Question

我需要将所有numpy文件合并到一个numpy文件中，然后将该文件转换为二进制文件。我首先尝试创建两个不同的列表文件，我将文件名称合并到两个不同的列表中。我的包含numpy文件的文件夹如下所示：

File1_Trace1
File2_Trace2
File2_Trace3

我尝试使用此代码：

import numpy as np
import sys
import os

# be sure to pass numbers on command line
if len(sys.argv)==3:
    start_index, stop_index = int(sys.argv[1]), int(sys.argv[2])
else:
# or just use
    start_index, stop_index = 1, 1000

path = r'Y:\My_Scripts'
File_path= r'Y:\My_Scripts\1000_Traces'
lines_npy = [line.rstrip('\n') for line in open(os.path.join(path, 'Traces.list_npy'))]
lines_bin = [line.rstrip('\n') for line in open(os.path.join(path, 'Traces.list_bin'))]
all_traces = np.zeros((stop_index - start_index + 1, 8000))
for x in range(start_index, stop_index):
    trace = np.load(os.path.join(path, lines_npy[x]))
    trace.astype('float32').tofile(os.path.join(File_path, lines_bin[x]))
    all_traces[x] = trace
    all_traces.astype('float32').tofile(os.path.join(path, 'TRACES.bin'))

它不起作用的问题，首先它问我两个列表文件。然后它让我误解了这个错误：

  trace = np.load(os.path.join(path, lines_npy[x]))
  IndexError: list index out of range

Answer 1

问题在于索引lines_npy; x是无效值，可能太大了。要跟踪这一点，您需要关注代码的这一部分：

# be sure to pass numbers on command line
if len(sys.argv)==3:
    start_index, stop_index = int(sys.argv[1]), int(sys.argv[2])
else:
# or just use
    start_index, stop_index = 1, 1000

path = r'Y:\My_Scripts'
File_path= r'Y:\My_Scripts\1000_Traces'
lines_npy = [line.rstrip('\n') for line in open(os.path.join(path, 'Traces.list_npy'))]

print(len(lines_npy)
print(start_index, stop_index)

for x in range(start_index, stop_index):
    print(lines_npy[x]))

默认stop_index表示您可以拥有1000个文件？这有意义吗？

如何将我的所有文件名都放在一个文件中？

1 个答案: