---
title: 'Python Challenge Day 10 - Total amount of points
'
disqus: hackmd
---

## Python Challenge Day 10 - Total amount of points
### DESCRIPTION

Our football team has finished the championship.

Our team's match results are recorded in a collection of strings. Each match is represented by a string in the format "x:y", where x is our team's score and y is our opponents score.

For example: ["3:1", "2:2", "0:1", ...]

Points are awarded for each match as follows:

if x > y: 3 points (win)
if x < y: 0 points (loss)
if x = y: 1 point (tie)
We need to write a function that takes this collection and returns the number of points our team (x) got in the championship by the rules given above.

### Notes:

our team always plays 10 matches in the championship
0 <= x <= 4
0 <= y <= 4

程式邏輯
---
目的：計算積分數
x 是球隊的得分，y 是對手的得分。
例如：["3:1", "2:2", "0:1", ...]

每場比賽的得分如下：

如果 x > y：3 分（獲勝）
如果 x < y：0 分（損失）
如果 x = y：1 分（平局）
我們需要編寫一個函數來接受這個集合併返回我們的團隊 (x) 根據上面給出的規則在錦標賽中獲得的積分數，且在錦標賽中打了 10 場比賽。
0 <= x <= 4
0 <= y <= 4

### Practice 1:

def points(games):
    count = 0
    for i in games:
        if int(i[0]) > int(i[2]):
            count += 3
        elif int(i[0]) == int(i[2]):
            count += 1
    return count

* 計算總積分，需要定義一個元素去放積分的加總 count=0
* 用for迴圈計算; for 和 in 是 Python 的關鍵字，這兩個關鍵字之間會放我們自訂的變數，而 in的後面則可接一個序列 (Sequence)，迴圈會依序從序列 (例如: 一連串清單資料，一組數字或一個名單) 裡取得元素，並將元素指派給前面自訂的變數，然後執行迴圈裡的內容。
例如：
["3:1"]
["元素0是3,元素1是：,元素2是1"]

### Practice 2:

def points(games):
    count = 0
    for score in games:
        res = score.split(':')
        if res[0]>res[1]:
            count += 3
        elif res[0] == res[1]:
            count += 1
    return count

* 用split分隔':'，讓比分成為各兩個元素

## 參考資源

:::info
* [Kata](https://www.codewars.com/kata/5bb904724c47249b10000131)
:::