Python Forum

Full Version: Checking the elements of a matrix with an elements of a list
You're currently viewing a stripped down version of our content. View the full version with proper formatting.
Pages: 1 2
Hello folks,
I need a small help , it is very easy question but since I am beginner I stucked a bit.
I have a matrix like that :
C=[[1 2 3 4 ]
[7 8 9 10]
[12 13 14 15]
[5 7 7 6]]
I need to check all the elements of third line of matrix [12 13 14 15]:

If first element is > b:
do this
with the same fashion from first to fourth which is 15 , I have to check. I could not select specifically third line of matrix to check the elements ? How can I do that ?
Thank you
Try this, should help you figure it out:

C=[[1, 2, 3, 4 ],
[7, 8, 9, 10],
[12, 13, 14, 15],
[5, 7, 7, 6]]

for row in C:
    print(row)
    for element in row:
        print(element)
        
print(C[3])
print(C[3][2])
Hello thank you ,
but how can I read the third line of matrix one by one 12,13,14,15?
what I am trying to do:
if 12>a:
do this
if 13>b:
do this
etc
That is shown in the code I shared. You need to try it and experiment.
Hello,
I have an easy question. I have an matrix like
C=[[1 2 3]
[4 5 6]
[7 8 9]
and I have a list L=[5 10 5]
I need to check if the elements of first line is less than 5 (L[0]), if the elements of second line is less than 10(L[1]) and if the elements of last line is less than 10(L[2])
And I need the combine this check with other condition :
Let's say
if a>b and "the controls above"

How can I write it ?
Thank you so much
how to you think it should be written? Give it a go.
I had thought :

if (a<=b and all(C[i][n]<=L[i]) for i in range(T)for n in range(N)):
Do this
but it did not work
>>> c=[[1, 2, 3], [4, 5, 6], [7, 8, 9]]
>>> sum(c[0]) < 5 and (sum(c[1]) < 10 and sum(c[2]) < 10)
False
>>> 
But we do not know the content of list L=[5 10 5] I just gave an example like that: I should use the index of L to for the right hand side of inequlaity. and the other question how will I build the if statement with the first condition ?

Than you
Numpy arrays are intended to be used as objects - looping is wasteful and inefficient. This is how you can compare by rows and by columns
Output:
In [89]: table = numpy.array(range(1, 10)).reshape(3, 3) In [90]: table Out[90]: array([[1, 2, 3], [4, 5, 6], [7, 8, 9]]) In [91]: compare_to = numpy.array([2, 5, 6]) In [92]: compare_to Out[92]: array([2, 5, 6]) In [93]: # Compare by columns In [94]: table < compare_to Out[94]: array([[ True, True, True], [False, False, False], [False, False, False]]) In [95]: # Compare by rows In [96]: table < compare_to.reshape(3, 1) Out[96]: array([[ True, False, False], [ True, False, False], [False, False, False]])
numpy is a complex package - going into it without learning Python is a bad idea, and using it without learning is an exercise in futility
Pages: 1 2