[Tutor] column heading

[dn]

On 26/06/20 3:46 PM, SUMAN KANTI ROY wrote:
> a=[1,2,3,4,5,6,7,8,9,10]
> arr=[[1,2],[2,23],[4,5]]
> dic={}
> for i in range(0,len(arr)):
>      if arr[i][0] not in dic:
>          dic[arr[i][0]]=arr[i][1]
> arr1=[]
> def find_key(val,dic):
>      for i in dic.keys():
>          if dic[i]==val:
>              return i
> new_arr=sorted(dic.values(),reverse=True)
> ans= [[0 for i in range(3)] for i in range(len(new_arr))]
> for i in range(len(new_arr)):
>      ans[i][0]=a[i]
>      ans[i][1]=find_key(new_arr[i],dic)
>      ans[i][2]=new_arr[i]
> print(ans)
> 
> 0/p=====[[1, 2, 23], [2, 4, 5], [3, 1, 2]]
> I want to make it 3 column matrix ,,,with column heading
> name['device',R.B.N','CQI']
> please guide me...


As @Alan has said, you are trying to write Java/C/? in Python. The 
Python language has its own idioms and techniques. I'm also going to 
criticise the choice of variable_names - and Python uses "lists" not 
"arrays"!

Here is some food for thought (and there are other ways of achieving the 
same):

- because the array is a list of 2-member lists you can use Python's 
dynamic conversions:

 >>> arr=[[1,2],[2,23],[4,5]]
 >>> dic = dict( arr )
 >>> dic
{1: 2, 2: 23, 4: 5}

- each element of a dictionary consists of a key and a value. They can 
be considered separately:

 >>> keys = dic.keys()
 >>> keys
dict_keys([1, 2, 4])
 >>> values = dic.values()

- two lists can be joined - except that the basic result is an iterator:

 >>> inverted_arr = list( zip( dic.values(), dic.keys() ) )
 >>> inverted_arr
[(2, 1), (23, 2), (5, 4)]

- and just to complete the picture, we do the list-to-dict 'trick' again:

 >>> inverted_dic = dict( inverted_arr )
 >>> inverted_dic
{2: 1, 23: 2, 5: 4}

- which has generated all of the constructs needed!

I'll leave it to you to decide if you need the inverted dict. Research 
the Python documentation (it's very good!), and note that zip() allows 
multiple lists to be zip-ped together, providing...

Can you now solve the problem without using any for-loops?
-- 
Regards =dn