TypeError: & # 39; str # 39;在python中不支持项赋值

Question

我正在进行数据分析，因为我想导航并显示时间，我使用codeskulptor(python)，我使用以下代码导航:

def keydown(key):
    global season, year, navtime
    if key == 37:
        navtime += 1
        season[2] = str(int(season[2]) - 3) # error
        if int(season[0] - 3) <= 0:
            year = str(int(year) - 1)
            season = '10-12' 
        else:
            season[0] = str(int(season[0] - 3))
    if key == 39:
        navtime -= 1
        season[2] = str(int(season[2]) + 3) # error
        if int(season[0] + 3) >= 12:
            year = str(int(year) + 1)
            season = '1-3'
        else:
            season[0] = str(int(season[0] + 3))

我之前已经定义了所有的变量，我提出了错误:TypeError: 'str' does not support item assignment在python中。我该如何解决这个问题?

我在这个项目中使用simplegui模块

Answer 1

将变量season设置为字符串:

season = '1-3'

，然后尝试分配给特定的索引:

season[2] = str(int(season[2]) - 3)

你会得到这个错误，因为字符串对象是不可变的

如果你想替换字符串中的字符，你需要构建一个新的字符串对象:

season = season[:-1] + str(int(season[2]) - 3)

替换最后一个字符和

season = str(int(season[0] - 3)) + season[1:]

替换第一个

也许你应该让season的列表的两个值:

season = [1, 3]