如何讓它在同一行打印輸出？

Question

我正在嘗試打開兩個文件並收集 2010 年至 2019 年的信息，並僅報告當年索賠的均值和標准差

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()

我有上面的代碼。 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

每次我更改縮進的 position 時，它都會給出不同的答案。 我希望平均值和標准偏差同時打印在同一行上，就像下面的 output 一樣。 我希望 output 如下所示。 我如何讓它像下面的 output 一樣打印出來？

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

編輯：抱歉，犯了一個錯誤。 我知道你是從哪里來的。 除了縮進之外，以下所有要點仍然適用。 要修復此錯誤，您應該使用zip ，它允許您一次迭代這兩個文件：

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)

這給了我們：

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

---

一些可以使您的代碼更易於閱讀並因此更易於調試的快速提示：使用上下文管理器（即with語句），並且不要在嵌套的 for 循環中重復使用迭代變量。

使用上下文管理器自動打開和關閉文件，通常更安全。 你可以這樣做：

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

此外，在兩個 for 循環中重復使用line不僅是不好的做法，而且將不可避免地導致程序重寫您仍然需要的變量。 最后，你的縮進是錯誤的：你只是在迭代所有的平均文件后打印，因此對於平均來說總是會得到相同的 output。 把這一切放在一起，我們有。

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)

哪個應該按預期工作。 參見上面的固定代碼

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)