Hello,
Can you help me with this please?
I have a list that contains elements to be created (in a 3D app), in the
list each element is a dictionary that contains data, like:
Elements = [
{'Width': 3.0, 'Depth': 3.0, 'Name': 'Access', 'Parent':
'Plot', 'Height': 3.0},
{'Width': 3.0, 'Depth': 3.0, 'Name': 'Circulation_01',
'Parent': 'Access', 'Height': 3.0},
{'Width': 3.0, 'Depth': 3.0, 'Name': 'Circulation_02',
'Parent': 'Access', 'Height': 3.0},
{'Width': 3.0, 'Depth': 3.0, 'Name': 'Circulation_03',
'Parent': 'Access', 'Height': 3.0},
{'Width': 2.0, 'Depth': 6.0, 'Name': 'Int_Circ_01',
'Parent': 'Circulation_01', 'Height': 3.0},
{'Width': 2.0, 'Depth': 5.0, 'Name': 'Int_Circ_02',
'Parent': 'Circulation_01', 'Height': 3.0},
{'Width': 2.0, 'Depth': 6.5, 'Name': 'Int_Circ_03',
'Parent': 'Circulation_02', 'Height': 3.0},
{'Width': 2.0, 'Depth': 5.0, 'Name': 'Int_Circ_04',
'Parent': 'Circulation_02', 'Height': 3.0},
{'Width': 2.0, 'Depth': 5.0, 'Name': 'Int_Circ_05',
'Parent': 'Circulation_03', 'Height': 3.0},
{'Width': 2.0, 'Depth': 5.0, 'Name': 'Int_Circ_06',
'Parent': 'Circulation_03', 'Height': 3.0},
]
so a for loop is used to iterate the list, like:
for element in elements:
create object with the desired width, depth, name, etc
The thing is that there can only be a "Circulation" by story, so I am
thinking in adding each created object to a built_Objects list and
appending the created object to the list, like:
for element in elements:
create element
append element['Name'] to built_Objects
My question is, how can I check how many times "Circulation" appears in
the built_Objects list? I think that if I get the number of times /N/
that "Circulation" appears I can multiply the next circulation elevation
/N/ times and avoid having two circulations in the same level. Is this a
correct reasoning?
I did a little research and found that count could help me, so I tried:
print Built_Elements.count('Circulation')
but well is not working. The result is 0, I guess that count is looking
for the exact term and not something similar
If you know the solution or a better way to do this please let me know.
Thanks in advance,
Carlos
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