问需要从python3中的文本文件中提取表格数据
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Stack Overflow用户

提问于 2019-01-16 18:30:50

回答 3查看 538关注 0票数 1

我有一个量子化学程序的输出，我希望从中提取表格数据，以便输入到我大约25年前编写的FORTRAN程序的Python端口中。

有些输出文件相当长，多达6000行，这就排除了使用电子表格进行处理的可能性。

典型的表格形式如下：

                             CARTESIAN COORDINATES

   1    C        0.011987266    -0.003842185     0.006578784
   2    H        1.097152909    -0.003956163     0.013339310
   3    H       -0.349612312     1.019316731     0.001903075
   4    H       -0.344276148    -0.517463019    -0.880495291
   5    H       -0.355315644    -0.513266496     0.891567896

我并不是要求有人为我编写Python代码，而是在迷宫般的可用代码中给我一些指导。

python-3.x

extract

tabular

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回答 3

Stack Overflow用户

回答已采纳

发布于 2019-01-16 19:02:12

我建议你调查一下np.genfromtxt。下面的代码片段将从存储在名为data.txt的文件中的问题中读取示例数据。

import numpy as np
data = np.genfromtxt('data.txt', skip_header=2, dtype=[('id', 'i8'),('label','S1'),('x','f8'),('y','f8'),('z','f8')])
print(data)

输出

 [(1, b'C',  0.01198727, -0.00384219,  0.00657878)
 (2, b'H',  1.09715291, -0.00395616,  0.01333931)
 (3, b'H', -0.34961231,  1.01931673,  0.00190307)
 (4, b'H', -0.34427615, -0.51746302, -0.88049529)
 (5, b'H', -0.35531564, -0.5132665 ,  0.8915679 )]

票数 0

Stack Overflow用户

发布于 2019-01-16 20:49:41

我会使用readline和split。

cc = 'CARTESIAN_COORDINATES.txt'

with open(cc) as data:
    lines = data.readlines()[2:] # skip first two lines
    for line in lines:
        ls = line.split()
        a, b, c, d, e = int(ls[0]), ls[1], float(ls[2]), float(ls[3]), float(ls[4])
        print(a, b, c, d, e)

输出：

1 C 0.011987266 -0.003842185 0.006578784
2 H 1.097152909 -0.003956163 0.01333931
3 H -0.349612312 1.019316731 0.001903075
4 H -0.344276148 -0.517463019 -0.880495291
5 H -0.355315644 -0.513266496 0.891567896

票数 1

Stack Overflow用户

发布于 2019-01-16 19:13:08

正则表达式是为从数据中提取数据而构建-如果您的表总是定义良好，则可以使用f.e.：https://regex101.com/r/QUT2o3/2提取它们

import re

regex = r"(\d+ +\w+ (?: +-?\d+\.\d+){3}.+?(?:\n|\Z){2})+"

test_str = ("                      CARTESIAN COORDINATES\n\n"
    "   1    C        0.011987266    -0.003842185     0.006578784\n"
    "   2    H        1.097152909    -0.003956163     0.013339310\n"
    "   3    H       -0.349612312     1.019316731     0.001903075\n"
    "   4    H       -0.344276148    -0.517463019    -0.880495291\n"
    "   5    H       -0.355315644    -0.513266496     0.891567896\n\n\n\n"
    "                      CARTESIAN COORDINATES\n\n"
    "   1    C        0.011987266    -0.003842185     0.006578784\n"
    "   2    H        1.097152909    -0.003956163     0.013339310\n"
    "   3    H       -0.349612312     1.019316731     0.001903075\n"
    "   4    H       -0.344276148    -0.517463019    -0.880495291\n"
    "   5    H       -0.355315644    -0.513266496     0.891567896\n\n\n"
    "                      CARTESIAN COORDINATES\n\n"
    "   1    C        0.011987266    -0.003842185     0.006578784\n"
    "   2    H        1.097152909    -0.003956163     0.013339310\n"
    "   3    H       -0.349612312     1.019316731     0.001903075\n"
    "   4    H       -0.344276148    -0.517463019    -0.880495291\n"
    "   5    H       -0.355315644    -0.513266496     0.891567896")

应用正则表达式：

matches = re.findall(regex, test_str, re.MULTILINE | re.DOTALL)
for m in matches:
    print('\n'.join(x.strip() for x in m.splitlines()))

输出：

1    C        0.011987266    -0.003842185     0.006578784
2    H        1.097152909    -0.003956163     0.013339310
3    H       -0.349612312     1.019316731     0.001903075
4    H       -0.344276148    -0.517463019    -0.880495291
5    H       -0.355315644    -0.513266496     0.891567896

1    C        0.011987266    -0.003842185     0.006578784
2    H        1.097152909    -0.003956163     0.013339310
3    H       -0.349612312     1.019316731     0.001903075
4    H       -0.344276148    -0.517463019    -0.880495291
5    H       -0.355315644    -0.513266496     0.891567896

1    C        0.011987266    -0.003842185     0.006578784
2    H        1.097152909    -0.003956163     0.013339310
3    H       -0.349612312     1.019316731     0.001903075
4    H       -0.344276148    -0.517463019    -0.880495291
5    H       -0.355315644    -0.513266496     0.891567896