【Ollama】ローカル環境でLLM推論が可能！特徴や使い方を徹底解説

ollama --version
# ollama version is 0.5.7

ollama serve --idle-timeout 600

FROM ./Llama-3-ELYZA-JP-8B-q4_k_m.gguf

TEMPLATE """{{ if .System }}<|start_header_id|>system<|end_header_id|>

{{ .System }}<|eot_id|>{{ end }}{{ if .Prompt }}<|start_header_id|>user<|end_header_id|>

{{ .Prompt }}<|eot_id|>{{ end }}<|start_header_id|>assistant<|end_header_id|>

{{ .Response }}<|eot_id|>"""

PARAMETER stop "<|start_header_id|>"

PARAMETER stop "<|end_header_id|>"

PARAMETER stop "<|eot_id|>"

PARAMETER stop "<|reserved_special_token"

ollama create elyza:jp8b -f Modelfile

ollama run elyza:jp8b

> ollama run elyza:jp8b<br>>>> Send a message (/? for help)

回答
世界の興行収入ランキングトップ10（Box Office Mojo調べ）に基づく情報です。

2023年現在、歴代で最も興行収入が高い映画は「アバター」（2009年）です。

興行収入：27億5600万ドル（約3034億円）

ジェームズ・キャメロン監督作品で、SF大作の金字塔と評されています。3D映像技術を駆使し、壮大な世界観と独特のストーリーが話題になりました。

2021年には、ディズニー映画「スター・ウォーズ／スカイウォーターの夜明け」が興行収入24億ドル超えしてランクインしましたが、「アバター」を超えるには至りませんでした。

回答
最初に宇宙へ行った人は、ユーリ・ガガリーンというソビエト連邦の飛行士です。彼は1961年4月12日、ソ連の宇宙船「ヴォストーク1号」に搭乗し、高度約200kmの地球の大気圏まで到達しました。

この時、ガガーリンは27歳でした。彼は世界は初の有人宇宙飛行を成功させましたが、その直後、機体の問題で大急ぎで帰還する必要があり、地球の重力に適応する時間もなく、僅か9分間という短い時間で地上に戻りました。

この歴史的事件は「ガガーリン宇宙飛行」と呼ばれ、世界中で大きな話題を呼びました。

回答
以下は、Pythonでクイックソートを実装した例です。

def quicksort(arr):

if len(arr) <= 1:

return arr

pivot = arr[len(arr) // 2]

left = [x for x in arr if x < pivot]

middle = [x for x in arr if x == pivot]

right = [x for x in arr if x > pivot]

return quicksort(left) + middle + quicksort(right)

この関数は、クイックソートの基本的なアルゴリズムに従っています。まず、配列の長さが1以下の場合にはそのまま返します。次に、配列の真ん中の要素をピボットとして選びます。左側の要素はピボットより小さいもの、右側の要素はピボットより大きいものに分けます。最後に、左側と右側の部分を再度クイックソートして、結果を連結します。

この関数を使用するには、以下のように呼び出します。

arr = [5, 2, 8, 3, 1, 6, 4]

arr = quicksort(arr)

print(arr)

ollama run deepseek-r1:1.5b

<think>
Alright, the user is asking for a Quicksort implementation in Python. I
need to provide a clear and concise solution.

I'll structure the code step by step, explaining each part so it's easy to
follow.

First, I'll write the function definition with parameters arr and return.

Next, the base case: if the array has one or zero elements, just return it
as is.

Then, I'll choose a pivot. For simplicity, I'm choosing the last element.

After that, partition the array into elements less than pivot, equal to,
and greater than.

I'll handle the equal part recursively by sorting the smaller subarray.

Finally, concatenate the sorted parts: less + equal (sorted) + greater.

I should test it with a simple example to ensure it works.

Make sure to add comments explaining each step so the user understands how
the code functions.
</think>

はい！ Quicksort は要素をative分のwayで、最eadyが選択し、要素が左右に分
classified します。 下の例を参考に writing your own implementation of
QuickSort in Python。

以下は Quicksort 的算法を Python で示した code です：

```python
def quicksort(arr):
    if len(arr) <= 1:
        return arr

    pivot = arr[-1]
    less = [x for x in arr[:-1] if x < pivot]
    greater = [x for x in arr[:-1] if x > pivot]
    equal = [x for x in arr[:-1] if x == pivot]

    equal_sorted = quicksort(equal)
    return quicksort(less) + equal_sorted + quicksort(greater)

# 試行
print(quicksort([3, 6, 8, 10, 1, 2, 1]))
```

### 実際の codedecription:
- **Pivot**: 最eadyの要素を用いています。この例では `arr[-1]` 即然是 `1`。
- **Partition**: 順数が `pivot` 下に分 classified します。
  - 最初から `pivot` 下の要素を `\ : arr[:-1]`
  - 左の要素は `x < pivot`
  - 右の要素は `x > pivot`

### 実際の operation:
- `less` : 探索が `pivot` 下に `$less = [x for x in arr[:-1] if x <
pivot]`
- `greater`: 探索が `pivot` 下に `$greater = [x for x in arr[:-1]
if x > pivot]`

### 等同の要素:
- `[x for x in arr[:-1] if x == pivot]`

### 記号解释:
- 等同の要素を recursively sort し、sorted 形式で再 \% 多数に分 classified
します。
- 最eady `quicksort(less)` と `quicksort(greater)`を回 \ `{return
quicksort(less) + equal_sorted + quicksort(greater)`

### 実際の result:
```
[1, 2, 3, 1, 6, 8, 10]
```

Quicksort は平均 `{O(n log n)}` 时间で実現します。最eadyの要素の候補が `\
: arr[:-1]` 形式で `[x for x in arr[:-1] if x < pivot]`を用いています。

ollama show --modelfile deepseek-r1:1.5b > Modelfile
ls -la

ollama create deepseek-r1-1.5b-jp -f ./Modelfile

ollama run deepseek-r1:1.5b-jp:latest

Prompt：<strong>クイックソートをPythonで書いてください</strong>

<think>
Alright, so I need to write a Python program that can perform a quick
sort. Hmm, quick sort is one of the most efficient sorting algorithms out
there, right? It works by selecting a pivot element and rearranging the
elements such that all elements less than the pivot come before it, and
those greater come after.

First, I should plan how to structure this program. The input will be an
array, so I'll start by reading that from standard input. That's easy
enough with the input() function in Python. But wait, what if the user
doesn't enter anything? Oh right, I should handle that case too, maybe
with a try-except block.

Once the input is read, the core of quick sort is partitioning the array.
So I need to select a pivot and split the array into elements less than or
equal to the pivot and those greater. I'll choose a random element as the
pivot because it makes the algorithm more efficient on average by reducing
the number of comparisons.

Next, I have to recursively apply the same process to the subarrays on
either side of the pivot until the entire array is sorted.

I should think about data types. The array elements can be integers or
maybe strings? Looking back at the user's message, it seems like they want
a numerical sort, so probably integers are intended.

Now, how to handle edge cases? For example, if all elements are the same,
the algorithm should still work fine without issues. Also, for even and
odd numbers of elements, quick sort handles them well as long as the
recursion isn't too deep. But Python's default recursion depth limit might
be a problem for large arrays; I'll need to consider that or switch to an
iterative approach if necessary.

Testing is important. I should write test cases where I can manually check
what the sorted array should look like and ensure my code produces it
correctly.

I also want to make sure that the function returns a new list each time,
so the original input isn't modified. That's already handled by returning
a new array at each step.

Another thing is how the pivot is chosen. For randomization, using the
random module's choice function would be better than just taking the first
element or an index from a range. Maybe something like selecting between 0
and len(arr)-1 with random.randint(0, len(arr)-1).

Wait, but if I'm modifying the list during partitioning, will that affect
subsequent calls? No, because each recursive call operates on its own
subarray, so it's safe.

I should also be cautious about using recursion in a way that might hit
Python's stack limit. If the input array is too long, say over 10^4
elements, this could cause a maximum recursion depth error or slow
performance. But perhaps for most cases, especially with small inputs, it
works fine.

Putting it all together, I'll structure my program like this:

1. Read the input using input() and handle any possible exceptions.
2. Define a recursive function to perform quick sort.
3. In the function:
  a. Select a random pivot index.
  b. Partition the array into elements less than or equal to the pivot
and those greater.
  c. Recursively apply quick sort on the subarrays.
4. Call the function with the input array and print the sorted result.

I think that's a solid plan. Now, I'll translate this into Python code,
making sure everything is in Japanese as per the user's request.
</think>

```
def 快速 SORT:
    # 誰ががとる？
    pivot = random.randrange(len(arr))
    arr[pivot], arr[0] = arr[0], arr[pivot]

    # 接 rounding 选gt;
    # 掛けgt;
    less, greater = [], []
    for num in arr:
        if num <= arr[0]:
            less.append(num)
        elif num > arr[0]:
            greater.append(num)

    return 快速 SORT(less) + [arr[0]] + 快速 SORT(greater)

# 引数を入力
arr = input().strip()
# 定 number が整 数?
try:
    if arr == '':
        print('')
        exit()
except ValueError as e:
    print(e)

# 快速 SORT Done?
print 快速 SORT(arr)
```