I created a list called a, when I inserted a at the beginning of the list this happened
>>> a = ['a', 'b', 'c', 'd']
>>> a.insert(0, a)
>>> a
[[...], 'a', 'b', 'c', 'd']
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vd743r
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4You've inserted a reference to `a` as the first element of `a`. What don't you understand? – Iain Shelvington Jul 18 '20 at 02:26
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`a` is a list. What did you expect to happen by inserting a list? – Some programmer dude Jul 18 '20 at 02:28
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1It's just the way Python displays a circular reference. If it didn't, then you'd have infinite recursion. – ggorlen Jul 18 '20 at 02:35
2 Answers
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Running this code:
a = ['a', 'b', 'c', 'd']
a.insert(0, a)
print(a)
print(a[0])
yields this result:
[[...], 'a', 'b', 'c', 'd']
[[...], 'a', 'b', 'c', 'd']
this leads me to believe this is how python handles printing recursive referencing. a contains a reference to a, which contains a reference to a... meaning there's no good way to actually print it, so it just prints [...]
if you want to include the values of a inside of a pointer to a, you might want to copy it
this code:
import copy
a = ['a', 'b', 'c', 'd']
a.insert(0, copy.deepcopy(a))
print(a)
results in this:
[['a', 'b', 'c', 'd'], 'a', 'b', 'c', 'd']
Warlax56
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