How to find the max of each row in 2d arrays?

Question

For example, I have this 2d array:

[
    [
     0.0,
     0.24320757858085434,
     0.14893361727523413,
     0.29786723455046826,
     0.18838778030301612,
     0.12160378929042717
    ],
    [
     0.23717478210768014,
     0.0,
     0.16770789675478251,
     0.20539938644228997,
     0.25981195646349819,
     0.1299059782317491
    ],
    [
     0.21681956134183847,
     0.250361664212574,
     0.0,
     0.23178986094050727,
     0.16390018248131957,
     0.13712873102376066
    ],
    [
     0.2933749527592357,
     0.20744741852633861,
     0.15681550844086434,
     0.0,
     0.18554661183269694,
     0.15681550844086434
    ],
    [
     0.20305810393286577,
     0.28716752453162431,
     0.12135042758887897,
     0.20305810393286577,
     0.0,
     0.18536584001376513
    ],
    [
     0.17877693623386351,
     0.19584032147389943,
     0.13848001934394774,
     0.23407395508684939,
     0.25282876786143976,
     0.0
    ]
]

which gives sets of probabilities. How can I find the best probability of each row? And also is there any way to find for example the 2nd, 3rd best probability without changing the elements' positions?

Possible duplicate of [How to get indices of a sorted array in Python](https://stackoverflow.com/questions/6422700/how-to-get-indices-of-a-sorted-array-in-python) — wwii, Apr 15 '18 at 13:53
This one is better: [How to find k biggest numbers from a list of n numbers assuming n > k](https://stackoverflow.com/q/17906949/2823755) - the accepted answer (`heapq` solution) looks promising. — wwii, Apr 15 '18 at 14:15
Did one of the below solutions help? If so, please consider accepting (green tick on left), so other users know. — jpp, Apr 17 '18 at 09:31

jpp · Answer 1 · 2018-04-15T15:48:19.103

You can do this easily with 3rd party library numpy. First create a numpy array:

A = np.array([[0.0, 0.24320757858085434, 0.14893361727523413, 0.29786723455046826, 0.18838778030301612, 0.12160378929042717], [0.23717478210768014, 0.0, 0.16770789675478251, 0.20539938644228997, 0.25981195646349819, 0.1299059782317491], [0.21681956134183847, 0.250361664212574, 0.0, 0.23178986094050727, 0.16390018248131957, 0.13712873102376066], [0.2933749527592357, 0.20744741852633861, 0.15681550844086434, 0.0, 0.18554661183269694, 0.15681550844086434], [0.20305810393286577, 0.28716752453162431, 0.12135042758887897, 0.20305810393286577, 0.0, 0.18536584001376513], [0.17877693623386351, 0.19584032147389943, 0.13848001934394774, 0.23407395508684939, 0.25282876786143976, 0.0]])

To return the maximum of each row:

res = A.max(axis=1)

For the second largest in each row, you can use numpy.sort. This sorts along an axis (not in place) and then extracts the 2nd largest (via -2).

res = np.sort(A, axis=1)[:, -2]

These are both vectorised calculations. You can perform these calculations using lists of lists, but this is inadvisable.

Thank you! What do you suggest using instead of lists of lists? Tupples? — btloseltwin, Apr 15 '18 at 13:47
`numpy` array, I showed in the first line of code how you can convert a list of lists into a `numpy` array. In this format, you can perform array-based calculations conveniently in a vectorised way. — jpp, Apr 15 '18 at 13:48
[link](https://i.imgur.com/fAkrbgJ.jpg) can u see why for example in the first row it gives me the 3rd bigger instead of the second? — btloseltwin, Apr 15 '18 at 14:55

score 3 · Accepted Answer · answered Apr 15 '18 at 13:58

@jpp's numpy solution is probably the way to go, for the reasons they gave, but if you wanted to do it from pure python, you could do the following:

#Get the maximum value for each list

[[max(i)] for i in my_list]

# [[0.29786723455046826], [0.2598119564634982], [0.250361664212574], 
# [0.2933749527592357], [0.2871675245316243], [0.25282876786143976]]

# Get the maximum 2 values for each list:

[sorted(i)[-2:] for i in my_list]

# Get the maximum 3 values for each list:

[sorted(i)[-3:] for i in my_list]

And so on. Note that this will not reorder the original list, as the sorting is occurring in the sublists being created in the list comprehension

eneski · Answer 3 · 2018-04-15T15:30:52.910

You can first sort each row in descending order, then select first or second largest elements depending on your need.

a = [
    [
     0.0,
     0.24320757858085434,
     0.14893361727523413,
     0.29786723455046826,
     0.18838778030301612,
     0.12160378929042717
    ],
    [
     0.23717478210768014,
     0.0,
     0.16770789675478251,
     0.20539938644228997,
     0.25981195646349819,
     0.1299059782317491
    ],
    [
     0.21681956134183847,
     0.250361664212574,
     0.0,
     0.23178986094050727,
     0.16390018248131957,
     0.13712873102376066
    ]
]

for i in range(0, len(a)):
    a[i].sort(reverse=True)

print "1st Largests:"
for row in a:
    print "\t" + str(row[0])

print "2nd Largests:"
for row in a:
    print "\t" + str(row[1])

PS: if you are worrying about efficiency, then what you need to look for is partitioning. Lomuto and Hoare partition schemes are two famous ones.

How to find the max of each row in 2d arrays?

3 Answers3