473,390 Members | 1,399 Online
Bytes | Software Development & Data Engineering Community
Post Job

Home Posts Topics Members FAQ

Join Bytes to post your question to a community of 473,390 software developers and data experts.

Link Dictionary

Hi,
I have two dictionaries

dict1={'B8': set(['I8', 'H8', 'B2', 'B7', 'F8', 'C9', 'G8', 'B4',
'B5', 'B6', 'C8', 'E8', 'D8', 'B3', 'A9', 'A8', 'C7', 'B9', 'A7',
'B1']), 'B9': set(['I9', 'H9', 'A7', 'F9', 'B3', 'B6', 'G9', 'B4',
'B5', 'C9', 'B7', 'E9', 'B1', 'B2', 'D9', 'A9', 'A8', 'C8', 'B8',
'C7']), 'D1': set(['F1', 'F2', 'F3', 'G1', 'I1', 'D2', 'H1', 'A1',
'D4', 'B1', 'D8', 'D9', 'D6', 'D7', 'C1', 'D5', 'E1', 'D3', 'E3',
'E2'])}

and
dict2=
{'I6': '0', 'H9': '9', 'I2': '0', 'E8': '0', 'H3': '0', 'H7': '0',
'I7': '3', 'I4': '0', 'H5': '0', 'F9': '0', 'G7': '5', 'G6': '9',
'G5': '0', 'E1': '7', 'G3': '2', 'G2': '0', 'G1': '0', 'I1': '0',
'C8': '0', 'I3': '5', 'E5': '0', 'I5': '1', 'C9': '0', 'G9': '0',
'G8': '0', 'A1': '0', 'A3': '3', 'A2': '0', 'A5': '2', 'A4': '0',
'A7': '6', 'A6': '0', 'C3': '1', 'C2': '0', 'C1': '0', 'E6': '0',
'C7': '4', 'C6': '6', 'C5': '0', 'C4': '8', 'I9': '0', 'D8': '0',
'I8': '0', 'E4': '0', 'D9': '0', 'H8': '0', 'F6': '8', 'A9': '0',
'G4': '6', 'A8': '0', 'E7': '0', 'E3': '0', 'F1': '0', 'F2': '0',
'F3': '6', 'F4': '7', 'F5': '0', 'E2': '0', 'F7': '2', 'F8': '0',
'D2': '0', 'H1': '8', 'H6': '3', 'H2': '0', 'H4': '2', 'D3': '8',
'B4': '3', 'B5': '0', 'B6': '5', 'B7': '0', 'E9': '8', 'B1': '9',
'B2': '0', 'B3': '0', 'D6': '2', 'D7': '9', 'D4': '1', 'D5': '0',
'B8': '0', 'B9': '1', 'D1': '0'}

Now I want to create a dict which would have both the keys and values
to be of the corresponding values of dict2.

Something like this:

Eg. The first key in dict1 i.e. B8 as 0 (0 is the value of B8 in
dict2) mapped as set(['0','0','0',...]).

Can anyone help me out with this.
-
Anush

Jun 11 '07 #1
8 1284
En Mon, 11 Jun 2007 17:24:26 -0300, anush shetty
<it************@gmail.comescribió:
Hi,
I have two dictionaries

dict1={'B8': set(['I8', 'H8', 'B2', 'B7', 'F8', 'C9', 'G8', 'B4',
'B5', 'B6', 'C8', 'E8', 'D8', 'B3', 'A9', 'A8', 'C7', 'B9', 'A7',
'B1']), 'B9': set(['I9', 'H9', 'A7', 'F9', 'B3', 'B6', 'G9', 'B4',
'B5', 'C9', 'B7', 'E9', 'B1', 'B2', 'D9', 'A9', 'A8', 'C8', 'B8',
'C7']), ...

and
dict2=
{'I6': '0', 'H9': '9', 'I2': '0', 'E8': '0', 'H3': '0', 'H7': '0',
'I7': '3', 'I4': '0', 'H5': '0', 'F9': '0', 'G7': '5', 'G6': '9', [...]
'B2': '0', 'B3': '0', 'D6': '2', 'D7': '9', 'D4': '1', 'D5': '0',
'B8': '0', 'B9': '1', 'D1': '0'}

Now I want to create a dict which would have both the keys and values
to be of the corresponding values of dict2.

Something like this:

Eg. The first key in dict1 i.e. B8 as 0 (0 is the value of B8 in
dict2) mapped as set(['0','0','0',...]).
Sets can't have duplicate elements, so set(['0','0','0'])==set(['0'])
And dictionaries can't have duplicate keys either.
What do you want to do exactly? How would your desired structure look like?

--
Gabriel Genellina

Jun 11 '07 #2
anush shetty wrote:
Hi,
I have two dictionaries

dict1={'B8': set(['I8', 'H8', 'B2', 'B7', 'F8', 'C9', 'G8', 'B4',
'B5', 'B6', 'C8', 'E8', 'D8', 'B3', 'A9', 'A8', 'C7', 'B9', 'A7',
'B1']), 'B9': set(['I9', 'H9', 'A7', 'F9', 'B3', 'B6', 'G9', 'B4',
'B5', 'C9', 'B7', 'E9', 'B1', 'B2', 'D9', 'A9', 'A8', 'C8', 'B8',
'C7']), 'D1': set(['F1', 'F2', 'F3', 'G1', 'I1', 'D2', 'H1', 'A1',
'D4', 'B1', 'D8', 'D9', 'D6', 'D7', 'C1', 'D5', 'E1', 'D3', 'E3',
'E2'])}

and
dict2=
{'I6': '0', 'H9': '9', 'I2': '0', 'E8': '0', 'H3': '0', 'H7': '0',
'I7': '3', 'I4': '0', 'H5': '0', 'F9': '0', 'G7': '5', 'G6': '9',
'G5': '0', 'E1': '7', 'G3': '2', 'G2': '0', 'G1': '0', 'I1': '0',
'C8': '0', 'I3': '5', 'E5': '0', 'I5': '1', 'C9': '0', 'G9': '0',
'G8': '0', 'A1': '0', 'A3': '3', 'A2': '0', 'A5': '2', 'A4': '0',
'A7': '6', 'A6': '0', 'C3': '1', 'C2': '0', 'C1': '0', 'E6': '0',
'C7': '4', 'C6': '6', 'C5': '0', 'C4': '8', 'I9': '0', 'D8': '0',
'I8': '0', 'E4': '0', 'D9': '0', 'H8': '0', 'F6': '8', 'A9': '0',
'G4': '6', 'A8': '0', 'E7': '0', 'E3': '0', 'F1': '0', 'F2': '0',
'F3': '6', 'F4': '7', 'F5': '0', 'E2': '0', 'F7': '2', 'F8': '0',
'D2': '0', 'H1': '8', 'H6': '3', 'H2': '0', 'H4': '2', 'D3': '8',
'B4': '3', 'B5': '0', 'B6': '5', 'B7': '0', 'E9': '8', 'B1': '9',
'B2': '0', 'B3': '0', 'D6': '2', 'D7': '9', 'D4': '1', 'D5': '0',
'B8': '0', 'B9': '1', 'D1': '0'}

Now I want to create a dict which would have both the keys and values
to be of the corresponding values of dict2.

Something like this:

Eg. The first key in dict1 i.e. B8 as 0 (0 is the value of B8 in
dict2) mapped as set(['0','0','0',...]).

Can anyone help me out with this.
-
Anush
new_dict = {}
for akey, aset in dict1.items():
new_dict[akey] = sum(int(dict2[k]) for k in aset)

James
Jun 11 '07 #3
On Jun 12, 2:10 am, James Stroud <jstr...@mbi.ucla.eduwrote:
anush shetty wrote:
Hi,
I have two dictionaries
dict1={'B8': set(['I8', 'H8', 'B2', 'B7', 'F8', 'C9', 'G8', 'B4',
'B5', 'B6', 'C8', 'E8', 'D8', 'B3', 'A9', 'A8', 'C7', 'B9', 'A7',
'B1']), 'B9': set(['I9', 'H9', 'A7', 'F9', 'B3', 'B6', 'G9', 'B4',
'B5', 'C9', 'B7', 'E9', 'B1', 'B2', 'D9', 'A9', 'A8', 'C8', 'B8',
'C7']), 'D1': set(['F1', 'F2', 'F3', 'G1', 'I1', 'D2', 'H1', 'A1',
'D4', 'B1', 'D8', 'D9', 'D6', 'D7', 'C1', 'D5', 'E1', 'D3', 'E3',
'E2'])}
and
dict2=
{'I6': '0', 'H9': '9', 'I2': '0', 'E8': '0', 'H3': '0', 'H7': '0',
'I7': '3', 'I4': '0', 'H5': '0', 'F9': '0', 'G7': '5', 'G6': '9',
'G5': '0', 'E1': '7', 'G3': '2', 'G2': '0', 'G1': '0', 'I1': '0',
'C8': '0', 'I3': '5', 'E5': '0', 'I5': '1', 'C9': '0', 'G9': '0',
'G8': '0', 'A1': '0', 'A3': '3', 'A2': '0', 'A5': '2', 'A4': '0',
'A7': '6', 'A6': '0', 'C3': '1', 'C2': '0', 'C1': '0', 'E6': '0',
'C7': '4', 'C6': '6', 'C5': '0', 'C4': '8', 'I9': '0', 'D8': '0',
'I8': '0', 'E4': '0', 'D9': '0', 'H8': '0', 'F6': '8', 'A9': '0',
'G4': '6', 'A8': '0', 'E7': '0', 'E3': '0', 'F1': '0', 'F2': '0',
'F3': '6', 'F4': '7', 'F5': '0', 'E2': '0', 'F7': '2', 'F8': '0',
'D2': '0', 'H1': '8', 'H6': '3', 'H2': '0', 'H4': '2', 'D3': '8',
'B4': '3', 'B5': '0', 'B6': '5', 'B7': '0', 'E9': '8', 'B1': '9',
'B2': '0', 'B3': '0', 'D6': '2', 'D7': '9', 'D4': '1', 'D5': '0',
'B8': '0', 'B9': '1', 'D1': '0'}
Now I want to create a dict which would have both the keys and values
to be of the corresponding values of dict2.
Something like this:
Eg. The first key in dict1 i.e. B8 as 0 (0 is the value of B8 in
dict2) mapped as set(['0','0','0',...]).
Can anyone help me out with this.
-
Anush

new_dict = {}
for akey, aset in dict1.items():
new_dict[akey] = sum(int(dict2[k]) for k in aset)

James

So there is no way I can represent all the duplicates right.

So is there any solution where I could take the values of dictionary
and represent them separately?

-
Anush

Jun 11 '07 #4
On Jun 12, 1:50 am, "Gabriel Genellina" <gagsl-...@yahoo.com.ar>
wrote:
En Mon, 11 Jun 2007 17:24:26 -0300, anush shetty
<itsanushshe...@gmail.comescribió:
Hi,
I have two dictionaries
dict1={'B8': set(['I8', 'H8', 'B2', 'B7', 'F8', 'C9', 'G8', 'B4',
'B5', 'B6', 'C8', 'E8', 'D8', 'B3', 'A9', 'A8', 'C7', 'B9', 'A7',
'B1']), 'B9': set(['I9', 'H9', 'A7', 'F9', 'B3', 'B6', 'G9', 'B4',
'B5', 'C9', 'B7', 'E9', 'B1', 'B2', 'D9', 'A9', 'A8', 'C8', 'B8',
'C7']), ...
and
dict2=
{'I6': '0', 'H9': '9', 'I2': '0', 'E8': '0', 'H3': '0', 'H7': '0',
'I7': '3', 'I4': '0', 'H5': '0', 'F9': '0', 'G7': '5', 'G6': '9', [...]
'B2': '0', 'B3': '0', 'D6': '2', 'D7': '9', 'D4': '1', 'D5': '0',
'B8': '0', 'B9': '1', 'D1': '0'}
Now I want to create a dict which would have both the keys and values
to be of the corresponding values of dict2.
Something like this:
Eg. The first key in dict1 i.e. B8 as 0 (0 is the value of B8 in
dict2) mapped as set(['0','0','0',...]).

Sets can't have duplicate elements, so set(['0','0','0'])==set(['0'])
And dictionaries can't have duplicate keys either.
What do you want to do exactly? How would your desired structure look like?

--
Gabriel Genellina
Need it for a verification of the puzzle. It has duplicate values. But
there is any way where i could represent them with duplicates

-
Anush

Jun 11 '07 #5
On Jun 11, 10:19 pm, anush shetty <itsanushshe...@gmail.comwrote:
On Jun 12, 2:10 am, James Stroud <jstr...@mbi.ucla.eduwrote:
anush shetty wrote:
Hi,
I have two dictionaries
dict1={'B8': set(['I8', 'H8', 'B2', 'B7', 'F8', 'C9', 'G8', 'B4',
'B5', 'B6', 'C8', 'E8', 'D8', 'B3', 'A9', 'A8', 'C7', 'B9', 'A7',
'B1']), 'B9': set(['I9', 'H9', 'A7', 'F9', 'B3', 'B6', 'G9', 'B4',
'B5', 'C9', 'B7', 'E9', 'B1', 'B2', 'D9', 'A9', 'A8', 'C8', 'B8',
'C7']), 'D1': set(['F1', 'F2', 'F3', 'G1', 'I1', 'D2', 'H1', 'A1',
'D4', 'B1', 'D8', 'D9', 'D6', 'D7', 'C1', 'D5', 'E1', 'D3', 'E3',
'E2'])}
and
dict2=
{'I6': '0', 'H9': '9', 'I2': '0', 'E8': '0', 'H3': '0', 'H7': '0',
'I7': '3', 'I4': '0', 'H5': '0', 'F9': '0', 'G7': '5', 'G6': '9',
'G5': '0', 'E1': '7', 'G3': '2', 'G2': '0', 'G1': '0', 'I1': '0',
'C8': '0', 'I3': '5', 'E5': '0', 'I5': '1', 'C9': '0', 'G9': '0',
'G8': '0', 'A1': '0', 'A3': '3', 'A2': '0', 'A5': '2', 'A4': '0',
'A7': '6', 'A6': '0', 'C3': '1', 'C2': '0', 'C1': '0', 'E6': '0',
'C7': '4', 'C6': '6', 'C5': '0', 'C4': '8', 'I9': '0', 'D8': '0',
'I8': '0', 'E4': '0', 'D9': '0', 'H8': '0', 'F6': '8', 'A9': '0',
'G4': '6', 'A8': '0', 'E7': '0', 'E3': '0', 'F1': '0', 'F2': '0',
'F3': '6', 'F4': '7', 'F5': '0', 'E2': '0', 'F7': '2', 'F8': '0',
'D2': '0', 'H1': '8', 'H6': '3', 'H2': '0', 'H4': '2', 'D3': '8',
'B4': '3', 'B5': '0', 'B6': '5', 'B7': '0', 'E9': '8', 'B1': '9',
'B2': '0', 'B3': '0', 'D6': '2', 'D7': '9', 'D4': '1', 'D5': '0',
'B8': '0', 'B9': '1', 'D1': '0'}
Now I want to create a dict which would have both the keys and values
to be of the corresponding values of dict2.
Something like this:
Eg. The first key in dict1 i.e. B8 as 0 (0 is the value of B8 in
dict2) mapped as set(['0','0','0',...]).
Can anyone help me out with this.
-
Anush
new_dict = {}
for akey, aset in dict1.items():
new_dict[akey] = sum(int(dict2[k]) for k in aset)
James

So there is no way I can represent all the duplicates right.

So is there any solution where I could take the values of dictionary
and represent them separately?

-
Anush
I use lists instead of your dict1 but preserve your mapping dict2:

data = [
['B8',
['I8', 'H8', 'B2', 'B7', 'F8',
'C9', 'G8', 'B4', 'B5', 'B6',
'C8', 'E8', 'D8', 'B3', 'A9',
'A8', 'C7', 'B9', 'A7', 'B1']],
['B9',
['I9', 'H9', 'A7', 'F9', 'B3',
'B6', 'G9', 'B4', 'B5', 'C9',
'B7', 'E9', 'B1', 'B2', 'D9',
'A9', 'A8', 'C8', 'B8', 'C7']],
['D1',
['F1', 'F2', 'F3', 'G1', 'I1',
'D2', 'H1', 'A1', 'D4', 'B1',
'D8', 'D9', 'D6', 'D7', 'C1',
'D5', 'E1', 'D3', 'E3', 'E2']]]

dict2= {
'I6': '0', 'H9': '9', 'I2': '0', 'E8': '0', 'H3': '0', 'H7': '0',
'I7': '3', 'I4': '0', 'H5': '0', 'F9': '0', 'G7': '5', 'G6': '9',
'G5': '0', 'E1': '7', 'G3': '2', 'G2': '0', 'G1': '0', 'I1': '0',
'C8': '0', 'I3': '5', 'E5': '0', 'I5': '1', 'C9': '0', 'G9': '0',
'G8': '0', 'A1': '0', 'A3': '3', 'A2': '0', 'A5': '2', 'A4': '0',
'A7': '6', 'A6': '0', 'C3': '1', 'C2': '0', 'C1': '0', 'E6': '0',
'C7': '4', 'C6': '6', 'C5': '0', 'C4': '8', 'I9': '0', 'D8': '0',
'I8': '0', 'E4': '0', 'D9': '0', 'H8': '0', 'F6': '8', 'A9': '0',
'G4': '6', 'A8': '0', 'E7': '0', 'E3': '0', 'F1': '0', 'F2': '0',
'F3': '6', 'F4': '7', 'F5': '0', 'E2': '0', 'F7': '2', 'F8': '0',
'D2': '0', 'H1': '8', 'H6': '3', 'H2': '0', 'H4': '2', 'D3': '8',
'B4': '3', 'B5': '0', 'B6': '5', 'B7': '0', 'E9': '8', 'B1': '9',
'B2': '0', 'B3': '0', 'D6': '2', 'D7': '9', 'D4': '1', 'D5': '0',
'B8': '0', 'B9': '1', 'D1': '0'}

dout = [[dict2[i],[dict2[k] for k in j]]
for i,j in data]
print dout

- Paddy.

Jun 11 '07 #6
>
dout = [[dict2[i],[dict2[k] for k in j]]
for i,j in data]
print dout

- Paddy.
Working from your original dict1:

dout = [[dict2[i], [dict2[k] for k in setvalue]]
for i,setvalue in dict1.iteritems()]

- Paddy

Jun 11 '07 #7
anush shetty wrote:
On Jun 12, 2:10 am, James Stroud <jstr...@mbi.ucla.eduwrote:
>>anush shetty wrote:
>>>Hi,
I have two dictionaries
>>dict1={'B8': set(['I8', 'H8', 'B2', 'B7', 'F8', 'C9', 'G8', 'B4',
'B5', 'B6', 'C8', 'E8', 'D8', 'B3', 'A9', 'A8', 'C7', 'B9', 'A7',
'B1']), 'B9': set(['I9', 'H9', 'A7', 'F9', 'B3', 'B6', 'G9', 'B4',
'B5', 'C9', 'B7', 'E9', 'B1', 'B2', 'D9', 'A9', 'A8', 'C8', 'B8',
'C7']), 'D1': set(['F1', 'F2', 'F3', 'G1', 'I1', 'D2', 'H1', 'A1',
'D4', 'B1', 'D8', 'D9', 'D6', 'D7', 'C1', 'D5', 'E1', 'D3', 'E3',
'E2'])}
>>>and
dict2=
{'I6': '0', 'H9': '9', 'I2': '0', 'E8': '0', 'H3': '0', 'H7': '0',
'I7': '3', 'I4': '0', 'H5': '0', 'F9': '0', 'G7': '5', 'G6': '9',
'G5': '0', 'E1': '7', 'G3': '2', 'G2': '0', 'G1': '0', 'I1': '0',
'C8': '0', 'I3': '5', 'E5': '0', 'I5': '1', 'C9': '0', 'G9': '0',
'G8': '0', 'A1': '0', 'A3': '3', 'A2': '0', 'A5': '2', 'A4': '0',
'A7': '6', 'A6': '0', 'C3': '1', 'C2': '0', 'C1': '0', 'E6': '0',
'C7': '4', 'C6': '6', 'C5': '0', 'C4': '8', 'I9': '0', 'D8': '0',
'I8': '0', 'E4': '0', 'D9': '0', 'H8': '0', 'F6': '8', 'A9': '0',
'G4': '6', 'A8': '0', 'E7': '0', 'E3': '0', 'F1': '0', 'F2': '0',
'F3': '6', 'F4': '7', 'F5': '0', 'E2': '0', 'F7': '2', 'F8': '0',
'D2': '0', 'H1': '8', 'H6': '3', 'H2': '0', 'H4': '2', 'D3': '8',
'B4': '3', 'B5': '0', 'B6': '5', 'B7': '0', 'E9': '8', 'B1': '9',
'B2': '0', 'B3': '0', 'D6': '2', 'D7': '9', 'D4': '1', 'D5': '0',
'B8': '0', 'B9': '1', 'D1': '0'}
>>>Now I want to create a dict which would have both the keys and values
to be of the corresponding values of dict2.
>>>Something like this:
>>>Eg. The first key in dict1 i.e. B8 as 0 (0 is the value of B8 in
dict2) mapped as set(['0','0','0',...]).
>>>Can anyone help me out with this.
-
Anush

new_dict = {}
for akey, aset in dict1.items():
new_dict[akey] = sum(int(dict2[k]) for k in aset)

James

So there is no way I can represent all the duplicates right.

So is there any solution where I could take the values of dictionary
and represent them separately?

-
Anush
Just leave off the sum and turn the generator into a list comprehension:

new_dict = {}
for akey, aset in dict1.items():
new_dict[akey] = [dict2[k] for k in aset]

James
Jun 11 '07 #8
>
Working from your original dict1:

dout = [[dict2[i], [dict2[k] for k in setvalue]]
for i,setvalue in dict1.iteritems()]

- Paddy

I dont think this works.

Jun 11 '07 #9

This thread has been closed and replies have been disabled. Please start a new discussion.

Similar topics

1
by: none | last post by:
or is it just me? I am having a problem with using a dictionary as an attribute of a class. This happens in python 1.5.2 and 2.2.2 which I am accessing through pythonwin builds 150 and 148...
4
by: brianobush | last post by:
# # My problem is that I want to create a # class, but the variables aren't known # all at once. So, I use a dictionary to # store the values in temporarily. # Then when I have a complete set, I...
1
by: Lothar Armbr?ster | last post by:
Hello out there, this morning I had to get some rows via flashback query. Since my undo retention is just 3h, the undo segments were already aged out. But I have a physical standby database that...
7
by: homecurr | last post by:
Is there a easy way to put a map in my web page? I just need a static map show a given address. I am thinking about a link like <img src="http://maps.yahoo.com/something?addr...
3
by: Robert L. Smith, Jr. | last post by:
I want to include an online dictionary, a binary file with its own unique format, into an executable at link time, using the linker in Microsoft Visual-C++ V6. Is it possible? What is the link...
1
by: john wright | last post by:
I have a dictionary oject I created and I want to bind a listbox to it. I am including the code for the dictionary object. Here is the error I am getting: "System.Exception: Complex...
8
by: akameswaran | last post by:
I wrote up a quick little set of tests, I was acutally comparing ways of doing "case" behavior just to get some performance information. Now two of my test cases had almost identical results which...
8
by: Brian L. Troutwine | last post by:
I've got a problem that I can't seem to get my head around and hoped somebody might help me out a bit: I've got a dictionary, A, that is arbitarily large and may contains ints, None and more...
3
by: JamesB | last post by:
I have a config screen in my app that updates a dictionary<key,value>. I want to store a copy of this before going into the config screen so if the user wants to cancel all their changes I can...
1
by: sachin2 | last post by:
I am using 3 types of dictionaries. 1) Dictionary<string, string > d = new Dictionary<string, string>(); 2) Dictionary<string, List<string>> d = new Dictionary<string, List<string>>(); 3)...
0
by: Charles Arthur | last post by:
How do i turn on java script on a villaon, callus and itel keypad mobile phone
0
by: ryjfgjl | last post by:
If we have dozens or hundreds of excel to import into the database, if we use the excel import function provided by database editors such as navicat, it will be extremely tedious and time-consuming...
0
by: emmanuelkatto | last post by:
Hi All, I am Emmanuel katto from Uganda. I want to ask what challenges you've faced while migrating a website to cloud. Please let me know. Thanks! Emmanuel
0
BarryA
by: BarryA | last post by:
What are the essential steps and strategies outlined in the Data Structures and Algorithms (DSA) roadmap for aspiring data scientists? How can individuals effectively utilize this roadmap to progress...
1
by: nemocccc | last post by:
hello, everyone, I want to develop a software for my android phone for daily needs, any suggestions?
1
by: Sonnysonu | last post by:
This is the data of csv file 1 2 3 1 2 3 1 2 3 1 2 3 2 3 2 3 3 the lengths should be different i have to store the data by column-wise with in the specific length. suppose the i have to...
0
by: Hystou | last post by:
Most computers default to English, but sometimes we require a different language, especially when relocating. Forgot to request a specific language before your computer shipped? No problem! You can...
0
Oralloy
by: Oralloy | last post by:
Hello folks, I am unable to find appropriate documentation on the type promotion of bit-fields when using the generalised comparison operator "<=>". The problem is that using the GNU compilers,...
0
by: Hystou | last post by:
Overview: Windows 11 and 10 have less user interface control over operating system update behaviour than previous versions of Windows. In Windows 11 and 10, there is no way to turn off the Windows...

By using Bytes.com and it's services, you agree to our Privacy Policy and Terms of Use.

To disable or enable advertisements and analytics tracking please visit the manage ads & tracking page.