for _ in range(NUM_OF_STRING):
sortedDict[_] = {str(string[_]), sortIndexArr[_]}
for __ in range(NUM_OF_STRING):
print(sortedDict[__])
This is the code, and if you run it,
{0, "['P', 'S', 'R', 'S', 'P', 'P', 'S', 'E', 'P', 'E', 'R', 'S', 'R', 'R', 'P', 'S', 'P', 'R']"}
{0, "['P', 'S', 'D', 'S', 'P', 'E', 'E', 'P', 'P', 'E', 'S', 'E', 'D', 'E', 'D', 'P', 'E', 'S']"}
{0, "['R', 'R', 'D', 'D', 'E', 'E', 'D', 'P', 'P', 'D', 'E', 'D', 'D', 'D', 'D']"}
{0, "['R', 'R', 'R', 'P', 'R', 'S', 'S', 'S', 'E', 'E', 'D', 'S', 'S', 'R', 'D', 'E', 'P', 'S', 'R']"}
{0, "['S', 'S', 'E', 'S', 'R', 'E', 'P', 'P', 'S', 'R', 'E', 'D', 'R', 'P', 'R', 'E', 'S', 'D', 'P']"}
{0, "['E', 'S', 'P', 'D', 'D', 'S', 'R', 'D', 'D', 'D', 'D', 'D', 'E', 'R']"}
{0, "['E', 'P', 'S', 'S', 'D', 'S', 'R', 'D', 'P', 'P', 'S', 'P', 'D', 'R', 'S']"}
{"['D', 'R', 'R', 'S', 'E', 'R', 'P', 'R', 'R', 'P', 'E', 'P', 'R', 'S', 'D', 'S', 'D', 'E']", 0}
{"['P', 'D', 'S', 'S', 'P', 'E', 'E', 'P', 'R', 'R', 'P', 'E', 'D', 'R', 'E', 'R', 'E', 'D', 'R']", 0}
{"['P', 'R', 'R', 'S', 'S', 'R', 'E', 'R', 'P', 'S', 'R', 'D', 'R', 'E', 'S', 'R', 'S', 'R', 'P', 'S']", 0}
In this way, the value of sortIndexArr[_]
appears on the left, and the value of string[_]
appears.
The value of string[_]
is multiple characters in the list
The value of sortIndexArr[_]
is a number.
There is no order because sortedDict[k]
is set
. (I'm sure it's inside.) So unlike list
or tuple
, print
comes out without order.
(Corrected) The answer below is an incorrect answer, so you can ignore it.
OrderedDict
https://www.geeksforgeeks.org/ordereddict-in-python/
You can use this.
However, since Python 3.7, the dictionary also remembers the insertion order, so the use of OrderedDict, which occupies more memory, is no longer necessary to use it only to guarantee the order in simple rotation.
https://docs.python.org/3/library/collections.html#ordereddict-objects
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