I have the following output sample:
[[-5.53759409e-01 -2.68382610e-01 4.06747784e+00]
[-1.66055379e+00 -8.08889466e-01 7.06720368e+01]
[ 2.92172488e-01 8.17347290e-01 3.18001189e+00]
[ 1.89072607e+00 -6.68502526e-01 9.08233869e+01]
[-1.31451627e+00 1.61831269e+00 5.41709058e+00]
[ 1.15886824e+00 3.31177259e-01 5.14391851e+00]
[ 1.87270676e+00 1.24100260e+00 2.64360316e+01]
[ 1.93323801e+00 -5.64255644e-02 7.28368451e+01]
[ 1.33014215e+00 1.96282476e+00 2.96295301e-01]]
The minimum function value at generation 10 is [0.2962953]
I have concatenated two arrays - the coordinate array (elements 0 and 1) and the function values (element 2) to form the above array.
However, I would like to not only display the minimum function value e.g 0.2962953 but also the coordinates associated with it, hence the row of the above array.
Any ideas how I would approach this?
In this case, I would need the bottom row of the above array and a way to highlight the coordinates and function value.
Problem fixed! Just used: printValues = array[np.argmin(array[:, 2]), (0,1)]
Related
why a[:,[x]] could create a column vector from an array? The [ ] represents what?
Could anyone explain to me the principle?
a = np.random.randn(5,6)
a = a.astype(np.float32)
print(a)
c = torch.from_numpy(a[:,[1]])
[[-1.6919796 0.3160475 0.7606999 0.16881375 1.325092 0.71536326]
[ 1.217861 0.35804042 0.0285245 0.7097111 -2.1760604 0.992101 ]
[-1.6351479 0.6607222 0.9375339 0.5308735 -1.9699149 -2.002803 ]
[-1.1895325 1.1744579 -0.5980689 -0.8906375 -0.00494479 0.51751447]
[-1.7642071 0.4681248 1.3938268 -0.7519176 0.5987852 -0.5138923 ]]
###########################################
tensor([[0.3160],
[0.3580],
[0.6607],
[1.1745],
[0.4681]])
The [ ] mean you are giving extra dimension. Try numpy shape method to see the diference.
a[:,1].shape
output :
(10,)
with [ ]
a[:,[1]].shape
output :
(10,1)
That syntax is for array slicing in numpy, where arrays are indexed as a[rows, columns, page, ... (higher-dimensions)]
Selecting for a specific row/column/page is done by giving a specific number or range of numbers. So when you use a[1,2], numpy gets the element from row 1, column 2.
You can select for several specific indices by giving the dimension multiple values. So a[[1,3],1] gets you both elements (1,1) and (1,3).
The : tells numpy to get everything from that specific array dimension. So when you use a[:,1], numpy gets every row in column 1. Alternatively, a[1,:] gets every column in row 1.
I want to combine two arrays which represent a curve where the variable is column 1, however the column 0 values do not always match:
import numpy as np
arr1= np.array([(12,1003),(17,900),(20,810)])
arr2= np.array([(10,1020),(17,902),(19,870),(21,750)])
I want to combine these into one array where the column 0 is combined and both column 1s are stacked with gaps where there is no value for the corresponding column 0 value, something like this:
arr3=np.array([((10,None,1020),(12,1003,None),(17,900,902),(19,None,870),(20,810,None),(21,None,750))])
The reason for this is that I want to be able to get mean values of the second column for each array but they are not at exactly the same column 0 value so the idea of creating this array is to then interpolate to replace all the None values, then create mean values from column 1 and 2 and have an extra column to represent that.
I have used numPy for everything else so far but obviously have got stuck with the np.column_stack function as it needs lists of the same length and also will be blind to stacking based on values from column o. Lastly I do not want to create a fit for the data as the actual data is non-linear and possibily not consistent so a fit will not work and interpolation seems like the most accurate method.
There may be an answer already but due to me not knowing how to describe it well I can't find it. Also I am relatively new to python so please don't make any assumptions about my knowledge other than it is very little.
Thank you.
will this help ??
import pandas
import numpy as np
arr1= np.array([(12,1003),(17,900),(20,810)])
arr2= np.array([(10,1020),(17,902),(19,870),(21,750)])
d1 = pandas.DataFrame(arr1)
d2 = pandas.DataFrame(arr2)
d1.columns = d2.columns = ['t','v']
d3 = pandas.DataFrame(np.array(d1.merge(d2, on='t',how='outer')))
print d3.values
# use d3.as_matrix() to convert to numpy array
output
[[ 12. 1003. nan]
[ 17. 900. 902.]
[ 20. 810. nan]
[ 10. nan 1020.]
[ 19. nan 870.]
[ 21. nan 750.]]
I have read data from a file and stored into a matrix (frag_coords):
frag_coords =
[[ 916.0907976 -91.01391344 120.83596334]
[ 916.01117655 -88.73389753 146.912555 ]
[ 924.22832597 -90.51682575 120.81734705]
...
[ 972.55384732 708.71316138 52.24644577]
[ 972.49089559 710.51583744 72.86369124]]
type(frag_coords) =
class 'numpy.matrixlib.defmatrix.matrix'
I do not have any issues when reordering the matrix by a specified column. For example, the code below works just fine:
order = np.argsort(frag_coords[:,2], axis=0)
My issue is that:
len(frag_coords[0]) = 1
I need to access the individual numbers of the first row individually, I've tried splitting it, transforming it into a list and everything seems to return the 3 numbers not as columns but rather as a single element with len=1. I need help please!
Your problem is that you're using a matrix instead of an ndarray. Are you sure you want that?
For a matrix, indexing the first row alone leads to another matrix, a row matrix. Check frag_coords[0].shape: it will be (1,3). For an ndarray, it would be (3,).
If you only need to index the first row, use two indices:
frag_coords[0,j]
Or if you store the row temporarily, just index into it as a row matrix:
tmpvar = frag_coords[0] # shape (1,3)
print(tmpvar[0,2]) # for column 2 of row 0
If you don't need too many matrix operations, I'd advise that you use np.arrays instead. You can always read your data into an array directly, but at a given point you can just transform an existing matrix with np.array(frag_coords) too if you wish.
I'm new to python so I need some help with this:
I have 2D array of numbers that represent density of a circle of material in a space and I want to locate the centre.
So I want to get the index of the numbers representing the diameter and then the middle index will be the centre. In this code I only store the value of the density : tempdiameter.append(cell)
I want the index of the cell itself. How can I do that.
Also I don't want to use lists for diameter. so how can I create a dynamic 1D np array?
Thanks
for row in x:
for cell in row:
if cell!=0:
tempdensity+=cell
tempdiameter.append(cell)
if tempdensity>maxdensity:
maxdensity=tempdensity
if len(tempdiameter)>=len(diameter):
diameter=tempdiameter
tempdensity=0
tempdiameter=[]
To get the row with the highest number of non-zero cells and the highest sum you can do
densities = x.sum(axis=1)
lengths = (x > 0).sum(axis=1)
center = x[(densities == densities.max()) & (lengths == lengths.max()]
Try to avoid using loops in numpy. Let me know if this isn't what you want and I'll try to answer it better. You should provide sample input/output when asking a question. You can also edit your question rather than adding comments.
Here's my problem: I'm dealing with output from different receivers, and they are listed by number in column 0 of my array. I'm trying to find the indices that correspond to certain receiver values that show up. For my code below, I was trying to find all indices that had a value of 6.
My problem is that for the output (print) I'm only getting [ ], as if there are no indices that correspond to values for receiver 6. I've seen the data file and know this to be incorrect. The data text file is a 4x22000ish array. Any help would be greatly appreciated. thanks.
from numpy import *
data = loadtxt("/home/***")
s,t,q = data[:,0], data[:,2], data[:,3]
t,q = loadtxt("/home/***", usecols = (2,3), unpack=True)
indices = []
for index, value in enumerate(data[:,0]):
if value == '6':
indices.append(index)
print indices
numpy.nonzero(data[:,0]==6)[0]
data[:,0]==6 returns an array of booleans, 1 when the condition is true, 0 when it is false
numpy.nonzero returns the index of nonzero elements inside of a container
you may also be interested to know that you can do things like
data[data[:,0]==6,2]
to grab all the elements from the 2nd column when the first column is zero