methods to organize the data in the Perl source code. Arrays are one type of storing data; hashes are another.
The difference is quite obvious here:
my #arr = (4, 3, 2, 1);
my $arr_count = #arr;
my $list_count = (4, 3, 2, 1);
print $arr_count, "\n"; # 4
print $list_count; # 1
At first sight, there are two identical lists here. Note, though, that only the one that is assigned to #arr variable is correctly counted by scalar assignment. The $list_count stores 1 - the result of evaluating expression with comma operator (which basically gives us the last expression in that enumeration - 1).
Note that there's a slight (but very important) difference between list operators/functions and array ones: the former are kind-of omnivores, as they don't change the source data, the latter are not. For example, you can safely slice and join your list, like this:
print join ':', (4,2,3,1)[1,2];
... but trying to 'pop' it will give you quite a telling message:
pop (4, 3, 2, 1);
### Type of arg 1 to pop must be array (not list)...
An array is a type of variable. It contains 0 or more scalars at monotonously increasing indexes. For example, the following creates array #a:
my #a;
Being variables, you can manipulate arrays. You can add elements, change the values of elements, etc.
"List" means many things. The two primary uses for it are to refer to list values and instances of the list operator.
A list value is an ordered collection of zero or more scalars on the stack. For example, the sub in the following code returns a list to be assigned to #a (an array).
my #a = f();
List values can't be manipulated; they are absorbed in whole by any operator to which they are passed. They are just how values are passed between subs and operators.
The list operator (,) is an N-ary operator* that evaluates each of its operands in turn. In list context, the list operator returns a list consisting of an amalgamation of the lists returned by its operands. For example, the list operator in the following snippet returns a list value consisting of all the elements of arrays #a and #b:
my #c = ( #a, #b );
(By the way, parens don't create lists. They're just there to override precedence.)
You cannot manipulate a list operator since it's code.
* — The docs say it's a binary operator (at least in scalar context), but it's not true.
Simple demonstration of difference.
sub getarray{ my #x = (2,4,6); return #x; }
sub getlist { return (2,4,6); }
Now, if you do something like this:
my #a = getarray();
my #b = getlist();
Then #a and #b will both contain the same value - the list (2,4,6). However, if you do this:
my $a = getarray();
my $b = getlist();
Then $a will contain the value 3, while $b will contain the value 6.
So yes, you can say that arrays are variables containing list values, but that doesn't tell the whole story, because arrays and literal lists behave quite differently at times.
Lists are comma-separated values (csv) or expressions (cse) . Arrays (and hashes) are containers.
One can initialize an array or hash by a list:
#a = ("profession", "driver", "salary", "2000");
%h = ("profession", "driver", "salary", "2000");
One can return a list:
sub f {
return "moscow", "tel-aviv", "madrid";
}
($t1, $t2, $t3) = f();
print "$t1 $t2 $t3\n";
($t1, $t2, $t3) is a list of scalar containers $t1, $t2, $t3.
Lists are a form of writing perl expressions (part of syntax) while arrays are data structures (memory locations).
The difference between lists and arrays confuses many. Perl itself got it wrong by misnaming its builtin function wantarray(): "This function should have been named wantlist() instead." There is an answer in perlfaq4, "What is the difference between a list and an array?", but it did not end my confusion.
I now believe these to be true:
An array in scalar context becomes a count of its elements.
The comma operator in scalar context returns the last element.
You can't make a reference to a list; \(2, 4, 6) returns a list of references to the scalars in the list. You can use [2, 4, 6] to make a reference to an anonymous array.
You can index a list (to get its nth element) without making an array if you make a list slice, so (2, 4, 6)[1] is 4.
But what if I want to count the elements in a list, or get the last element of an array? Should I convert between arrays and lists somehow?
You can always convert a list to an array with [...] syntax. One way to count the elements in a list is to make an anonymous array, then immediately dereference it in scalar context, like so:
sub list { return qw(carrot stick); }
my $count = #{[list()]};
print "Count: $count\n"; # Count: 2
Another way is to use the list assignment operator, like so:
sub list { return qw(carrot stick); }
my $count = (()=list());
print "Count: $count\n"; # Count: 2
There is no array in this code, but the list assignment operator returns the number of things being assigned. I assign them to an empty list of variables. In code golf, I write ()=$str=~/reg/g to count the regular expression matches in some string.
You need not convert an array to a list, because an array in list context is already a list. If you want the last element of an array, just say $array[-1].
The comma operator would return the last element of a list, but I can't use it to get the last element of an array. If I say ((),#array) in scalar context, then #array is in scalar context and I get the count.
You need not make an array to index a list. You can make an anonymous array, as in [list()]->[1], or you can make a list slice, as in (list())[1]. I had trouble with list slices because they have different syntax. A list slice needs parentheses! In C or Python or Ruby, func()[1] would do the array index on the function's return value, but in Perl, func()[1] is a syntax error. You must say (func())[1].
For example, I want to print the 3rd highest number in array. Because I'm lazy, I sort the array and take the 3rd last element:
my #array = (112, 101, 114, 108, 32, 104, 97, 99, 107);
print +(sort { $a <=> $b } #array)[-3], "\n"; # prints 108
The unary + prevents the print() function stealing my parentheses.
You can use a list slice on an array, as in (#array)[1]. This works because an array is a list. The difference between lists and arrays is that arrays can do $array[1].
An Array Vs A List
A list is a different kind of data structure from an array.
The biggest difference is in the idea of direct access Vs sequential access. Arrays allow both; direct and sequential access, while lists allow only sequential access. And this is because the way that these data structures are stored in memory.
In addition, the structure of the list doesn’t support numeric index like an array is. And, the elements don’t need to be allocated next to each other in the memory like an array is.
Arrays
An array is an ordered collection of items, where each item inside the array has an index.
here my answer about sigils and context
but main difference is this:
arrays have a scalar-context-value like count of elements.
lists have a scalar-context-value like LAST element in list.
so, you need to know about goat-operator: =()=.
Usage?
perl -e '$s =()= qw(a b c); print $s' # uh? 3? (3 elements, array context)
perl -e '$s = qw(a b cLastElementThatYouSee); print $s' # uh? cLastElementThatYouSee? (list context, last element returned)
as you see, =()= change context to array
I am a little confused about the usage of cells and arrays in MATLAB and would like some clarification on a few points. Here are my observations:
An array can dynamically adjust its own memory to allow for a dynamic number of elements, while cells seem to not act in the same way:
a=[]; a=[a 1]; b={}; b={b 1};
Several elements can be retrieved from cells, but it doesn't seem like they can be from arrays:
a={'1' '2'}; figure; plot(...); hold on; plot(...); legend(a{1:2});
b=['1' '2']; figure; plot(...); hold on; plot(...); legend(b(1:2));
%# b(1:2) is an array, not its elements, so it is wrong with legend.
Are these correct? What are some other different usages between cells and array?
Cell arrays can be a little tricky since you can use the [], (), and {} syntaxes in various ways for creating, concatenating, and indexing them, although they each do different things. Addressing your two points:
To grow a cell array, you can use one of the following syntaxes:
b = [b {1}]; % Make a cell with 1 in it, and append it to the existing
% cell array b using []
b = {b{:} 1}; % Get the contents of the cell array as a comma-separated
% list, then regroup them into a cell array along with a
% new value 1
b{end+1} = 1; % Append a new cell to the end of b using {}
b(end+1) = {1}; % Append a new cell to the end of b using ()
When you index a cell array with (), it returns a subset of cells in a cell array. When you index a cell array with {}, it returns a comma-separated list of the cell contents. For example:
b = {1 2 3 4 5}; % A 1-by-5 cell array
c = b(2:4); % A 1-by-3 cell array, equivalent to {2 3 4}
d = [b{2:4}]; % A 1-by-3 numeric array, equivalent to [2 3 4]
For d, the {} syntax extracts the contents of cells 2, 3, and 4 as a comma-separated list, then uses [] to collect these values into a numeric array. Therefore, b{2:4} is equivalent to writing b{2}, b{3}, b{4}, or 2, 3, 4.
With respect to your call to legend, the syntax legend(a{1:2}) is equivalent to legend(a{1}, a{2}), or legend('1', '2'). Thus two arguments (two separate characters) are passed to legend. The syntax legend(b(1:2)) passes a single argument, which is a 1-by-2 string '12'.
Every cell array is an array! From this answer:
[] is an array-related operator. An array can be of any type - array of numbers, char array (string), struct array or cell array. All elements in an array must be of the same type!
Example: [1,2,3,4]
{} is a type. Imagine you want to put items of different type into an array - a number and a string. This is possible with a trick - first put each item into a container {} and then make an array with these containers - cell array.
Example: [{1},{'Hallo'}] with shorthand notation {1, 'Hallo'}