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Does string slicing perform copy in memory?

I'm wondering if :

a = "abcdef"
b = "def"
if a[3:] == b:
    print("something")

does actually perform a copy of the "def" part of a somewhere in memory, or if the letters checking is done in-place ?

Note : I'm speaking about a string, not a list (for which I know the answer)

over 4 years ago · Santiago Trujillo
2 Respuestas
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0

Possible talking point (feel free to edit adding information).

I have just written this test to verify empirically what the answer to the question might be (this cannot and does not want to be a certain answer).

import sys

a = "abcdefg"

print("a id:", id(a))
print("a[2:] id:", id(a[2:]))
print("a[2:] is a:", a[2:] is a)

print("Empty string memory size:", sys.getsizeof(""))
print("a memory size:", sys.getsizeof(a))
print("a[2:] memory size:", sys.getsizeof(a[2:]))

Output:

a id: 139796109961712
a[2:] id: 139796109962160
a[2:] is a: False
Empty string memory size: 49
a memory size: 56
a[2:] memory size: 54

As we can see here:

  • the size of an empty string object is 49 bytes
  • a single character occupies 1 byte (Latin-1 encoding)
  • a and a[2:] ids are different
  • the occupied memory of each a and a[2:] is consistent with the memory occupied by a string with that number of char assigned
over 4 years ago · Santiago Trujillo Denunciar

0

String slicing makes a copy in CPython.

Looking in the source, this operation is handled in unicodeobject.c:unicode_subscript. There is evidently a special-case to re-use memory when the step is 1, start is 0, and the entire content of the string is sliced - this goes into unicode_result_unchanged and there will not be a copy. However, the general case calls PyUnicode_Substring where all roads lead to a memcpy.

To empirically verify these claims, you can use a stdlib memory profiling tool tracemalloc:

# s.py
import tracemalloc

tracemalloc.start()
before = tracemalloc.take_snapshot()
a = "." * 7 * 1024**2  # 7 MB of .....   # line 6, first alloc
b = a[1:]                                # line 7, second alloc
after = tracemalloc.take_snapshot()

for stat in after.compare_to(before, 'lineno')[:2]:
    print(stat)

You should see the top two statistics output like this:

/tmp/s.py:6: size=7168 KiB (+7168 KiB), count=1 (+1), average=7168 KiB
/tmp/s.py:7: size=7168 KiB (+7168 KiB), count=1 (+1), average=7168 KiB

This result shows two allocations of 7 meg, strong evidence of the memory copying, and the exact line numbers of those allocations will be indicated.

Try changing the slice from b = a[1:] into b = a[0:] to see that entire-string-special-case in effect: there should be only one large allocation now, and sys.getrefcount(a) will increase by one.

In theory, since strings are immutable, an implementation could re-use memory for substring slices. This would likely complicate any reference-counting based garbage collection process, so it may not be a useful idea in practice. Consider the case where a small slice from a much larger string was taken - unless you implemented some kind of sub-reference counting on the slice, the memory from the much larger string could not be freed until the end of the substring's lifetime.

For users that specifically need a standard type which can be sliced without copying the underlying data, there is memoryview. See What exactly is the point of memoryview in Python for more information about that.

over 4 years ago · Santiago Trujillo Denunciar
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