运行 平均问题
Running average issue
如果您能就这个问题给我任何提示，我将不胜感激。我该如何开始，知道吗？
我是运行一个产生能量的软件。我有 N 个时间步（块），并且在每个时间步中它生成几行。第一列是每个时间步中生成的行数的计数器。因此，我将有一个如下所示的文本文件，它是一个巨大文件的简化示例：
#Block: 3 3 0
1 -0.3746468365E+04 0.9420348015E+00 -0.3745526330E+04 0.1636348407-151 316.8679 0.2626532250-312
2 -0.3746707738E+04 0.1149418891E+01 -0.3745558319E+04 0.1220432713E+00 386.6247 0.2626532250-312
3 -0.3746757823E+04 0.1195378991E+01 -0.3745562444E+04 0.1636348407-151 402.0841 0.2626532250-312
#Block: 4 4 1
1 -0.3746625102E+04 0.6261182789E+00 -0.3745998983E+04 0.1636348407-151 210.6045 0.2626532250-312
2 -0.3746723956E+04 0.7190568481E+00 -0.3746004899E+04 0.1636348407-151 241.8658 0.2626532250-312
3 -0.3746758909E+04 0.7514702248E+00 -0.3746007439E+04 0.1636348407-151 252.7685 0.2626532250-312
4 -0.3746707738E+04 0.7067911904E+00 -0.3746000947E+04 -0.3205844292E+00 237.7401 0.2626532250-312
5 -0.3746680579E+04 0.6897161187E+00 -0.3745990863E+04 0.1636348407-151 231.9966 0.2626532250-312
#Block: 5 5 1
1 -0.3746625102E+04 0.6261182789E+00 -0.3745998983E+04 0.1636348407-151 210.6045 0.2626532250-312
2 -0.3746723956E+04 0.7190568481E+00 -0.3746004899E+04 0.1636348407-151 241.8658 0.2626532250-312
3 -0.3746758909E+04 0.7514702248E+00 -0.3746007439E+04 0.1636348407-151 252.7685 0.2626532250-312
4 -0.3746707738E+04 0.7067911904E+00 -0.3746000947E+04 -0.3205844292E+00 237.7401 0.2626532250-312
5 -0.3746680579E+04 0.6897161187E+00 -0.3745990863E+04 0.1636348407-151 231.9966 0.2626532250-312
我想计算这些数据第 3 列的 运行 平均值并保存在 "average text file" 中。因此，
文本文件中的第一个值应该是块 1 的第 3 列的平均值。
（前 3 个值=（3.5+7.1+5.4）/3）
文本文件中的第二个值应该是块 1 和块 2 的第 3 列的平均值。
（前8个值=（3.5+7.1+5.4+2.5+1.1+5.4+4.4+1.4）/8）
文本文件中的第三个值应该是块 1、块 2 和块 3 的第 3 列的平均值。
（前 12 个值=（3.5+7.1+5.4+2.5+1.1+5.4+4.4+1.4+3.5+1.1+8.1+9.4）/12）
我有N块。实际上，我正在计算所有块的第 3 列的平均值作为时间步长的函数。
这是我试过的方法
with open('xxx.txt','r') as file:
groups = [] # Will contain the final data
current_group = [] # Temporary
line = file.readline()
while line != "":
if line.startswith("#Block"):
# Store the current group and start a new one
groups.append(current_group)
current_group = []
else:
# Add the number to the current group
current_group.append(float(line.split()[2]))
line = file.readline()
averages = list()
for i in xrange(len(groups)):
flatten_list = list(itertools.chain(*groups[:i+1]))
print (flatten_list)
averages.append(sum(flatten_list) / len(flatten_list))
with open('output.txt', 'a+') as output_f:
output_f.writelines(averages)
到目前为止你做得很好，因为你已经设法将每个块的第 3 列提取为 groups 列表中的列表，现在，在你的 while 循环之后你可以做
averages = list()
for i in range(len(groups)):
flatten_list = list(itertools.chain(*groups[:i+1]))
print flatten_list
averages.append(sum(flatten_list) / len(flatten_list))
with open('output.txt', 'a+') as output_f:
output_f.writelines(averages)
完整代码将是
with open('xxx.txt','r') as file:
groups = [] # Will contain the final data
current_group = [] # Temporary
line = file.readline()
while line != "":
if line.startswith("#Block"):
# Store the current group and start a new one
groups.append(current_group)
current_group = []
else:
# Add the number to the current group
current_group.append(float(line.split()[2]))
line = file.readline()
averages = list()
for i in range(len(groups)):
flatten_list = list(itertools.chain(*groups[:i+1]))
print (flatten_list)
averages.append(sum(flatten_list) / len(flatten_list))
with open('output.txt', 'a+') as output_f:
output_f.writelines(averages)
输入文件名是'input.txt'。
输出文件名为 'result.txt'.
with open('INPUT.TXT', 'r') as fout:
lst_of_num=[]
for line in fout:
if line[0]=='#':
#print line
if lst_of_num:
#print sum(lst_of_num) / float(len(lst_of_num))
with open("result.txt", "a+") as myfile:
myfile.write(str(sum(lst_of_num) / float(len(lst_of_num)))+'\n')
#lst_of_num=[]
else:
lst_of_num.append(float(line.split()[2]))
with open("result.txt", "a+") as myfile:
myfile.write(str(sum(lst_of_num) / float(len(lst_of_num)))+'\n')
更新为 Python 3 和新输入
import itertools
history = []
with open('sample.txt') as f:
group = []
for line in f:
line = line.strip()
if 'Block' in line:
if group:
history.append(group)
group = []
elif line:
group.append(float(line.split()[2]))
history.append(group)
with open('output.txt', 'w') as f:
for i in range(len(history)):
l = list(itertools.chain(*history[:i+1]))
print(sum(l) / len(l), file=f)
并输出：
$ cat output.txt
1.0956108945
0.84749816805
0.790241385023077
如果您能就这个问题给我任何提示，我将不胜感激。我该如何开始，知道吗？
我是运行一个产生能量的软件。我有 N 个时间步（块），并且在每个时间步中它生成几行。第一列是每个时间步中生成的行数的计数器。因此，我将有一个如下所示的文本文件，它是一个巨大文件的简化示例：
#Block: 3 3 0
1 -0.3746468365E+04 0.9420348015E+00 -0.3745526330E+04 0.1636348407-151 316.8679 0.2626532250-312
2 -0.3746707738E+04 0.1149418891E+01 -0.3745558319E+04 0.1220432713E+00 386.6247 0.2626532250-312
3 -0.3746757823E+04 0.1195378991E+01 -0.3745562444E+04 0.1636348407-151 402.0841 0.2626532250-312
#Block: 4 4 1
1 -0.3746625102E+04 0.6261182789E+00 -0.3745998983E+04 0.1636348407-151 210.6045 0.2626532250-312
2 -0.3746723956E+04 0.7190568481E+00 -0.3746004899E+04 0.1636348407-151 241.8658 0.2626532250-312
3 -0.3746758909E+04 0.7514702248E+00 -0.3746007439E+04 0.1636348407-151 252.7685 0.2626532250-312
4 -0.3746707738E+04 0.7067911904E+00 -0.3746000947E+04 -0.3205844292E+00 237.7401 0.2626532250-312
5 -0.3746680579E+04 0.6897161187E+00 -0.3745990863E+04 0.1636348407-151 231.9966 0.2626532250-312
#Block: 5 5 1
1 -0.3746625102E+04 0.6261182789E+00 -0.3745998983E+04 0.1636348407-151 210.6045 0.2626532250-312
2 -0.3746723956E+04 0.7190568481E+00 -0.3746004899E+04 0.1636348407-151 241.8658 0.2626532250-312
3 -0.3746758909E+04 0.7514702248E+00 -0.3746007439E+04 0.1636348407-151 252.7685 0.2626532250-312
4 -0.3746707738E+04 0.7067911904E+00 -0.3746000947E+04 -0.3205844292E+00 237.7401 0.2626532250-312
5 -0.3746680579E+04 0.6897161187E+00 -0.3745990863E+04 0.1636348407-151 231.9966 0.2626532250-312
我想计算这些数据第 3 列的 运行 平均值并保存在 "average text file" 中。因此，
文本文件中的第一个值应该是块 1 的第 3 列的平均值。 （前 3 个值=（3.5+7.1+5.4）/3）
文本文件中的第二个值应该是块 1 和块 2 的第 3 列的平均值。 （前8个值=（3.5+7.1+5.4+2.5+1.1+5.4+4.4+1.4）/8）
文本文件中的第三个值应该是块 1、块 2 和块 3 的第 3 列的平均值。 （前 12 个值=（3.5+7.1+5.4+2.5+1.1+5.4+4.4+1.4+3.5+1.1+8.1+9.4）/12）
我有N块。实际上，我正在计算所有块的第 3 列的平均值作为时间步长的函数。
这是我试过的方法
with open('xxx.txt','r') as file:
groups = [] # Will contain the final data
current_group = [] # Temporary
line = file.readline()
while line != "":
if line.startswith("#Block"):
# Store the current group and start a new one
groups.append(current_group)
current_group = []
else:
# Add the number to the current group
current_group.append(float(line.split()[2]))
line = file.readline()
averages = list()
for i in xrange(len(groups)):
flatten_list = list(itertools.chain(*groups[:i+1]))
print (flatten_list)
averages.append(sum(flatten_list) / len(flatten_list))
with open('output.txt', 'a+') as output_f:
output_f.writelines(averages)
到目前为止你做得很好，因为你已经设法将每个块的第 3 列提取为 groups 列表中的列表，现在，在你的 while 循环之后你可以做
averages = list()
for i in range(len(groups)):
flatten_list = list(itertools.chain(*groups[:i+1]))
print flatten_list
averages.append(sum(flatten_list) / len(flatten_list))
with open('output.txt', 'a+') as output_f:
output_f.writelines(averages)
完整代码将是
with open('xxx.txt','r') as file:
groups = [] # Will contain the final data
current_group = [] # Temporary
line = file.readline()
while line != "":
if line.startswith("#Block"):
# Store the current group and start a new one
groups.append(current_group)
current_group = []
else:
# Add the number to the current group
current_group.append(float(line.split()[2]))
line = file.readline()
averages = list()
for i in range(len(groups)):
flatten_list = list(itertools.chain(*groups[:i+1]))
print (flatten_list)
averages.append(sum(flatten_list) / len(flatten_list))
with open('output.txt', 'a+') as output_f:
output_f.writelines(averages)
输入文件名是'input.txt'。 输出文件名为 'result.txt'.
with open('INPUT.TXT', 'r') as fout:
lst_of_num=[]
for line in fout:
if line[0]=='#':
#print line
if lst_of_num:
#print sum(lst_of_num) / float(len(lst_of_num))
with open("result.txt", "a+") as myfile:
myfile.write(str(sum(lst_of_num) / float(len(lst_of_num)))+'\n')
#lst_of_num=[]
else:
lst_of_num.append(float(line.split()[2]))
with open("result.txt", "a+") as myfile:
myfile.write(str(sum(lst_of_num) / float(len(lst_of_num)))+'\n')
更新为 Python 3 和新输入
import itertools
history = []
with open('sample.txt') as f:
group = []
for line in f:
line = line.strip()
if 'Block' in line:
if group:
history.append(group)
group = []
elif line:
group.append(float(line.split()[2]))
history.append(group)
with open('output.txt', 'w') as f:
for i in range(len(history)):
l = list(itertools.chain(*history[:i+1]))
print(sum(l) / len(l), file=f)
并输出：
$ cat output.txt
1.0956108945
0.84749816805
0.790241385023077