如何让它在同一行打印输出？

Question

我正在尝试打开两个文件并收集 2010 年至 2019 年的信息，并仅报告当年索赔的均值和标准差

mean_file = open('data/mean.txt', 'r')
std_file = open('data/std.txt', 'r')
count = 2009

for line in std_file:
    std = float(line)
    for line in mean_file:
        mean = float(line)
    count += 1
    print('Year',count, 'Mean:', mean, 'Standard Deviation:', std)

mean_file.close()
std_file.close()

我有上面的代码。 我得到的output是

Year 2019 Mean: 217557.4 Standard Deviation: 82296.33
Year 2019 Mean: 217557.4 Standard Deviation: 77808.0
Year 2019 Mean: 217557.4 Standard Deviation: 66939.77
Year 2019 Mean: 217557.4 Standard Deviation: 65486.56
Year 2019 Mean: 217557.4 Standard Deviation: 59126.12
Year 2019 Mean: 217557.4 Standard Deviation: 58712.14
Year 2019 Mean: 217557.4 Standard Deviation: 55465.54
Year 2019 Mean: 217557.4 Standard Deviation: 44621.54
Year 2019 Mean: 217557.4 Standard Deviation: 47821.1
Year 2019 Mean: 217557.4 Standard Deviation: 43170.7

每次我更改缩进的 position 时，它都会给出不同的答案。 我希望平均值和标准偏差同时打印在同一行上，就像下面的 output 一样。 我希望 output 如下所示。 我如何让它像下面的 output 一样打印出来？

Year 2010 Mean: 455692.98 Standard Deviation: 82296.33
Year 2011 Mean: 409110.4 Standard Deviation: 77808.0
Year 2012 Mean: 372226.67 Standard Deviation: 66939.77
Year 2013 Mean: 341826.79 Standard Deviation: 65486.56
Year 2014 Mean: 306567.67 Standard Deviation: 59126.12
Year 2015 Mean: 276956.5 Standard Deviation: 58712.14
Year 2016 Mean: 263900.21 Standard Deviation: 55465.54
Year 2017 Mean: 243116.25 Standard Deviation: 44621.54
Year 2018 Mean: 220894.98 Standard Deviation: 47821.1
Year 2019 Mean: 217557.4 Standard Deviation: 43170.7

Answer 1

编辑：抱歉，犯了一个错误。 我知道你是从哪里来的。 除了缩进之外，以下所有要点仍然适用。 要修复此错误，您应该使用zip ，它允许您一次迭代这两个文件：

with open('std.txt', 'r') as std_file:
    with open('means.txt', 'r') as mean_file:
        for line in zip(std_file,mean_file):
            std = float(line[0])
            mean = float(line[1])
            count += 1
            print('Year',count, 'Mean:', mean, 'Standard Deviation:', std)

这给了我们：

Year 2010 Mean: 455692.98 Standard Deviation: 82296.33
Year 2011 Mean: 409110.4 Standard Deviation: 77808.0
Year 2012 Mean: 372226.67 Standard Deviation: 66939.77
Year 2013 Mean: 341826.79 Standard Deviation: 65486.56
Year 2014 Mean: 306567.67 Standard Deviation: 59126.12
Year 2015 Mean: 276956.5 Standard Deviation: 58712.14
Year 2016 Mean: 263900.21 Standard Deviation: 55465.54
Year 2017 Mean: 243116.25 Standard Deviation: 44621.54
Year 2018 Mean: 220894.98 Standard Deviation: 47821.1
Year 2019 Mean: 217557.4 Standard Deviation: 43170.7

---

一些可以使您的代码更易于阅读并因此更易于调试的快速提示：使用上下文管理器（即with语句），并且不要在嵌套的 for 循环中重复使用迭代变量。

使用上下文管理器自动打开和关闭文件，通常更安全。 你可以这样做：

with open('data/std.txt', 'r') as std_file:
    with open('data/mean.txt', 'r') as mean_file:
        #do stuff

此外，在两个 for 循环中重复使用line不仅是不好的做法，而且将不可避免地导致程序重写您仍然需要的变量。 最后，你的缩进是错误的：你只是在迭代所有的平均文件后打印，因此对于平均来说总是会得到相同的 output。 把这一切放在一起，我们有。

count = 2009
with open('data/std.txt', 'r') as std_file:
    with open('data/mean.txt', 'r') as mean_file:
        for line_std in std_file:
            std = float(line_std)
            for line_mean in mean_file:
                mean = float(line_mean)
                count += 1
                print('Year',count, 'Mean:', mean, 'Standard Deviation:', std)

哪个应该按预期工作。 参见上面的固定代码

Answer 2

for sline, mline in zip(std_file, mean_file):
   std = float(line)
   mean = float(line)
   count = count+1
   print('Year',count, 'Mean:', mean, 'Std:', std)