导入函数参数numpy

时间:2019-02-06 12:32:59

标签: python numpy genfromtxt

我正在尝试导入文本文件(.xyz),该文件如下所示:

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
#

文本文件具有一个标头(前11行),该标头包含一些如下所示的数值,该数据也分为三列,其中一列具有数值,但也包含文字:数据”。我还想将数值“ 0”更改为“无数据”。

我可以跳过Header,但是主要的问题是我要告诉代码有三列,并且“没有数据”的位置表示0。 这是我到目前为止所使用的,

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)

2 个答案:

答案 0 :(得分:2)

您可以添加invalid_raise = False来跳过有问题的行或usecols=np.arange(0, 3),但是我将采用以下方法:

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:

要删除文件的最后一行,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']]

答案 1 :(得分:0)

这是另一种方法。首先,读取所有行,并将每一行放入列表的元素中。这全部由readlines()完成。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以形成单个字符串。由该字符串组成一个numpy数组,并将其重整为正确的格式

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.]]