How to remove \n from a list element?


Question

I'm trying to get Python to a read line from a .txt file and write the elements of the first line into a list. The elements in the file were tab- separated so I used split("\t") to separate the elements. Because the .txt file has a lot of elements I saved the data found in each line into a separate list.

The problem I currently have is that it's showing each list like this:

['Name1', '7.3', '6.9', '6.6', '6.6', '6.1', '6.4', '7.3\n']

How can I remove \n from the last element of the list and make it just '7.3'?

1
62
6/17/2015 1:27:01 PM

Accepted Answer

If you want to remove \n from the last element only, use this:

t[-1] = t[-1].strip()

If you want to remove \n from all the elements, use this:

t = map(lambda s: s.strip(), t)

You might also consider removing \n before splitting the line:

line = line.strip()
# split line...
124
10/3/2010 11:16:38 AM

From Python3 onwards

map no longer returns a list but a mapObject, thus the answer will look something like

>>> map(lambda x:x.strip(),l)
<map object at 0x7f00b1839fd0>

You can read more about it on What’s New In Python 3.0.

map() and filter() return iterators. If you really need a list, a quick fix is e.g. list(map(...))

So now what are the ways of getting trough this?


Case 1 - The list call over map with a lambda

map returns an iterator. list is a function that can convert an iterator to a list. Hence you will need to wrap a list call around map. So the answer now becomes,

>>> l = ['Name1', '7.3', '6.9', '6.6', '6.6', '6.1', '6.4', '7.3\n']
>>> list(map(lambda x:x.strip(),l))
['Name1', '7.3', '6.9', '6.6', '6.6', '6.1', '6.4', '7.3']

Very good, we get the output. Now we check the amount of time it takes for this piece of code to execute.

$ python3 -m timeit "l = ['Name1', '7.3', '6.9', '6.6', '6.6', '6.1', '6.4', '7.3\n'];list(map(lambda x:x.strip(),l))"
100000 loops, best of 3: 2.22 usec per loop

2.22 microseconds. That is not so bad. But are there more efficient ways?


Case 2 - The list call over map withOUT a lambda

lambda is frowned upon by many in the Python community (including Guido). Apart from that it will greatly reduce the speed of the program. Hence we need to avoid that as much as possible. The toplevel function str.strip. Comes to our aid here.

The map can be re-written without using lambda using str.strip as

>>> list(map(str.strip,l))
['Name1', '7.3', '6.9', '6.6', '6.6', '6.1', '6.4', '7.3']

And now for the times.

$ python3 -m timeit "l = ['Name1', '7.3', '6.9', '6.6', '6.6', '6.1', '6.4', '7.3\n'];list(map(str.strip,l))"
1000000 loops, best of 3: 1.38 usec per loop

Fantastic. You can see the efficiency differences between the two ways. It is nearly 60% faster. Thus the approach without using a lambda is a better choice here.


Case 3 - Following Guidelines, The Regular way

Another important point from What’s New In Python 3.0 is that it advices us to avoid map where possible.

Particularly tricky is map() invoked for the side effects of the function; the correct transformation is to use a regular for loop (since creating a list would just be wasteful).

So we can solve this problem without a map by using a regular for loop.

The trivial way of solving (the brute-force) would be:-

>>> l = ['Name1', '7.3', '6.9', '6.6', '6.6', '6.1', '6.4', '7.3\n']
>>> final_list = []
>>> for i in l:
...     final_list.append(i.strip())
... 
>>> final_list
['Name1', '7.3', '6.9', '6.6', '6.6', '6.1', '6.4', '7.3']

The timing setup

def f():
    l = ['Name1', '7.3', '6.9', '6.6', '6.6', '6.1', '6.4', '7.3\n']
    final_list = []
    for i in l:
         final_list.append(i.strip())
import timeit
print(min(timeit.repeat("f()","from __main__ import f")))

And the result.

1.5322505849981098

As you can see the brute-force is a bit slower here. But it is definitely more readable to a common programmer than a map clause.


Case 4 - List Comprehensions

A list comprehension here is also possible and is the same as in Python2.

>>> [i.strip() for i in l]
['Name1', '7.3', '6.9', '6.6', '6.6', '6.1', '6.4', '7.3']

Now for the timings:

$ python3 -m timeit "l = ['Name1', '7.3', '6.9', '6.6', '6.6', '6.1', '6.4', '7.3\n'];[i.strip() for i in l]"
1000000 loops, best of 3: 1.28 usec per loop

As you can see the list-comprehension is more effective than map (even that without a lambda). Hence the thumb rule in Python3 is to use a list comprehension instead of map


Case 5 - In-Place mechanisms and Space Efficiency (T-M-T)

A final way is to make the changes in-place within the list itself. This will save a lot of memory space. This can be done using enumerate.

>>> l = ['Name1', '7.3', '6.9', '6.6', '6.6', '6.1', '6.4', '7.3\n']
>>> for i,s in enumerate(l):
...     l[i] = s.strip()
... 
>>> l
['Name1', '7.3', '6.9', '6.6', '6.6', '6.1', '6.4', '7.3']

The timing result would be 1.4806894720022683. But however this way is space effective.


Conclusion

A comparitive list of timings (Both Python 3.4.3 and Python 3.5.0)

----------------------------------------------------
|Case| method          | Py3.4 |Place| Py3.5 |Place|
|----|-----------------|-------|-----|-------|-----|
| 1  | map with lambda | 2.22u | 5   | 2.85u | 5   |
| 2  | map w/o lambda  | 1.38u | 2   | 2.00u | 2   |
| 3  | brute-force     | 1.53u | 4   | 2.22u | 4   |
| 4  | list comp       | 1.28u | 1   | 1.25u | 1   |
| 5  | in-place        | 1.48u | 3   | 2.14u | 3   |
----------------------------------------------------

Finally note that the list-comprehension is the best way and the map using lambda is the worst. But again --- ONLY IN PYTHON3


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