First of all, I apologize for the title of the question, which is not clarifying.
I have a list with the names of several files in a folder, of the type: file1_A1, file1_A2, file2_A1, file2_A2, ...
How can I create two lists from this list, one containing all the elements finished with _A1 and one containing all the _A2?
Thank you!
If we assume that what you have is a list of chains in this way:
nombres = ['archivo1_A1', 'archivo1_A2', 'archivo2_A1', 'archvio2_A2']
The simplest thing is to use str.endswith :
lista_A1 = [nombre for nombre in nombres if nombre.endswith('_A1')]
lista_A2 = [nombre for nombre in nombres if nombre.endswith('_A2')]
If all your files end in '_A1' or '_A2' it could be done in a single for .
If you need something more complex because you work with more complex file names, for example, with extension ('file1_A1.py', 'file1_A1.exe', etc) it would be best to use regular expressions.
Surely there are better ways, for now it occurs to me is to use list comprehension, a very powerful language technique that allows to transform / filter any list quickly:
lista = [ "archivo1_A1",
"archivo1_A2",
"archivo2_A1",
"archvio2_A2"
]
a1 = [a for a in lista if a[-2:] == "A1" ]
a2 = [a for a in lista if a[-2:] == "A2" ]
print("Archivos A1 = {}".format(a1))
print("Archivos A2 = {}".format(a2))
And the exit:
Archivos A1 = ['archivo1_A1', 'archivo2_A1']
Archivos A2 = ['archivo1_A2', 'archvio2_A2']
The structure:
[a for a in lista if a[-2:] == "A1" ]
What it says is:
[]
a of the original list a for a in lista
A1 ( if a[-2:] == "A1" )