First of all, if sorting a dictionary by value is the frequent operation then you should consider using different data structure. But if your use case is to access data faster then dictionary is useful.

To sort a dictionary by value, you can use `sorted()`

function. Here are some of the implementations –

**Code Example 1:**

```
>>> x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
>>> {k: v for k, v in sorted(x.items(), key=lambda item: item[1])}
{0: 0, 2: 1, 1: 2, 4: 3, 3: 4}
```

**Code Example 2:**

```
>>> x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
>>> dict(sorted(x.items(), key=lambda item: item[1]))
{0: 0, 2: 1, 1: 2, 4: 3, 3: 4}
```

**Code Example 3:**

```
>>> x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
>>> L = [(k,v) for (k,v) in x.items()]
>>> sorted(L, key=lambda n: n[1])
```

**Code Example 4:**

```
>>> x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
>>> L = ((k,v) for (k,v) in x.items())
>>> sorted(L, key=lambda n: n[1])
```

**Code Example 5:**

```
>>> x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
>>> def sk(n): return n[1]
>>> sorted(x.items(), key=sk)
```

**Code Example 6:**

```
>>> from operator import itemgetter
>>> sorted(x.items(), key=itemgetter(1))
```

*Code Example 6 * is the best in terms of execution time. Check out live demo to get execution time of all 6 approaches.

## Live Demo

Demo might not work in private window (incognito)

## Comments