﻿ Python: Find the indices of all occurrences of target in the uneven matrix - w3resource

# Python: Find the indices of all occurrences of target in the uneven matrix

## Python Programming Puzzles: Exercise-18 with Solution

An irregular/uneven matrix, or ragged matrix, is a matrix that has a different number of elements in each row. Ragged matrices are not used in linear algebra, since standard matrix transformations cannot be performed on them, but they are useful as arrays in computing.

Write a Python program to find the indices of all occurrences of target in the uneven matrix.

```Input:
[([1, 3, 2, 32, 19], [19, 2, 48, 19], [], [9, 35, 4], [3, 19]), 19]
Output:
[[0, 4], [1, 0], [1, 3], [4, 1]]

Input:
[([1, 2, 3, 2], [], [7, 9, 2, 1, 4]),2]
Output:
[[0, 1], [0, 3], [2, 2]]
```

Pictorial Presentation:

Sample Solution-1:

Python Code:

``````def test(M, T):
return [[i, j] for i, row in enumerate(M) for j, n in enumerate(row) if n == T]
M = [[1, 3, 2, 32, 19], [19, 2, 48, 19], [], [9, 35, 4], [3, 19]]
T = 19
print("Matrix:")
print(M)
print("Target value:")
print(T)
print("Indices of all occurrences of the target value in the said uneven matrix:")
print(test(M,T))

M = [[1, 2, 3, 2], [], [7, 9, 2, 1, 4]]
T = 2
print("\nMatrix:")
print(M)
print("Target value:")
print(T)
print("Indices of all occurrences of the target value in the said uneven matrix:")
print(test(M,T))
``````

Sample Output:

```Matrix:
[[1, 3, 2, 32, 19], [19, 2, 48, 19], [], [9, 35, 4], [3, 19]]
Target value:
19
Indices of all occurrences of the target value in the said uneven matrix:
[[0, 4], [1, 0], [1, 3], [4, 1]]

Matrix:
[[1, 2, 3, 2], [], [7, 9, 2, 1, 4]]
Target value:
2
Indices of all occurrences of the target value in the said uneven matrix:
[[0, 1], [0, 3], [2, 2]]
```

Flowchart:

## Visualize Python code execution:

The following tool visualize what the computer is doing step-by-step as it executes the said program:

Sample Solution-2:

Python Code:

``````#License: https://bit.ly/3oLErEI

def test(M, T):
indices = []
for i, num in enumerate(M[0]):
if num == T:
indices.append([0, i])
# Search for element starting from the first column and storing the indices of the found elements in a list
for row, row_num in zip(M[1:], range(1, len(M))):
for i, num in enumerate(row):
if num == T:
indices.append([row_num, i])
return indices

M = [[1, 3, 2, 32, 19], [19, 2, 48, 19], [], [9, 35, 4], [3, 19]]
T = 19
print("Matrix:")
print(M)
print("Target value:")
print(T)
print("Indices of all occurrences of the target value in the said uneven matrix:")
print(test(M,T))

M = [[1, 2, 3, 2], [], [7, 9, 2, 1, 4]]
T = 2
print("\nMatrix:")
print(M)
print("Target value:")
print(T)
print("Indices of all occurrences of the target value in the said uneven matrix:")
print(test(M,T))
``````

Sample Output:

```Matrix:
[[1, 3, 2, 32, 19], [19, 2, 48, 19], [], [9, 35, 4], [3, 19]]
Target value:
19
Indices of all occurrences of the target value in the said uneven matrix:
[[0, 4], [1, 0], [1, 3], [4, 1]]

Matrix:
[[1, 2, 3, 2], [], [7, 9, 2, 1, 4]]
Target value:
2
Indices of all occurrences of the target value in the said uneven matrix:
[[0, 1], [0, 3], [2, 2]]
```

Flowchart:

## Visualize Python code execution:

The following tool visualize what the computer is doing step-by-step as it executes the said program:

Python Code Editor :

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## Python: Tips of the Day

Find the index of an item in a list?

Given a list ["foo", "bar", "baz"] and an item in the list "bar", how do I get its index (1) in Python?

```>>> ["foo", "bar", "baz"].index("bar")
1
```

Caveats follow

Note that while this is perhaps the cleanest way to answer the question as asked, index is a rather weak component of the list API, and I can't remember the last time I used it in anger. It's been pointed out to me in the comments that because this answer is heavily referenced, it should be made more complete. Some caveats about list.index follow. It is probably worth initially taking a look at the documentation for it:

```list.index(x[, start[, end]])
```

Linear time-complexity in list length

An index call checks every element of the list in order, until it finds a match. If your list is long, and you don't know roughly where in the list it occurs, this search could become a bottleneck. In that case, you should consider a different data structure. Note that if you know roughly where to find the match, you can give index a hint. For instance, in this snippet, l.index(999_999, 999_990, 1_000_000) is roughly five orders of magnitude faster than straight l.index(999_999), because the former only has to search 10 entries, while the latter searches a million:

```>>> import timeit
>>> timeit.timeit('l.index(999_999)', setup='l = list(range(0, 1_000_000))', number=1000)
9.356267921015387
>>> timeit.timeit('l.index(999_999, 999_990, 1_000_000)', setup='l = list(range(0, 1_000_000))', number=1000)
0.0004404920036904514
```

Only returns the index of the first match to its argument

A call to index searches through the list in order until it finds a match, and stops there. If you expect to need indices of more matches, you should use a list comprehension, or generator expression.

```>>> [1, 1].index(1)
0
>>> [i for i, e in enumerate([1, 2, 1]) if e == 1]
[0, 2]
>>> g = (i for i, e in enumerate([1, 2, 1]) if e == 1)
>>> next(g)
0
>>> next(g)
2
```

Most places where I once would have used index, I now use a list comprehension or generator expression because they're more generalizable. So if you're considering reaching for index, take a look at these excellent Python features.

Throws if element not present in list

A call to index results in a ValueError if the item's not present.

```>>> [1, 1].index(2)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: 2 is not in list
```

If the item might not be present in the list, you should either

• Check for it first with item in my_list (clean, readable approach), or
• Wrap the index call in a try/except block which catches ValueError (probably faster, at least when the list to search is long, and the item is usually present.)

Ref: https://bit.ly/2ALwXwe