Reputation: 329
I need to find the identification number of a big number of files while iterating throught them.
The file names are loaded onto a list and look like:
ID322198.nii
ID9828731.nii
ID23890.nii
FILEID988312.nii
So the best way to approach this would be to find the number that sits between ID
and .nii
Because number of digits varies I can't simply select [-10:-4]
of thee file name. Any ideas?
Upvotes: 0
Views: 76
Reputation: 288
for name in files:
name = name.replace('.nii', '')
id_num = name.replace(name.rstrip('0123456789'), '')
How this works:
# example
name = 'ID322198.nii'
# remove '.nii'. -> name1 = 'ID322198'
name1 = name.replace('.nii', '')
# strip all digits from the end. -> name2 = 'ID'
name2 = name1.rstrip('0123456789')
# remove 'ID' from 'ID322198'. -> id_num = '322198'
id_num = name1.replace(name2, '')
Upvotes: 0
Reputation: 615
to find the position of ID
and .nii
, you can use python's index()
function
for line in file:
idpos =
nilpos =
data =
or as a list of ints:
[ int(line[line.index("ID")+1:line.index(".nii")]) for line in file ]
Upvotes: 1
Reputation: 500
It seems like you could filter the digits out, like this:
digits = ''.join(d for d in filename if d.isdigit())
That will work nicely as long as there are no other digits in the filename (e.g backups with a .1 suffix or something).
Upvotes: 0
Reputation: 177
Using rindex
:
s = 'ID322198.nii'
s = s[s.rindex('D')+1 : s.rindex('.')]
print(s)
Returns:
322198
Then apply this sintax to a list of strings.
Upvotes: 0