将str转换为浮动python
将str转换为浮动python
提问于 2018-05-15 22:22:01
为什么会抛出一个错误：ValueError: could not convert string to float:
frequencies.append(float(l[start+1:stop1].strip()))
losses.append(float(l[stop1+5:stop2].strip()))float()命令不会将值解析为float类型吗？我哪里错了？频率和损耗都是列表
代码如下：
def Capture():
impedance = 0
losses = []
frequencies = []
Xtalk = []
start = 0
stop1 = 0
stop2 =0
for filename in glob.glob(os.path.join(user_input, '*.txt')):
with open(filename, 'r') as f:
for l in f.readlines():
if l.startswith(' Impedance'):
v = l[12:-7]
impedance = float(v)
if l.startswith(' Xtalk'):
Xtalk.append(l[7:].strip())
if l.startswith(' Loss per inch'):
start = l.find('@')
stop1 = l.find('GHz', start)
stop2 = l.find('dB', start)
frequencies.append(float(l[start+1:stop1].strip()))
losses.append(float(l[stop1+5:stop2].strip()))
print(impedance, frequencies, losses, Xtalk)它基本上从一个文本文件中提取值，并将它们打印到控制台上，文本文件如下所示：
Impedance = 71.28 ohms
Begin Post processing
Frequency multiplier = 1Hz
number of ports = 12
Start Frequency = 0
End Frequency = 40000000000
Number of Frequency points = 4001
Touchstone Output file = C:\Users\Aravind_Sampathkumar\Desktop\IMLC\BO\Output_TW_3.5-TS_3-core_h_2.xml_5000mil.s12p
Output format = Real - Imaginary
Loss per inch @ 2.500000e+00 GHz = -0.569 dB
Loss per inch @ 5 GHz = -0.997 dB
Xtalk #1 (Conductor 1 2):
Step response Next= -0.56 mV
Step response Fext peak @ 5 inches= 0.11 mV
Xtalk #2 (Conductor 5 6):
Step response Next= -0.56 mV
Step response Fext peak @ 5 inches= 0.11 mV
Finished post processing回答 1
Stack Overflow用户
发布于 2018-05-15 22:28:24
首先确定变量的格式是什么。
使用" float ()“解析器无法将带有逗号的字符串转换为浮点型
a = "1,2345"
float(a)
Traceback (most recent call last):
File "<input>", line 1, in <module>
ValueError: could not convert string to float: '1,2345'
a = "1.2345"
float(a)
1.2345页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/50352653
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