﻿ Python: Create set difference - w3resource

Python: Create set difference

Python sets: Exercise-8 with Solution

Write a Python program to create set difference.

Sample Solution-1:

Using difference() Method

Python Code:

``````setc1 = set(["green", "blue"])
setc2 = set(["blue", "yellow"])
print("Original sets:")
print(setc1)
print(setc2)
r1 = setc1.difference(setc2)
print("\nDifference of setc1 - setc2:")
print(r1)
r2 = setc2.difference(setc1)
print("\nDifference of setc2 - setc1:")
print(r2)
setn1 = set([1, 1, 2, 3, 4, 5])
setn2 = set([1, 5, 6, 7, 8, 9])
print("\nOriginal sets:")
print(setn1)
print(setn2)
r1 = setn1.difference(setn2)
print("\nDifference of setn1 - setn2:")
print(r1)
r2 = setn2.difference(setn1)
print("\nDifference of setn2 - setn1:")
print(r2)
```
```

Sample Output:

```Original sets:
{'green', 'blue'}
{'yellow', 'blue'}

Difference of setc1 - setc2:
{'green'}

Difference of setc2 - setc1:
{'yellow'}

Original sets:
{1, 2, 3, 4, 5}
{1, 5, 6, 7, 8, 9}

Difference of setn1 - setn2:
{2, 3, 4}

Difference of setn2 - setn1:
{8, 9, 6, 7}
```

Pictorial Presentation:

Visualize Python code execution:

The following tool visualize what the computer is doing step-by-step as it executes the said program:

Sample Solution-2:

Using - Operator

Python Code:

``````setc1 = set(["green", "blue"])
setc2 = set(["blue", "yellow"])
print("Original sets:")
print(setc1)
print(setc2)
r1 = setc1 - setc2
print("\nDifference of setc1 - setc2:")
print(r1)
r2 = setc2 - setc1
print("\nDifference of setc2 - setc1:")
print(r2)
setn1 = set([1, 1, 2, 3, 4, 5])
setn2 = set([1, 5, 6, 7, 8, 9])
print("\nOriginal sets:")
print(setn1)
print(setn2)
r1 = setn1 - setn2
print("\nDifference of setn1 - setn2:")
print(r1)
r2 = setn2 - setn1
print("\nDifference of setn2 - setn1:")
print(r2)
```
```

Sample Output:

```Original sets:
{'blue', 'green'}
{'yellow', 'blue'}

Difference of setc1 - setc2:
{'green'}

Difference of setc2 - setc1:
{'yellow'}

Original sets:
{1, 2, 3, 4, 5}
{1, 5, 6, 7, 8, 9}

Difference of setn1 - setn2:
{2, 3, 4}

Difference of setn2 - setn1:
{8, 9, 6, 7}
```

Visualize Python code execution:

The following tool visualize what the computer is doing step-by-step as it executes the said program:

Sample Solution-3:

Multiple sets in the difference() method

Python Code:

``````setn1 = set([1, 1, 2, 3, 4, 5])
setn2 = set([1, 5, 6, 7, 8, 9])
setn3 = set([3, 4, 5, 3, 9, 10])
setn4 = set([5, 7, 9, 10, 12, 14])
print("\nOriginal sets:")
print(setn1)
print(setn2)
print(setn3)
print(setn4)
print("\nDifference of setn1 - setn2,setn3,setn4:")
r1 = setn1.difference(setn2,setn3,setn4)
print(r1)
print("\nDifference of setn4 - setn1,setn2,setn3:")
r2 = setn4.difference(setn1,setn2,setn3)
print(r2)
```
```

Sample Output:

```Original sets:
{1, 2, 3, 4, 5}
{1, 5, 6, 7, 8, 9}
{3, 4, 5, 9, 10}
{5, 7, 9, 10, 12, 14}

Difference of setn1 - setn2,setn3,setn4:
{2}

Difference of setn4 - setn1,setn2,setn3:
{12, 14}
```

Visualize Python code execution:

The following tool visualize what the computer is doing step-by-step as it executes the said program:

Python Code Editor:

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Python: Tips of the Day

Find the index of an item in a list?

Given a list ["foo", "bar", "baz"] and an item in the list "bar", how do I get its index (1) in Python?

```>>> ["foo", "bar", "baz"].index("bar")
1
```

Caveats follow

Note that while this is perhaps the cleanest way to answer the question as asked, index is a rather weak component of the list API, and I can't remember the last time I used it in anger. It's been pointed out to me in the comments that because this answer is heavily referenced, it should be made more complete. Some caveats about list.index follow. It is probably worth initially taking a look at the documentation for it:

```list.index(x[, start[, end]])
```

Linear time-complexity in list length

An index call checks every element of the list in order, until it finds a match. If your list is long, and you don't know roughly where in the list it occurs, this search could become a bottleneck. In that case, you should consider a different data structure. Note that if you know roughly where to find the match, you can give index a hint. For instance, in this snippet, l.index(999_999, 999_990, 1_000_000) is roughly five orders of magnitude faster than straight l.index(999_999), because the former only has to search 10 entries, while the latter searches a million:

```>>> import timeit
>>> timeit.timeit('l.index(999_999)', setup='l = list(range(0, 1_000_000))', number=1000)
9.356267921015387
>>> timeit.timeit('l.index(999_999, 999_990, 1_000_000)', setup='l = list(range(0, 1_000_000))', number=1000)
0.0004404920036904514
```

Only returns the index of the first match to its argument

A call to index searches through the list in order until it finds a match, and stops there. If you expect to need indices of more matches, you should use a list comprehension, or generator expression.

```>>> [1, 1].index(1)
0
>>> [i for i, e in enumerate([1, 2, 1]) if e == 1]
[0, 2]
>>> g = (i for i, e in enumerate([1, 2, 1]) if e == 1)
>>> next(g)
0
>>> next(g)
2
```

Most places where I once would have used index, I now use a list comprehension or generator expression because they're more generalizable. So if you're considering reaching for index, take a look at these excellent Python features.

Throws if element not present in list

A call to index results in a ValueError if the item's not present.

```>>> [1, 1].index(2)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: 2 is not in list
```

If the item might not be present in the list, you should either

• Check for it first with item in my_list (clean, readable approach), or
• Wrap the index call in a try/except block which catches ValueError (probably faster, at least when the list to search is long, and the item is usually present.)

Ref: https://bit.ly/2ALwXwe