Question

我使用Python来解析来自以下csv文件的数据 -

{::[name]str1_str2_str3[0]},1,U0.00 - Sensor1 Not Ready\nTry Again,1,0,12

{::[name]str1_str2_str3[0]},2,U0.01 - Sensor2 Not Ready\nTry Again,1,0,12

{::[name]str1_str2_str3[0]},3,U0.02 - \n,1,0,12

{::[name]str1_str2_str3[0]},4,U0.03 - Sensor4 Not Ready\nTry Again,1,0,12

{::[name]str1_str2_str3[0]},5,U0.04 - \n,1,0,12

在column1中，我正在解析0中的值[ ]。然后是column2和column3中的值，我正在解析子字符串“Sensor1 Not Ready”，然后按如下方式打印到另一个文件 -

SENSOR1_NOT_READY 0,1
SENSOR2_NOT_READY 0,2

依旧......

现在，当我打印解析后的值时，我得到以下内容 -

SENSOR1_NOT_READY 0,1
SENSOR2_NOT_READY 0,2
SENSOR2_NOT_READY 0,3
SENSOR4_NOT_READY 0,4
SENSOR4_NOT_READY 0,5

我想跳过第3列中没有数据的行（例如 - 3文件中的行5和csv。我该怎么做？

预期产出 -

SENSOR1_NOT_READY 0,1
SENSOR2_NOT_READY 0,2
SENSOR4_NOT_READY 0,4

以下是我的Python脚本 -

with open('filename.csv','rb') as csvfile:
    reader = csv.reader(csvfile)
    for row in reader:
        tag_name = row[0]
        bit_num = row[1]
        error_name = row[2]

        # Regular expressions
        term0 = '\[(\d)\].*'
        term1 = '(\d+)'
        term2 = r'.*-\s([\w\s]+)\\n'

        capture0 = list(re.search(term0, tag_name).groups())
        capture1 = list(re.search(term1, bit_num).groups())
        temp = re.search(term2, error_name)
        if temp:
            result = list(temp.groups())
        else:
            None
        result[-1] = '_'.join(result[-1].split()).upper()
        capture2 = ','.join(result)

        tp = (capture0[0], capture1[0], capture2) # Tuple
        f.write('{2} {0},{1},\n'.format(tp[0], tp[1], tp[2]))

Answer 1

构建一个搜索“正常”行的正则表达式。也许像"^U0.0[1-5] - \n$"这样的东西？然后在打印错误之前使用类似if not re.search(x):的内容。

with open('filename.csv','rb') as csvfile:
    reader = csv.reader(csvfile)
    for row in reader:
        tag_name = row[0]
        bit_num = row[1]
        error_name = row[2]

        # Regular expressions
        term0 = '\[(\d)\].*'
        term1 = '(\d+)'
        term2 = r'.*-\s([\w\s]+)\\n'
        term3 = '^U0.0[1-5] - \n$'

        capture0 = list(re.search(term0, tag_name).groups())
        capture1 = list(re.search(term1, bit_num).groups())
        temp = re.search(term2, error_name)
        if temp:
            result = list(temp.groups())
        else:
            None
        result[-1] = '_'.join(result[-1].split()).upper()
        capture2 = ','.join(result)

        tp = (capture0[0], capture1[0], capture2) # Tuple
        if not re.search(temp3,error_name):
            f.write('{2} {0},{1},\n'.format(tp[0], tp[1], tp[2])) #I assume this is the print line?

将解析后的数据从csv文件打印到新文件

1 个答案: