Question

这是我到目前为止所拥有的。

from os import path
dirt = path.dirname(__file__)
#########################################################
def save(hs):
    f = open(path.join(dirt, "testscores.txt"), "w")
    for i in range(3):
        f.write(str(hs[i]))
        if i != len(hs):
            f.write("\n")
    f.close()
def load():
    global hs
    f = open(path.join(dirt, "testscores.txt"), "r")
    hs = f.read().splitlines()
    f.close()
#########################################################
try:
    load()
except:
    hs = [5000, 2000, 300]
hst = int(input("score?"))
hs.append(hst)
print(hs)
hs.sort(key = int, reverse = True)
hs = hs[:3]
save(hs)

我一直遇到列表索引超出范围，sort（）无法正常工作或临时高分被写入4次的情况。

Answer 1

列表索引超出范围

说您的"testscores.txt"为空。调用load()时，您会得到hs = []。然后输入1个新分数，您得到hs = [100]。然后调用save()，它会尝试获取显然超出范围的hs[1]和hs[2]。

尝试使用with statement，而不是open和close。也不要使用global hs，而是返回hs

def load():
    with open("file path", "r") as f:
        hs = f.read().splitlines()
    return hs

try:
    hs = load()

对于save()，为避免索引超出范围，您要检查输入列表的长度。 hs = hs[:3]是不必要的。理想情况下，该功能应该能够处理所有输入，因此让该功能进行检查。

def save(hs):
    # How many scores are we saving
    # hm_score equates to 3 if more than 3 scores are in the list
    # equates to 0, 1, or 2 if less than 3 scores are in the list
    hm_score = min(len(hs), 3)

    # Trim hs to 3 if more than 3 scores are in the list
    hs = hs[:hm_score]

    with open("file path", 'w') as f:
        # Join all the scores in hs with a line break
        text = '\n'.join(hs)
        f.write(text)
    # File is closed automatically outside the with statement

sort（）不起作用

应该没错。也可以考虑将sort()放在save()内，以使环境更清洁。

我正在尝试将前三名高分保存到单独行的txt文件中

1 个答案: