如何用Python格式化输出？

Question

我很难用Python格式化一些代码： 我的代码在这里：

keys = ['(Lag)=(\d+\.?\d*)','\t','(Autocorrelation Index): (\d+\.?\d*)',       '(Autocorrelation Index): (\d+\.?\d*)',     '(Semivariance): (\d+\.?\d*)']

import re
string1 = ''.join(open("dummy.txt").readlines())
found = []
for key in keys:
found.extend(re.findall(key, string1))
for result in found:
    print '%s  =  %s' % (result[0],result[1])
raw_input()

到目前为止，我得到的输出是：

Lag = 1

Lag = 2

Lag = 3

Autocorrelation Index = #value

......

......

Semivariance = #value

但我想要的输出是：

^{pr2}$

如果在一个CSV文件或一个txt文件中可以输出，那就太好了！在

我认为这是一种输出循环的方法，但我对循环不是很在行。在

我的更新代码（旧版本）

基于@mutzmatron答案

keys = ['(Lag)=(\d+\.?\d*)',
    '(Autocorrelation Index): (\d+\.?\d*)',
    '(Semivariance): (\d+\.?\d*)']

import re
string1 = open("dummy.txt").readlines().join()
found = []
for key in keys:
    found.extend(re.findall(key, string1))
raw_input()
for result in found:
    print '%s  =  %s' % (result[0], result[1])

raw_input()

尚未编译！我使用的是IDLE python2.6，不知道错误消息，因为我不知道提示符中的pause命令！在

原始问题

我对python完全陌生，有一个问题。我正在处理一个大的文本文件。 以下是其中的一个片段：

Band: WDRVI20((0.2*b4-b3)/((0.2*b4)+b3))
Basic Statistics:
  Min: -0.963805
  Max: 0.658219
  Mean: 0.094306
  Standard Deviation: 0.131797
Spatial Statistics, ***Lag=1***:
  Total Number of Observations (Pixels): 769995
  Number of Neighboring Pairs: 1538146
  Moran's I:
    ***Autocorrelation Index: 0.8482564597***
    Expected Value, if band is uncorrelated: -0.000001
    Standard Deviation of Expected Value (Normalized): 0.000806
    Standard Deviation of Expected Value (Randomized): 0.000806
    Z Significance Test (Normalized): 1052.029088
    Z Significance Test (Randomized): 1052.034915
  Geary's C:
    ***Autocorrelation Index: 0.1517324729***
    Expected Value, if band is uncorrelated: 1.000000
    Standard Deviation of Expected Value (Normalized): 0.000807
    Standard Deviation of Expected Value (Randomized): 0.000809
    Z Significance Test (Normalized): 1051.414163
    Z Significance Test (Randomized): 1048.752451
  ***Semivariance: 0.0026356529***
Spatial Statistics, Lag=2:
  Total Number of Observations (Pixels): 769995
  Number of Neighboring Pairs: 3068924
  Moran's I:
 Autocorrelation Index: 0.6230691635
   Expected Value, if band is uncorrelated: -0.000001
   Standard Deviation of Expected Value (Normalized): 0.000571
   Standard Deviation of Expected Value (Randomized): 0.000571
 Z Significance Test (Normalized): 1091.521976
 Z Significance Test (Randomized): 1091.528022
  Geary's C:
Autocorrelation Index: 0.3769372504
  Expected Value, if band is uncorrelated: 1.000000
  Standard Deviation of Expected Value (Normalized): 0.000574
  Standard Deviation of Expected Value (Randomized): 0.000587
 Z Significance Test (Normalized): 1085.700399
 Z Significance Test (Randomized): 1061.931158
Semivariance: 0.0065475488

我需要提取星***值之间的信息（例如：Autocorrelation Index，Semivariance值）并进行处理，可能会将其写入不同的文本文件或excel文件。我可以吗？我们将非常感谢您的帮助。在

Answer 1

填充要查找的键列表（regular expressions）。例如

keys = ['(Lag)=(\d+\.?\d*)',
        '(Autocorrelation Index): (\d+\.?\d*)',
        '(Semivariance): (\d+\.?\d*)']

然后用正则表达式来搜索它们

^{pr2}$

然后你应该有一个你想要的条目列表，用它你可以做你下一步需要做的事情！在

结果：

Lag  =  1
Autocorrelation Index  =  0.8482564597
Autocorrelation Index  =  0.1517324729
Semivariance  =  0.0026356529

CSV

要输出到CSV，请使用csv模块

import csv
outfile = open('fileout.csv', 'w')
wrt = csv.writer(outfile)
wrt.writerows(found)
outfile.close()

Answer 2

为了按节格式化数据，可能最简单的方法是对数据段进行如下处理

keys =['(Lag)=(\d+\.?\d*)',
    '(Autocorrelation Index): (\d+\.?\d*)',
    '(Semivariance): (\d+\.?\d*)']

import re
string1 = ''.join(open("dummy.txt").readlines())

sections = string1.split('Spatial Statistics')

output = []
heads = []

for isec, sec in enumerate(sections):
    found = []
    output.append([])
    for key in keys:
        found.extend(re.findall(key, sec))
    for result in found:
        print '%s  =  %s' % (result[0],result[1])
        output[-1].append(result[1])
    if len(found) > 0 & len(heads) == 0:
        heads = [result[0] for result in found]    

fout = open('output.csv', 'w')
wrt = csv.writer(fout)
wrt.writerow(heads)
wrt.writerows(outputs)
fout.close()