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The subject requires the implementation of a simple function that loops the left of an arrayA is stored inN>0) Integer to move each integer loop to the left without allowing the use of a different arrayM≥0) a position that willThe data in a is determined by (A?0??A?1???A?N−1?? ) Transformation to (A?M???A?N−1??A?0??A?1???A?m−1?? (The first M-number loop moves to the M-position of the last face ). If you still need to consider how many times the progra

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#创建phone={'Alice':'12234','beth ' :'352235'}#dict使用items=[('name','gumby'), ('age' ,}d=dict (items) d=dict (name='gumby', age=42)#基本字典操作lend (d) d[k]d[k]= in dx={}x[42]='foobae' x{:'foobar'}people={'Alice':{'Phone':'123','Addr':'Foo drive34'},'Beth':{'Phone':'23234','Addr':'3123'}}labels={'Phone':'Phone number','Addr':'Address'}name=input ('Name:') Request=input ("phone Number (p) or address (a)?:")ifrequest=='P': key='Phone'ifrequest=='a': key='Addr'ifNameinchPeople:Print("%s '%s ' is%s."% (Nam

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() in [+]: Print aa[1, 2, 34, 45, 332, 2342]Quick way to copy a list: The first method, is the slice assignment, the second through the sorted function;in []: x = [1,454,2,34,3]in]: y = x[:] or y = sorted (x) give Yin the value after x slice [max]: Y.sort () in [Max]: xout[73]: [1, 454, 2, 3]in, yout[74]: [1, 2, 3, 34, 454]///See, is so capricious;The function of the 9.tuple function is basically the same as the list function, which takes a sequence as an argument and converts it to a tuple;In