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python - 在 music21 中创建音符图像

转载 作者:太空狗 更新时间:2023-10-30 01:22:08 31 4
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运行时出现错误:

from music21 import *

n1 = note.Note('C4', quarterLength=1)
n2 = note.Note('A4', quarterLength=1)
s = stream.Stream()
s.append(n1)
s.append(n2)
s.show('lily.svg')

Traceback (most recent call last):
File "C:\Python34\test.py", line 7, in <module>
s.show('lily.svg')
File "C:\Python34\lib\site-packages\music21\base.py", line 2206, in show
return formatWriter.show(self, regularizedConverterFormat, app=app, subformats=subformats, **keywords)
File "C:\Python34\lib\site-packages\music21\converter\subConverters.py", line 277, in show
returnedFilePath = self.write(obj, fmt, subformats=subformats, **keywords)
File "C:\Python34\lib\site-packages\music21\converter\subConverters.py", line 245, in write
conv = lily.translate.LilypondConverter()
File "C:\Python34\lib\site-packages\music21\lily\translate.py", line 147, in __init__
self.setupTools()
File "C:\Python34\lib\site-packages\music21\lily\translate.py", line 177, in setupTools
versionString = versionString.split()[-1]
IndexError: list index out of range

我已经安装了 scipy 和 mathplotlib,所以 music21 不会再提示它们不可用了。我在 Windows 7 上运行 Python 3.4。

如果我改为使用 s.show('musicxml.png') 来获取我的图像,我会收到错误消息:

Traceback (most recent call last):
File "C:\Python34\test.py", line 7, in <module>
s.show('musicxml.png')
File "C:\Python34\lib\site-packages\music21\base.py", line 2206, in show
return formatWriter.show(self, regularizedConverterFormat, app=app, subformats=subformats, **keywords)
File "C:\Python34\lib\site-packages\music21\converter\subConverters.py", line 147, in show
returnedFilePath = self.write(obj, fmt, subformats=subformats, **keywords)
File "C:\Python34\lib\site-packages\music21\converter\subConverters.py", line 637, in write
fp = self.runThroughMusescore(fp, **keywords)
File "C:\Python34\lib\site-packages\music21\converter\subConverters.py", line 606, in runThroughMusescore
elif not os.path.exists(musescoreFile):
File "C:\Python34\lib\genericpath.py", line 19, in exists
os.stat(path)
TypeError: stat: can't specify None for path argument

我需要做什么才能获取图像(最好是 svg)?

最佳答案

百合花

我有同样的错误。我已经通过以下方式为 music21 配置了 LilyPond:

  1. 将 LilyPond 文件夹移动到没有空格的路径(从 C:\Program Files (x86)\LilyPond\usr\binC:\LilyPond\usr\bin ).我在 music21 代码中看到它在执行 lilypond 命令时没有在路径周围放置必要的配额,所以不得不这样解决问题。
  2. Created configuration file in music21 and set lilypondPath

    us = environment.UserSettings()
    us.create()
    us['lilypondPath'] = 'C:/LilyPond/usr/bin/lilypond.exe'

    您可以检查是否设置正确:

    print us['lilypondPath']
  3. 好吧,这可能不是必需的,但在我尝试的过程中我重启了好几次所有东西,所以如果一切都不能立即奏效,你可以在最后尝试一下。

MUSESCORE

  1. 为了以防万一,安装了Musescore到没有空格的路径 (
  2. 向环境添加了两次 musescore 路径(发现了这种设置环境变量的新方法),一次为 "musescoreDirectPNGPath":

    environment.set("musescoreDirectPNGPath", "C:/MuseScore2/bin/MuseScore.exe")

    然后作为“musicxmlPath”:

    environment.set("musicxmlPath", "C:/MuseScore2/bin/MuseScore.exe")

  3. 经过多次尝试、调试等。我了解到,如果我们想使用 Musescore,传递文件名扩展名“.xml”而不是“.png”很重要:

    stream_name.show('musicalxml.xml')

    Musescore 无法打开.png 文件,但可以打开.xml 文件。

最后,我可以添加一些无需打开 lilypond 或 musescore 即可生成文件的代码。希望有人觉得有用

莉莉庞德:

# music21object - stream or score or any object that can be showed
conv = music21.converter.subConverters.ConverterLilypond()
scorename = 'myScoreName'
filepath = 'C:/path/to/musical_scores/' + scorename
conv.write(music21object, fmt = 'lilypond', fp=filepath, subformats = ['pdf'])

音乐评分:

from music21.converter.subConverters import ConverterMusicXML
conv_musicxml = ConverterMusicXML()
scorename = 'myScoreName.xml'
filepath = 'C:/path/to/musical_scores/' + scorename
out_filepath = conv_musicxml.write(music21object, 'musicxml', fp=filepath, subformats=['png'])

注意,分数名称有“.x​​ml”扩展名。

不幸的是,它没有将文件保存在指定的文件路径中。 Musescore 将“-1”添加到文件名,但可以获取此更改后的文件路径(如上面代码中的 out_filepath)并稍后重命名为我们想要的。

关于python - 在 music21 中创建音符图像,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/25879764/

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