# Get the last element of a string of type in a list

Let's say I have a list of different types:

i.e.

``````[7, 'string 1', 'string 2', [a, b c], 'string 3', 0, (1, 2, 3)]
```

```

Is there a Pythonic way to return 'string 3'?

+3

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Once you have a given type, you can use several kinds of concepts to get what you need.

``````[el for el in lst if isinstance(el, given_type)][-1]
# Gives the last element if there is one, else IndexError
```

```

or

``````next((el for el in reversed(lst) if isinstance(el, given_type)), None)
# Gives the last element if there is one, else None
```

```

If this is something you do often enough, you can include it in a function:

``````def get_last_of_type(type_, iterable):
for el in reversed(iterable):
if isinstance(el, type_):
return el
return None
```

```
+12

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I think the easiest way is to grab the last element of the filtered list.

``````filter(lambda t: type(t) is type(''), your_list)[-1]
```

```

or

``````[el for el in your_list if type(el) is type('')][-1]
```

```
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``````>>> l = [7, 'string 1', 'string 2', 8, 'string 3', 0, (1, 2, 3)]
>>> from itertools import dropwhile
>>> next(dropwhile(lambda x: not isinstance(x, str), reversed(l)), None)
'string 3'
```

```

If you cannot use imports for some reason, the polyfill for `itertools.dropwhile`

looks like this:

``````def dropwhile(predicate, iterable):
# dropwhile(lambda x: x<5, [1,4,6,4,1]) --> 6 4 1
iterable = iter(iterable)
for x in iterable:
if not predicate(x):
yield x
break
for x in iterable:
yield x
```

```
+2

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try it

``````lst = [7, 'string 1', 'string 2', [a, b c], 'string 3', 0, (1, 2, 3)]
filtered_string = [ x for x in lst if type(x) is str]
```

```

now you can get any of its index for the last index

``````filtered_string[-1]
```

```
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