Find Indices of Elements That Match a Given Condition

const arr = ["00", "000", "", "0", "000"]
const maxLength = arr  .map(e => e.length)  .reduce((a, b) => a > b ? a : b)  const result = arr  .map((val, i) => val.length === maxLength ? i : false) // log the index of the item if the item's length is 1  .filter(i => i !== false) // remove non-truthy elements  console.dir(result)

find first element and index matching condition in list

You could do it in a list comprehension. This is basically the same as your code but condensed into one line, and it builds a list of results that match the criteria.

The first way gets all the matches

mylist = [-1,-2,3,4,5,6]

results = [(i, el) for i, el in enumerate(mylist) if el > 0]

Another way would be to use a generator expression which is probably faster, and just unpack it. This gets the first one.

*next((i, el) for i, el in enumerate(mylist) if el > 0))

This loops the list and checks the condition, then puts the index and element into a tuple. Doing this inside parentheses turns it into a generator, which is much faster because it hasn't actually got to hold everything in memory, it just generates the responses as you need them. Using next() you can iterate through them. As we only use next() once here it just generates the first match. Then we unpack it with *

As there are two other valid answers here I decided to use timeit module to time each of them and post the results. For clarity I also timed the OP's method. Here is what I found:

import timeit
# Method 1 Generator Expression
print(timeit.timeit('next((i, el) for i, el in enumerate([-1,-2,3,4,5,6]) if el > 0)', number=100000))
0.007089499999999999

# Method 2 Getting index of True
print(timeit.timeit('list(x > 0 for x in [-1,-2,3,4,5,6]).index(True)', number=100000))
0.008104599999999997

# Method 3 filter and lambda
print(timeit.timeit('myidx , myel = list(filter(lambda el: el[1] > 0, enumerate([-1,-2,3,4,5,6])))[0]', number=100000))
0.0155314

statement = """
for idx, el in enumerate([-1,-2,3,4,5,6]):
    if el > 0:
        myidx, myel = idx, el
        break
"""

print(timeit.timeit(statement, number=100000))
0.04074070000000002

Get all the indexes from the array on matching the condition

You don't need to convert the array to List to do this operation, all you can do is:

int[] indexes = IntStream.range(0, updateResultArray.length)
        .filter(i -> updateResultArray[i] > 0)
        .toArray();

get indexes with condition

You can use enumerate in a conditional list comprehension to get the index locations.

my_list = [5, 8, 2, 7, 8, 8, 2, 4]
search_value = 8
>>> [i for i, n in enumerate(my_list) if n == search_value]
[1, 4, 5]

If the search value is not in the list, then an empty list will be returned (not exactly None, but still a falsey).

Using pandas, you can use boolean indexing to get the matches, then extract the index to a list:

df[df[0] == search_value].index.tolist()

Using an empty list will satisfy the condition for None (they both evaluate to False). If you really need None, then use the suggestion of @cᴏʟᴅsᴘᴇᴇᴅ.

How to find the indices of all elements matching from one list to another in Dart/Flutter

Try this:

main() {
  var firstList = ["Travel", "Shopping", "Transport"];
  var secondList = [
    "Travel",
    "Shopping",
    "Shopping",
    "Travel",
    "Transport",
    "Transport",
    "Travel",
    "Travel"
  ];
  var thirdList = [];
  for (var i = 0; i < secondList.length; i++) {
    if (secondList[i] == "Travel") {
      thirdList.add(i);
    }
  }
  print(thirdList);
}

EDIT: For it to work for every item in firstList

Try this:

main() {
  var firstList = ["Travel", "Shopping", "Transport"];
  var secondList = [
    "Travel",
    "Shopping",
    "Shopping",
    "Travel",
    "Transport",
    "Transport",
    "Travel",
    "Travel"
  ];
  var thirdList = [];
  for (var i = 0; i < firstList.length; i++) {
    var sublist = [];
    for (var j = 0; j < secondList.length; j++) {
      if (secondList[j] == firstList[i]) {
        sublist.add(j);
      }
    }
    thirdList.add(sublist);
  }
  print(thirdList); // [[0, 3, 6, 7], [1, 2], [4, 5]]
}

How do I find an index of an element in an array matching a condition and starting the search from a particular point in the array?

Using select would return all values where the block returns true for example:

p arr = ["abc", "d", "efg", "h", "abcde", "k"]
# => ["abc", "d", "efg", "h", "abcde", "k"]
p arr.each_index.select{|i| i >= 2 and arr[i].length == 1}
# => [3, 5]

Instead use detect if you want to return only the first value where the block returns true:

p arr = ["abc", "d", "efg", "h", "abcde", "k"]
# => ["abc", "d", "efg", "h", "abcde", "k"]
p arr.each_index.detect{|i| i >= 2 and arr[i].length == 1}
# => 3

---

Related Topics