I have a json data like :
[
[
"2020-05-07T16:30:00.000+0530",
1,
29,
693,
0,
7,
3663,
7413
],
[
"2020-05-07T15:30:00.000+0530",
0,
16,
996,
3,
13,
4452,
10106
]
]
Using JQ, I want to add the corresponding elements of the both array and result a new array. In case of date string the value from one of the array will be fine. The Expected output is
[
"2020-05-07T16:30:00.000+0530",
1,
45,
1689,
3,
20,
8115,
17519
]
Please can u suggest the solution?
Pair corresponding elements using transpose, and create a new array with sums of them.
transpose | [.[0][0]] + map(add)[1:]
demo at jqplay.org
Related
I'm trying to flatten an image that's been converted to a 3D numpy array into three separate 1D arrays, representing RGB channels.
The image array is shaped (HEIGHT, WIDTH, RGB), and I've tried in vain to use both index slicing and unzipping to just return the 3rd dimension values.
Ideally, three separate arrays represent each RGB channel,
Example:
print(image)
[
[ [56, 6, 3], [23, 32, 53], [27, 33, 56] ],
[ [57, 2, 3], [23, 246, 49], [29, 253, 58] ]
]
red_channel, green_channel, blue_channel = get_third(image)
print(red_channel)
[56, 23, 27, 57, 23, 29]
I've thought of just using a nested for loop to iterate over the first two dimensions and then add each RGB array to a list or what not, but its my understanding that this would be both inefficient and a bit of an eyesore.
Thanks in advance!
EDIT
Clarification: By unzipping I mean using the star operator (*) within the zip function, like so:
zip(*image)
Also to clarify, I don't intend to retain the width and height, I just want to essentially only flatten and return the 3D dimension of the array.
red_channel, green_channel, blue_channel = np.transpose(np.reshape(image, (-1, 3)))
I have three equally dimensioned numpy arrays.
I would like to store the data from all three in an array of the same dimensions and size.
To do this, I would like to store three bytes of information per item in the array. I assume this would be a list.
e.g.
>>>red = np.array([[150,25],[37,214]])
>>>green = np.array([[190,27],[123,231]])
>>>blue = np.array([[10,112],[123,119]])
insert combination magic to make a combined array called RGB
>>>RGB
array([(150,190,10),(25,27,112)],[(37,123,123),(214,231,119)])
For a start, each is 2x2. Combined in a list with array, same construction as in making red, produces a 3x2x2.
In [344]: red = np.array([[150,25],[37,214]])
In [345]: green = np.array([[190,27],[123,231]])
In [346]: blue = np.array([[10,112],[123,119]])
In [347]: np.array([red,green,blue])
Out[347]:
array([[[150, 25],
[ 37, 214]],
[[190, 27],
[123, 231]],
[[ 10, 112],
[123, 119]]])
In [348]: _.shape
Out[348]: (3, 2, 2)
That's not the order you want, but we can easily reshape, and if needed transpose.
The target, with an added set of []
In [350]: np.array([[(150,190,10),(25,27,112)],[(37,123,123),(214,231,119)]])
Out[350]:
array([[[150, 190, 10],
[ 25, 27, 112]],
[[ 37, 123, 123],
[214, 231, 119]]])
In [351]: _.shape
Out[351]: (2, 2, 3)
so try moving the 3 shape to the end with transpose:
In [352]: np.array([red,green,blue]).transpose(1,2,0)
Out[352]:
array([[[150, 190, 10],
[ 25, 27, 112]],
[[ 37, 123, 123],
[214, 231, 119]]])
===========================
I should have suggested stack. This a newish version of concatenate that lets us join arrays on different new dimensions. With axis=0 it behaves like np.array. But to join on the last, to put the rgb dimension last use:
In [467]: np.stack((red,green,blue),axis=-1)
Out[467]:
array([[[150, 190, 10],
[ 25, 27, 112]],
[[ 37, 123, 123],
[214, 231, 119]]])
In [468]: _.shape
Out[468]: (2, 2, 3)
Note that this expression does not assume anything about the shape of red, etc, except that they are equal. So it will work with 3d arrays as well.
This question already has answers here:
How to condense summable metrics to a unique identifier in a ruby table
(3 answers)
Closed 6 years ago.
I have the following defined table. The index for each element in each row corresponds to the same field.
[[123.0, 23,"id1",34, "abc"],
[234.1,43, "id2", 24,"jsk"],
[423.5,53, "id1",1,"xyz"],
[1.4, 5, "id2",0,"klm"]]
In the above example I need to group and sum an output that sums each of the summable elements on the index for the unique identifier in the 3rd column. The result should look like this:
[[546.5,76, "id1",35],
[235.5,48, "id2",24]]
What's the best way to do this?
This is essentially the same as the solution to your previous question.
data = [ [ 123.0, 23, "id1", 34, "abc" ],
[ 234.1, 43, "id2", 24, "jsk" ],
[ 423.5, 53, "id1", 1, "xyz" ],
[ 1.4, 5, "id2", 0, "klm" ] ]
sums = Hash.new {|h,k| h[k] = [0, 0, 0] }
data.each_with_object(sums) do |(val0, val1, id, val2, _), sums|
sums[id][0] += val0
sums[id][1] += val1
sums[id][2] += val2
end
# => { "id1" => [ 546.5, 76, 35 ],
# "id2" => [ 235.5, 48, 24 ] }
The main difference is that instead of giving the Hash a default value of 0, we're giving it a default proc that initializes missing keys with [0, 0, 0]. (We can't just do Hash.new([0, 0, 0]) because then every value would be a reference to a single Array instance, rather than each value having its own Array.) Then, inside the block, we add each value (val0 et al) to the corresponding elements of sums[id].
If you wanted an Array of Arrays instead of a Hash with the id at index 2, then at the end, you would have to add something like this:
.map {|id, vals| vals.insert(2, id) }
However, a Hash with the ids as keys makes more sense as a data structure.
I have an array with ascii characters values like this
arryAsc[] = [97, 100, 97, 115, 100]
Now I want to make an array arrSplit[] and want to save these values in array like this.
arrSplit[] = [9, 7, 1, 0, 0, 9, 7, 1, 1, 5, 1, 0, 0]
If anyone can help thanks in advance.
You can simply do like this :
var a:Array = [97, 100, 97, 115, 100];
var b:Array = a.join('').split('');
trace(b); // gives : 9,7,1,0,0,9,7,1,1,5,1,0,0
Hope that can help.
Loop through the array, casting each element as a String, then use split() with an empty string as delimiter to break apart each string into an array of characters. You can use concat to combine all the smaller arrays back into your larger arrSplit array.
I am new in python. I have asked another question How to arrange three lists in such a way that the sum of corresponding elements if greater then appear first? Now the problem is following:
I am working with a large text file, in which there are 419040 rows and 6 columns containing floats. Among them I am taking first 3 columns to generate those three lists. So the lists I am actually working with has 419040 entries in each. While I was running the python code to extract the three columns into three lists the python shell was not responding, I suspected the large number of entries for this, I used this code:
file=open("file_location","r")
a=[]
b=[]
c=[]
for lines in file:
x=lines.split(" ")
a.append(float(x[0]))
b.append(float(x[1]))
c.append(float(x[2]))
Note: for small file this code was running perfectly.
To avoid this problem I am using the following code:
import numpy as np
a = []
b = []
c = []
a,b,c = np.genfromtxt('file_location',usecols = [0,1,2], unpack=True)
So when I am running the code given in answers to my previous question the same problem is happening. So what will be the corresponding code using numpy? Or, any other solutions?
If you're going to use numpy, then I suggest using ndarrays, rather than lists. You can use loadtxt since you don't have to handle missing data. I assume it'll be faster.
a = np.loadtxt('file.txt', usecols=(0, 1, 2))
a is now a two-dimensional array, stored as an np.ndarray datatype. It should look like:
>>> a
array([[ 1, 20, 400],
[ 5, 30, 500],
[ 3, 50, 100],
[ 2, 40, 300],
[ 4, 10, 200]])
However, you now need to re-do what you did in the previous question, but using numpy arrays rather than lists. This can be easily achieved like so:
>>> b = a.sum(axis=1)
>>> b
Out[21]: array([535, 421, 342, 214, 153])
>>> i = np.argsort(b)[::-1]
>>> i
Out[26]: array([0, 1, 2, 3, 4])
>>> a[i, :]
Out[27]:
array([[ 5, 30, 500],
[ 1, 20, 400],
[ 2, 40, 300],
[ 4, 10, 200],
[ 3, 50, 100]])
The steps involved are described in a little greater detail here.