lista = list.sort(lista) This should be lista.sort() The .sort() method is in-place, and returns None. If you want something not in-place, which returns a value, you could use sorted_list = sorted(lista) Aside #1: please don’t call your lists list. That clobbers the builtin list type. Aside #2: I’m not sure what this line is meant … Read more
This is called the placement new operator. It allows you to supply the memory the data will be allocated in without having the new operator allocate it. For example: Foo * f = new Foo(); The above will allocate memory for you. void * fm = malloc(sizeof(Foo)); Foo *f = new (fm) Foo(); The above … Read more
I’d write it this way: a[i:j] = sorted(a[i:j]) It is not in-place sort either, but fast enough for relatively small segments. Please note, that Python copies only object references, so the speed penalty won’t be that huge compared to a real in-place sort as one would expect.
If you want to apply mathematical operations to a numpy array in-place, you can simply use the standard in-place operators +=, -=, /=, etc. So for example: >>> def foo(a): … a += 10 … >>> a = numpy.arange(10) >>> a array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9]) >>> foo(a) >>> a … Read more
This post addresses the question about the absence of the equivalent of sed’s “-i” option, and in particular the situation described: I have a bunch of files and writing each one to a separate file wouldn’t be easy. There are several options, at least if you are working in a Mac or Linux or similar … Read more
You can use the -i flag correctly by providing it with a suffix to add to the backed-up file. Extending your example: sed -i.bu ‘s/oldword/newword/’ file1.txt Will give you two files: one with the name file1.txt that contains the substitution, and one with the name file1.txt.bu that has the original content. Mildly dangerous If you … Read more
This is a copy of the explanation on github. There is no guarantee that an inplace operation is actually faster. Often they are actually the same operation that works on a copy, but the top-level reference is reassigned. The reason for the difference in performance in this case is as follows. The (df1-df2).dropna() call creates … Read more
Knuth left this as an exercise (Vol 3, 5.2.5). There do exist in-place merge sorts. They must be implemented carefully. First, naive in-place merge such as described here isn’t the right solution. It downgrades the performance to O(N2). The idea is to sort part of the array while using the rest as working area for … Read more
sorted() returns a new sorted list, leaving the original list unaffected. list.sort() sorts the list in-place, mutating the list indices, and returns None (like all in-place operations). sorted() works on any iterable, not just lists. Strings, tuples, dictionaries (you’ll get the keys), generators, etc., returning a list containing all elements, sorted. Use list.sort() when you … Read more
To remove the line and print the output to standard out: sed ‘/pattern to match/d’ ./infile To directly modify the file – does not work with BSD sed: sed -i ‘/pattern to match/d’ ./infile Same, but for BSD sed (Mac OS X and FreeBSD) – does not work with GNU sed: sed -i ” ‘/pattern … Read more