我是Python的新手,通常用于Java。我目前正在尝试解析Praat输出的文本文件,该文本文件始终采用相同的格式,并且通常看起来像这样,并具有一些其他功能:
-- Voice report for 53. Sound T1_1001501_vowels --
Date: Tue Aug 7 12:15:41 2018
Time range of SELECTION
From 0 to 0.696562 seconds (duration: 0.696562 seconds)
Pitch:
Median pitch: 212.598 Hz
Mean pitch: 211.571 Hz
Standard deviation: 23.891 Hz
Minimum pitch: 171.685 Hz
Maximum pitch: 265.678 Hz
Pulses:
Number of pulses: 126
Number of periods: 113
Mean period: 4.751119E-3 seconds
Standard deviation of period: 0.539182E-3 seconds
Voicing:
Fraction of locally unvoiced frames: 5.970% (12 / 201)
Number of voice breaks: 1
Degree of voice breaks: 2.692% (0.018751 seconds / 0.696562 seconds)
我想输出如下内容:
0.696562,212.598,211.571,23.891,171.685,265.678,126,113,4.751119E-3,0.539182E-3,5.970,1,2.692
所以从本质上讲,我想从每行中打印出一个仅包含冒号和其后的空格之间的数字的字符串,并用逗号分隔。我知道这可能是一个愚蠢的问题,但是我无法在Python中弄清楚。任何帮助将非常感激!
答案 0 :(得分:1)
好的,这很简单,您需要进行一些调整才能为您工作。
import re
with open("file.txt", "r") as f:
lines = [s.strip() for s in f.readlines()]
numbers_list = []
for _ in lines :
numbers_list.append(re.findall(r'\d+', _))
print(numbers_list)
其中file.txt是您的文件。
答案 1 :(得分:1)
也许:
for line in text.splitlines():
line=line.strip()
head,sepa,tail=line.partition(":")
if sepa:
parts=tail.split(maxsplit=1)
if parts and all( ch.isdigit() or ch in ".eE%-+" for ch in parts[0]):
num=parts[0].replace("%"," ")
try:
print(float(num.strip()))
except ValueError:
print("invalid number:",num)
出局:
0.696562
212.598
211.571
23.891
171.685
265.678
126.0
113.0
0.004751119
0.000539182
5.97
1.0
2.692
答案 2 :(得分:0)
谢谢大家的帮助!我实际上想出了这个解决方案:
import csv
input = 't2_5.txt'
input_name = input[:-4]
def parse(filepath):
data = []
with open(filepath, 'r') as file:
file.readline()
file.readline()
file.readline()
for line in file:
if line[0] == ' ':
start = line.find(':') + 2
end = line.find(' ', start)
if line[end - 1] == '%':
end -= 1
number = line[start:end]
data.append(number)
with open(input_name + '_output.csv', 'wb') as csvfile:
wr = csv.writer(csvfile)
wr.writerow(data)
parse(input)