如何使用python重建和更改数据集的结构?

时间:2019-06-17 00:47:25

标签: python

我有一个数据集,我需要从该数据集重构一些数据为新样式

我的数据集如下所示(存储在名为train1.txt的文件中):

  

2342728,2414939,2397722,2386848,2398737,2367906,2384003,2399896,2359702,2414293,2411228,2416802,2322710,2387437,2397274,2344681,2396522,2386676,2413824,2328225,2413833,2335374,2328594,497 ,2384001,2372746,2386538,2348518,2380037,2374364,2352054,2377990,2367915,2412520,2348070,2356469,2353541,2413446,2391930,2366968,2364762,2347618,2396550,2370538,2393212,2364244、2387901、4752 ,2331890,2341328,2413686,2359209,2342027,2414843,2378401,2367772,2357576,2416791,2398673,2415237,2383922,2371110,2365017,2406357,2383444,2385709,2392694,2378109,2394742,2318516,2354062,2380081 ,2328407,2396727,2316901,2400923,2360206,971,2350695,2341332,2357275,2369945,2325241、2408952、2322395、2415137、2372785、2382132、2323580、2368945、2413009,2348581、2365287、2408766、2382349、2355549, ,2374616,2344619,2362449,2380907,2327352,2347183,2384375,2368019,2365927,2370027,2343649,2415694、233503 5,2389182,2354073,2363977,2346358,2373500,2411328,2348913,2372324,2368727,2323717,2409571,2403981,2353188,2343362,285721,2376836,2368107,2404464,2417233,2382750,2366329,675,2360991,2341475, 2346242,2391969,2345287,2321367,2416019,2343732,2384793,2347111,2332212,138,2342178,2405886,2372686,2365963,2342468

我需要转换为以下样式(我需要将新文件存储为train.txt):

2342728
2414939
2397722
2386848
2398737
2367906
2384003
2399896
2359702
2414293
And other numbers ….

我的python版本是2.7.13 我的操作系统是Ubuntu 14.04 LTS 感谢您的帮助。 非常感谢。

3 个答案:

答案 0 :(得分:1)

我建议使用正则表达式(正则表达式)。这可能有点矫kill过正,但是从长远来看,知道正则表达式非常强大。

import re
def return_no_commas(string):
    regex = r'\d*'
    matches = re.findall(regex, string)
    for match in matches:
        print(match)


numbers = """
2342728, 2414939, 2397722, 2386848, 2398737, 2367906, 2384003, 2399896, 2359702, 2414293, 2411228, 2416802, 2322710, 2387437, 2397274, 2344681, 2396522, 2386676, 2413824, 2328225, 2413833, 2335374, 2328594, 497966, 2384001, 2372746, 2386538, 2348518, 2380037, 2374364, 2352054, 2377990, 2367915, 2412520, 2348070, 2356469, 2353541, 2413446, 2391930, 2366968, 2364762, 2347618, 2396550, 2370538, 2393212, 2364244, 2387901, 4752, 2343855, 2331890, 2341328, 2413686, 2359209, 2342027, 2414843, 2378401, 2367772, 2357576, 2416791, 2398673, 2415237, 2383922, 2371110, 2365017, 2406357, 2383444, 2385709, 2392694, 2378109, 2394742, 2318516, 2354062, 2380081, 2395546, 2328407, 2396727, 2316901, 2400923, 2360206, 971, 2350695, 2341332, 2357275, 2369945, 2325241, 2408952, 2322395, 2415137, 2372785, 2382132, 2323580, 2368945, 2413009, 2348581, 2365287, 2408766, 2382349, 2355549, 2406839, 2374616, 2344619, 2362449, 2380907, 2327352, 2347183, 2384375, 2368019, 2365927, 2370027, 2343649, 2415694, 2335035, 2389182, 2354073, 2363977, 2346358, 2373500, 2411328, 2348913, 2372324, 2368727, 2323717, 2409571, 2403981, 2353188, 2343362, 285721, 2376836, 2368107, 2404464, 2417233, 2382750, 2366329, 675, 2360991, 2341475, 2346242, 2391969, 2345287, 2321367, 2416019, 2343732, 2384793, 2347111, 2332212, 138, 2342178, 2405886, 2372686, 2365963, 2342468
"""

return_no_commas(numbers)

让我解释一下所有功能。

import re

只导入正则表达式。我写的正则表达式是

regex = r'\d*'

开头的“ r”表示这是一个正则表达式,它只查找任何数字(即“ \ d”部分),并表示可以重复任意次数(即“ *”部分)。然后我们打印出所有匹配项。

我将您的数字保存在名为数字的字符串中,但是您可以轻松地在文件中读取并使用这些内容。

您会得到类似的东西:

2342728


2414939


2397722


2386848


2398737


2367906


2384003


2399896


2359702


2414293


2411228


2416802


2322710


2387437


2397274


2344681


2396522


2386676


2413824


2328225


2413833


2335374


2328594


497966


2384001


2372746


2386538


2348518


2380037


2374364


2352054


2377990


2367915


2412520


2348070


2356469


2353541


2413446


2391930


2366968


2364762


2347618


2396550


2370538


2393212

答案 1 :(得分:0)

在我看来,您的原始数据用逗号分隔。但是,您希望数据用换行符(\n)分隔。这很容易做到。

def covert_comma_to_newline(rfilename, wfilename):
    """
    rfilename -- name of file to read-from
    wfilename -- name of file to write-to
    """
    assert(rfilename != wfilename)
    # open two files, one in read-mode
    # the other in write-mode
    rfile = open(rfilename, "r")
    wfile = open(wfilename, "w")

    # read the file into a string
    rstryng = rfile.read()

    lyst = rstryng.split(",")
    # EXAMPLE:
    #     rstryng == "1,2,3,4"
    #     lyst    == ["1", "2", "3", "4"]

    # remove leading and trailing whitespace
    lyst = [s.strip() for s in lyst]

    wstryng = "\n".join(lyst)
    wfile.writelines(wstryng)
    rfile.close()
    wfile.close()
    return


covert_comma_to_newline("train1.txt", "train.txt")
# open and check the contents of `train.txt`

答案 2 :(得分:0)

由于其他人已经添加了答案,因此我将使用numpy来添加答案。 如果您可以使用numpy,那么就很简单:

 data = np.genfromtxt('train1.txt', dtype=int, delimiter=',')

如果要使用列表而不是numpy数组,则

data.tolist()

[2342728,
 2414939,
 2397722,
 2386848,
 2398737,
 2367906,
 2384003,
 2399896,
 ....
]