You can use list comprehension:

the_list = [['blue'], ['blue', 'red', 'black'], ['green', 'yellow'],['orange'], ['white', 'gray']]
filters = ['blue', 'white']
final_l = [i for i in the_list if not any(b in filters for b in i)]

Output:

[['green', 'yellow'], ['orange']]

Or, using filter:

final_l = filter(lambda x:not any(b in filters for b in x), the_list)

The following snippet will print [[(5, 6), (7, 2)]]:

a=[[(1,2),(7,-5),(7,4)],[(5,6),(7,2)],[(8,2),(20,7),(1,4)]]

def f(sub):
  return all(map(lambda (x,y): (0 < x < 10 and 0 < y < 10), sub))

a = filter(f, a)
print a

Import necessary packages and intialize data:

import pandas as pd
import re

my_list = [['Example1 (purple)',
  ' Example2 (blue)',
  ' Example3 (orange)',
  ' Example4 (yellow)',
  ' Example5 (red)',
  ' Example6 (pink)',
  ' Example7 (sky)'],
 ['Example8 (purple)',
  ' Example9 (blue)',
  ' Example10 (orange)',
  ' Example11 (sky)',
  ' Example12 (green)',
  ' Example13 (green)',
  ' Example14 (yellow)',
  ' Example15 (red)',
  ' Example16 (pink)',
  ' Example17 (pink)',
  ' Example18 (green)',
  ' Example19 (sky)']]

Flatten the list, so its not nested lists in lists. (This is why you got an error that lists don't have split. If you do [x.split() for x in my_list] it will give an error because the elements made up of my_list are lists)

Define a flatlist function and flatten list:

flat_list = lambda l: [item for sublist in l for item in sublist]
flat = flat_list(my_list)

Create an empty dataframe

df = pd.DataFrame({})

Extract out the elements of the single flat list. this strips it of whitespace, then splits it by the space, taking the 0th element for the "Example1", and then strips it again to remove whitespace. do it again but take the 1st element for the color. Wrap it in () and separate by a comma to return it as a tuple.

splitout = [(x.strip().split(' ')[0].strip(), x.strip().split(' ')[1]) for x in pd.Series(flat)]

set the two dataframe columns. the first is just grabbing the first element of the splitout which is always Example, the second uses re.sub to remove the () from the color

df['Example'] = [x[0] for x in splitout]
df['Color'] = [re.sub('[/(/)]', '', x[1]) for x in splitout]

      Example   Color
0    Example1  purple
1    Example2    blue
2    Example3  orange
3    Example4  yellow
4    Example5     red
5    Example6    pink
6    Example7     sky
7    Example8  purple
8    Example9    blue
9   Example10  orange
10  Example11     sky
11  Example12   green
12  Example13   green
13  Example14  yellow
14  Example15     red
15  Example16    pink
16  Example17    pink
17  Example18   green
18  Example19     sky

Then you can pivot into a larger dataframe with colors for columns:

pd.pivot_table(df.assign(v=1), index='Example', columns='Color', values='v')

Color      blue  green  orange  pink  purple  red  sky  yellow
Example                                                       
Example1    NaN    NaN     NaN   NaN     1.0  NaN  NaN     NaN
Example10   NaN    NaN     1.0   NaN     NaN  NaN  NaN     NaN
Example11   NaN    NaN     NaN   NaN     NaN  NaN  1.0     NaN
Example12   NaN    1.0     NaN   NaN     NaN  NaN  NaN     NaN
Example13   NaN    1.0     NaN   NaN     NaN  NaN  NaN     NaN
Example14   NaN    NaN     NaN   NaN     NaN  NaN  NaN     1.0
Example15   NaN    NaN     NaN   NaN     NaN  1.0  NaN     NaN
Example16   NaN    NaN     NaN   1.0     NaN  NaN  NaN     NaN
Example17   NaN    NaN     NaN   1.0     NaN  NaN  NaN     NaN
Example18   NaN    1.0     NaN   NaN     NaN  NaN  NaN     NaN
Example19   NaN    NaN     NaN   NaN     NaN  NaN  1.0     NaN
Example2    1.0    NaN     NaN   NaN     NaN  NaN  NaN     NaN
Example3    NaN    NaN     1.0   NaN     NaN  NaN  NaN     NaN
Example4    NaN    NaN     NaN   NaN     NaN  NaN  NaN     1.0
Example5    NaN    NaN     NaN   NaN     NaN  1.0  NaN     NaN
Example6    NaN    NaN     NaN   1.0     NaN  NaN  NaN     NaN
Example7    NaN    NaN     NaN   NaN     NaN  NaN  1.0     NaN
Example8    NaN    NaN     NaN   NaN     1.0  NaN  NaN     NaN
Example9    1.0    NaN     NaN   NaN     NaN  NaN  NaN     NaN

whole code:

import pandas as pd
import re

flat_list = lambda l: [item for sublist in l for item in sublist]
flat = flat_list(my_list)

splitout = [(x.strip().split(' ')[0].strip(), x.strip().split(' ')[1]) for x in pd.Series(flat)]

df = pd.DataFrame({})
df['Example'] = [x[0] for x in splitout]
df['Color'] = [re.sub('[/(/)]', '', x[1]) for x in splitout]

pivot = pd.pivot_table(df.assign(v=1), index='Example', columns='Color', values='v')

I would try something like this:

a = [[1,2,3],[2,3,4],[5,6,7],[7,8,9]]
b = [1,2]

result = [lst for lst in a if any(x in lst for x in b)]

I would suggest using a collections.defaultdict, as a multi-value dictionary:

from collections import defaultdict
d = defaultdict(list)
for x in L:
    d[x[2]].append(x[:2])

Now you can use d['G'] to get what you wanted, but also d['H'] to get the result for 'H'!

Edit: Source append multiple values for one key in Python dictionary

Well, this is not a best practice, but you can create a lambda function: use parenthesis to group conditions:

f = lambda e: filter(None, [f(y) for y in e]) if isinstance(e, list) else (e if e > 10 else None)

my_list = [[1], [2, [3, 12, [4, 11, 12]]], [5, 6, 13, 14], [15]]

>>> f(my_list)
[[[12, [11, 12]]], [13, 14], [15]]

For python 3 users:

f = lambda e: list(filter(None, [f(y) for y in e])) if isinstance(e, list) else (e if e > 10 else None)

I see you have a list and if you don't want to use additional packages like numpy you can do something like this. But it involve looping. Also, it is not modifying the original list.

the_nested = [[51502, 2, 0,  0, 1, 1],
              [8046 ,2 ,0 , 1 ,1 ,2],
              [40701, 1 ,1  ,1 ,1 ,1]]

res = []

_ = [res.append(x[1:]) for x in the_nested]

# Output : 
[[2, 0, 0, 1, 1], [2, 0, 1, 1, 2], [1, 1, 1, 1, 1]] 

You need a nested list comprehension for filtering a nested list.

[[y for y in x if not y.startswith('opt')] for x in inp]