python - Python3 - sorting a list -


i don't understand why 1 , 5 printed 3 times. (i know use list.sort() method.)

lst = [4,5,5,1,6,1] copy = lst.copy() sort = [] in range(len(lst)):     min_ = min(copy)     k in copy:         if k == min_:             sort.append(min_)     copy.remove(min_) print(sort)

list.remove() removes first occurrence of value, not such values. result you'll first add both 1 values sort, remove 1 of 2 1 values, add remaining 1 value sort again:

>>> lst = [4,5,5,1,6,1] >>> copy = lst.copy() >>> sort = [] >>> min_ = min(copy) >>> min_ 1 >>> k in copy: ...     if k == min_: ...         sort.append(min_) ...  >>> sort [1, 1] >>> copy.remove(min_) >>> copy [4, 5, 5, 6, 1] >>> min_ = min(copy) >>> min_ 1 >>> k in copy: ...     if k == min_: ...         sort.append(min_) ...  >>> sort [1, 1, 1]

you use list comprehension remove values, creating new copy excludes value remove filtering:

copy = [v v in copy if v != min_]

this not efficient, of course.

note next problem you'll run you'll have emptied out copy before have completed range(len(lst)) iterations. replace loop while copy: loop instead.

alternatively, add first match of min_ value sort:

for in range(len(lst)):     min_ = min(copy)     k in copy:         if k == min_:             sort.append(min_)             break     copy.remove(min_)

the break ends for loop early. of course, don't have loop find minimum value, min() call already did, can drop altogether:

for in range(len(lst)):     min_ = min(copy)     sort.append(min_)     copy.remove(min_)

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