导入函数参数numpy

Question

我正在尝试导入一个文本文件 (.xyz)，这个文件看起来像这样：

1 9 1 6 "Thu Feb 13 13:12:30 2014     "
0 0 0 0 0 0
38 38 915 915
"CJE                                                                              "
"2                                      "
"110321-025-01D-1ST                    
0 0 1 .1 73.7972 17 50
1 0 7 1 60 0 0 0 0
0 "                           "
1 0
#
38 38 No Data
39 38 No Data
40 38 No Data
41 38 3
42 38 No Data
43 38 4
44 38 4
45 38 5
#

文本文件有一个标题（前 11 行），其中包含一些数值，如下所示，数据也分为三列，其中一列有数值，还有写的字符：“No数据”。我还想将数值 0 的“无数据”更改。

我可以跳过标题，但主要问题是我有它告诉代码有三列，并且“没有数据”意味着 0。 这是我到现在用的，

import numpy as np
data = np.genfromtxt('180228_Test V2-4_0grad.xyz',
                 skip_header=11,
                 skip_footer=1,
                 names=True,

                 dtype=None,
                 delimiter=' ')
print(data)

Answer 1

您可以添加invalid_raise = False 以跳过违规行或添加usecols=np.arange(0, 3)，但我会采用以下方法：

list.txt：

1 9 1 6 "Thu Feb 13 13:12:30 2014     "
0 0 0 0 0 0
38 38 915 915
"CJE                                                                              "
"2                                      "
"110321-025-01D-1ST                    
0 0 1 .1 73.7972 17 50
1 0 7 1 60 0 0 0 0
0 "                           "
1 0
#
38 38 No Data
39 38 No Data
40 38 No Data
41 38 3
42 38 No Data
43 38 4
44 38 4
45 38 5

然后：

logFile = "list.txt"

# opening the file
with open(logFile) as f:

    #reading the lines after slicing it i.e. 11
    content = f.readlines()[11:]

# you may also want to remove empty lines
content = [l.strip() for l in content if l.strip()]

# for each line in content
for line in content:

     # if the line has No Data in it
     if line.find("No Data"):

         # Replacing the No Data with 0 using replace() method
         line = line.replace("No Data", "0")
     print(line)

输出：

38 38 0
39 38 0
40 38 0
41 38 3
42 38 0
43 38 4
44 38 4
45 38 5

编辑：

将它们添加到 3 列矩阵中：

_list = []
# for each line in content
for line in content:

     # if the line has No Data in it
     if line.find("No Data"):

         # Replacing the No Data with 0 using replace() method
         line = line.replace("No Data", "0")
     # print(line)
     # list comprehension for splitting on the basis of space and appending to the list
     _list.append([e for e in line.split(' ') if e])

print(_list)

输出：

[['38', '38', '0'], ['39', '38', '0'], ['40', '38', '0'], ['41', '38', '3'],
 ['42', '38', '0'], ['43', '38', '4'], ['44', '38', '4'], ['45', '38', '5']]

编辑 2：

要删除文件中的最后一行，您可以使用切片 content[:-1]::

logFile = "list.txt"

# opening the file
with open(logFile) as f:

    #reading the lines after slicing it i.e. 11
    content = f.readlines()[11:]

_list = []
# for each line in content
for line in content[:-1]:

     # if the line has No Data in it
     if line.find("No Data"):
         # Replacing the No Data with 0 using replace() method
         line = line.replace("No Data", "0")
     # list comprehension for splitting on the basis of space and appending to the list
     _list.append([e for e in line.strip().split(' ') if e])


print(_list)

输出：

[['38', '38', '0'], ['39', '38', '0'], ['40', '38', '0'], ['41', '38', '3'],
 ['42', '38', '0'], ['43', '38', '4'], ['44', '38', '4'], ['45', '38', '5']]

Answer 2

这是一种不同的方法。首先，读取所有行并将每一行放入列表的一个元素中。这都是由 readlines() 完成的。然后，忽略前 11 个句子。然后，对于行列表中的每一行，将“No Data”替换为 0。然后，将所有行粘合在一起形成一个字符串。由这个字符串组成一个 numpy 数组，并重新调整为正确的格式

import numpy as np

#Open the file and read the lines as a list of lines
with open('/home/we4sea/PycharmProjects/Noonreport-processing/GUI/test.txt','r') as f:
    file = f.readlines()

#Skip the first 11 lines
file = file[11:]

#Create new list where the replaced lines are placed
replaced = []

#Replace "No Data" with 0
for line in file:
    replaced.append(line.replace('No Data', '0'))

#Concatenate list to a single string
file = ''.join(replaced)

#Create numpy array from it and reshape to the correct format
data = np.fromstring(file, sep=' ').reshape(-1,3)

#Print the data
print(data)

输出：

[[38. 38.  0.]
 [39. 38.  0.]
 [40. 38.  0.]
 [41. 38.  3.]
 [42. 38.  0.]
 [43. 38.  4.]
 [44. 38.  4.]
 [45. 38.  5.]]