如何使用np.fromregex？

Question

我有点熟悉np.fromregex格式. 我阅读了教程，并试图实现它来读取数据文件。你知道吗

当使用简单的python列表理解读取文件时，它会给出所需的结果： [400, 401, 405, 408, 412, 414, 420, 423, 433]。你知道吗

但是，当np.fromregex给出另一个格式答案时： [(400,) (401,) (405,) (408,) (412,) (414,) (420,) (423,) (433,)]。你知道吗

如何更改代码，使regex的答案与简单的python for循环相同。你知道吗

谢谢。你知道吗

我知道这是一个简单的问题，但我花了很多时间去寻找 解决方案和它可能对其他人也有好处，并节省一些时间。你知道吗

相关链接：
https://docs.scipy.org/doc/numpy/reference/generated/numpy.fromregex.html
np.fromregex with string as dtype

from __future__ import print_function, division, with_statement, unicode_literals
import numpy as np
import re

data = """
DMStack failed for: lsst_z1.0_400.fits
DMStack failed for: lsst_z1.0_401.fits
DMStack failed for: lsst_z1.0_405.fits
DMStack failed for: lsst_z1.0_408.fits
DMStack failed for: lsst_z1.0_412.fits
DMStack failed for: lsst_z1.0_414.fits
DMStack failed for: lsst_z1.0_420.fits
DMStack failed for: lsst_z1.0_423.fits
DMStack failed for: lsst_z1.0_433.fits
"""

ifile = 'a.txt'
with open(ifile, 'w') as fo:
    fo.write(data.lstrip())


# regex
regexp = r".*_(\d+?).fits"

# This works fine
ans = [int(re.findall(regexp, line)[0]) for line in open(ifile)]
print(ans)

# using fromregex
dt = [('num', np.int32)]
x = np.fromregex(ifile, regexp, dt)
print(x)

更新
当我使用将来的导入时，上面的代码失败了。错误日志如下：

Traceback (most recent call last):
  File "a.py", line 31, in <module>
    x = np.fromregex(ifile, regexp, dt)
  File "/Users/poudel/miniconda2/lib/python2.7/site-packages/numpy/lib/npyio.py", line 1452, in fromregex
    dtype = np.dtype(dtype)
TypeError: data type not understood


$ which python
python is /Users/poudel/miniconda2/bin/python 

$ python -c "import numpy; print(numpy.__version__)"
1.14.0

Answer 1

import numpy as np
import cStringIO
import re

data = """
DMStack failed for: lsst_z1.0_400.fits
DMStack failed for: lsst_z1.0_401.fits
DMStack failed for: lsst_z1.0_405.fits
DMStack failed for: lsst_z1.0_408.fits
DMStack failed for: lsst_z1.0_412.fits
DMStack failed for: lsst_z1.0_414.fits
DMStack failed for: lsst_z1.0_420.fits
DMStack failed for: lsst_z1.0_423.fits
DMStack failed for: lsst_z1.0_433.fits
"""

# ifile = cStringIO.StringIO()
# ifile.write(data)
ifile = 'a.txt'
with open(ifile, 'w') as fo:
    fo.write(data.lstrip())


# regex
regexp = r".*_(\d+?).fits"

# This works fine
ans = [int(re.findall(regexp, line)[0]) for line in open(ifile)]
print(ans)

# using fromregex
dt = [('num', np.int32)]
x = np.fromregex(ifile, regexp, dt)
y=[]
for i in x:
    y = y + [i[0]]
print y

"""
[400, 401, 405, 408, 412, 414, 420, 423, 433]
[400, 401, 405, 408, 412, 414, 420, 423, 433]
"""

我不知道这样做没有一个循环。你知道吗

Answer 2

感谢@zipa和@hpaulj，终于有了这段代码 python2和将来的陈述。它也适用于python3。你知道吗

我们需要使用dt = [(str('num'), np.int32)]，而不是dt = [('num', np.int32)]。你知道吗

#!python
# -*- coding: utf-8 -*-#
#
# Imports
from __future__ import print_function, division, with_statement, unicode_literals
import numpy as np
import re

data = """
DMStack failed for: lsst_z1.0_400.fits
DMStack failed for: lsst_z1.0_401.fits
DMStack failed for: lsst_z1.0_405.fits
DMStack failed for: lsst_z1.0_408.fits
DMStack failed for: lsst_z1.0_412.fits
DMStack failed for: lsst_z1.0_414.fits
DMStack failed for: lsst_z1.0_420.fits
DMStack failed for: lsst_z1.0_423.fits
DMStack failed for: lsst_z1.0_433.fits
"""

ifile = 'a.txt'
with open(ifile, 'w') as fo:
    fo.write(data.lstrip())


# regex
regexp = r".*_(\d+?).fits"
dt = [(str('num'), np.int32)]
x = np.fromregex(ifile, regexp, dt)
print(x['num'])

Answer 3

只要选择小组，你就会得到你想要的：

dt = [('num', np.int32)]
x = np.fromregex(ifile, regexp, dt)
print(x['num'])
#[400 401 405 408 412 414 420 423 433]