## Returning the differences between lists of lists

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I have been having difficulty trying to create a method to return the differences between lists of lists. This is basically what I am looking for:

```s1r1=[[1,2,3,4,5,6],[2,4,6,8]]
uniq=[1,3,5,8]
```

I have been trying to use this method

```list(set(s1r1)-set(s1r1))
```

but it has only been returning the number '8'. I then tried using this format

```list(set(a).symmetric_difference(set(b))
```

and it is now saying that I have a syntax error on the last line of my script, even if there is nothing present. Does anyone have any insight?

Using ^ operator for symmetric_difference

```a=set([1,2,3,4,5,6])
b=set([2,4,6,8])
a^b
```

Output

{1, 3, 5, 8}

Python, compute list difference, Use set if you don't care about items order or repetition. Use list comprehensions if you do: >>> def diff(first, second): second = set(second) return [item for item in  How to find the set difference between two lists (LINQ) (C#) 07/20/2015; 2 minutes to read +5; In this article. This example shows how to use LINQ to compare two lists of strings and output those lines that are in names1.txt but not in names2.txt.

This will work for any number of lists:

```s1r1=[[1,2,3,4,5,6],[2,4,6,8]]

uniq = [[j for j in k if not all(j in i for i in s1r1)] for k in s1r1]
```

Yields:

```[[1, 3, 5], ]
```

And to return it as a single list:

```print([j for i in uniq for j in i])
```

Yields:

```[1, 3, 5, 8]
```

Python, Python code t get difference of two lists. # Using set(). def Diff(li1, li2):. return ( list ( set (li1) - set (li2))). # Driver Code. li1 = [ 10 , 15 , 20 , 25 , 30 , 35 , 40 ]. The image above demonstrates an array formula in cell B11 that extracts values that only exist in List 1 (B3:B7) and not in List 2 (D3:D8). The same formula is used in cell B15, however, with different cell references.

You don't need to cast to `list` to do a symmetric difference (disregard the `In:` `Out:` they're from my iPython terminal)

```In : s1r1 = [[1,2,3,4,5,6], [2,4,6,8]]

In : a = set(s1r1)

In : b = set(s1r1)

In : ans = a.union(b) - a.intersection(b)

In : print ans
set([8, 1, 3, 5])

In : ans = a.symmetric_difference(b)

In : print ans
set([1, 3, 5, 8])
```

Symmetric difference is just the opposite of an intersection

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```>>> a = set(s1r1)
>>> b = set(s1r1)
>>> uniq = a.union(b) - a.intersection(b)
>>> uniq
{8, 1, 3, 5}
```

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```>>> s1r1 = [[1,2,3,4,5,6],[2,4,6,8]]
>>> list(set(s1r1).symmetric_difference(s1r1))
['1', '5', '8', '3']
```

Alternatively you the `^` operator for symmetric difference.

```>>> list (set(s1r1) ^ set(s1r1))
[1, 3, 5, 8]
```

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