根据部分字符串的值重新排列/重组字符串列表

Question

I have a list of.csv files that I'm trying to rearrange in a specific way. Let's say my original list looks like this:

["Component 1 - Cable 1.csv", "Component 1 - Cable 2.csv", "Component 2 - Cable 1.csv", "Component 3 -Cable 1.csv"]

I would like the list to look like this instead, but can't figure out how to achieve it:

[["Component 1 - Cable 1.csv", "Component 1 - Cable 2.csv"], ["Component 2 - Cable 1.csv"], ["Component 3 - Cable 1.csv"]]

Answer 1

Use itertools.groupby :

from itertools import groupby

res = []
for _, g in groupby(l, key=lambda x:x.split('-')[0]):
    res.append(list(g))

Or one-liner:

res = [list(g) for _, g in groupby(l, key=lambda x:x.split('-')[0])]

Output of both:

[['Component 1 - Cable 1.csv', 'Component 1 - Cable 2.csv'],
 ['Component 2 - Cable 1.csv'],
 ['Component 3 -Cable 1.csv']]

Answer 2

As an alternative to itertools.groupby you can use a dictionary which may be easier for newbies as follows.

l = ["Component 1 - Cable 1.csv", "Component 1 - Cable 2.csv", "Component 2 - Cable 1.csv", "Component 3 -Cable 1.csv"]

# Option 1: Use of dictionary
newdata = {}
for x in l:
  group =  x.split('-')[0]
  if group not in newdata:  # Add list to dictionary for missing keys
    newdata[group] = []
  newdata[group].append(x)
print (list(newdata.values()))

# Option 2: Simply by using set default to remove conditional key check
newdata = {}
for x in l:
  group = x.split('-')[0]  # group based upon string before hyphen
  newdata.setdefault(group, []).append(x)  # removes need for conditional by providing a default value for missing keys

print (list(newdata.values()))

# Option 3: Simplify further by using default dict
from collections import defaultdict
newdata = defaultdict(list)         # creates dictionary where missing keys return a empty list
for x in l:
  group = x.split('-')[0]  # group based upon string before hyphen
  newdata[group].append(x)  # accumulate groupings together as value
                                           # in dictionary
print (list(newdata.values()))

All three options output

[['Component 1 - Cable 1.csv', 'Component 1 - Cable 2.csv'], ['Component 2 - Cable 1.csv'], ['Component 3 -Cable 1.csv']]

Answer 3

Assuming that you have missed a bracket between first and second elements of the output list and your expected output is:

[['Component 1 - Cable 1.csv'], ['Component 1 - Cable 2.csv'], ['Component 2 - Cable 1.csv'], ['Component 3 -Cable 1.csv']]

you can simply use:

l = ["Component 1 - Cable 1.csv", "Component 1 - Cable 2.csv", "Component 2 - Cable 1.csv", "Component 3 -Cable 1.csv"]
l = [[i] for i in l]

Here i is just some variable and you are applying a function over all elements in the given list.