I have a list of path like ['aaa/aaa.csv', 'aaa/bbb.csv', 'aaa/ccc.csv'].
How can it be converted to dictionary like {'aaa':['aaa.csv', 'bbb.csv', 'ccc.csv'] and so on with first folder in path is equal to others?
I tried this code, but got confused what to do next.
list_split = [i.split('/') for i in list]
dic = {}
list_temp = []
for item in list_split:
list_temp.append(item)
if len(list_temp) < 2:
pass
else:
for itemm in list_temp:
pass
dic = {}
lst = ['aaa/aaa.csv', 'aaa/bbb.csv', 'aaa/ccc.csv']
for item in lst:
slash = item.find('/')
key = item[:slash]
val = item[slash+1:]
if dic.has_key(key):
dic[key].append(val)
else:
dic[key] = [val]
>>> dic
{'aaa': ['aaa.csv', 'bbb.csv', 'ccc.csv']}
for key,val in [x.split("/") for x in lst]:original_list = ['aaa/aaa.csv', 'aaa/bbb.csv', 'aaa/ccc.csv', 'x/1.csv', 'y/2.csv'] # i added a couple more items to (hopefully) prove it works
dic = {}
for item in original_list:
path = item.split('/')
if path[0] not in dic:
dic[path[0]] = []
dic[path[0]].append('/'.join(path[1:]))
You can try this:
>>> L = ['aaa/aaa.csv', 'aaa/bbb.csv', 'aaa/ccc.csv']
>>> list_split = [tuple(i.split('/')) for i in L]
>>> newdict = {}
>>> for (key,item) in list_split:
if key not in newdict:
newdict[key]=[]
newdict[key].append(item)
Output:
{'aaa': ['aaa.csv', 'bbb.csv', 'ccc.csv']}
You can also use a defaultdict for this:
from collections import defaultdict
paths = ['aaa/aaa.csv', 'aaa/bbb.csv', 'aaa/ccc.csv', 'bbb/ccc.csv', 'aaa/bbb/ccc.csv']
dict = defaultdict(list)
for *key, value in [x.split('/') for x in paths]:
dict['/'.join(key)].append(value)
print(dict.items())
print(dict['aaa'])
This will also work for nested directories by putting the full path as the key.
path_dict = {}
for path in path_list:
if '/' in path:
path_dict[path.split('/')[0]] = path.split('/')[1:]
split, you may run into trouble if you encounter multiply nested directories, e.g./aaa/bbb/ccc.csv