使用存储的列名作为变量名提取 PyTables 表的最有效方法
以下代码提供了我需要的功能；但是，对于包含 200 个变量、64000 行的表来说，需要 10 秒多一点的时间。是否有更有效的方法来创建与列名称匹配的变量命名空间？
strExec = "a = table[:]"
for colobj in table.description._f_walk(type="Col"):
colName = colobj._v_pathname
strExec = strExec + '\n' + colName + " = a['" + colName + "']"
exec(strExec)
该代码将在分析环境中执行，并占最终用户等待时间的很大一部分；因此，我想确认这是根据 PyTable 列名称动态建立命名空间的最佳方法。
The following code provides my needed functionality; but, takes a bit more than 10 seconds for a table that includes 200 variables with 64000 rows. Is there a more efficient way to create a variable namespace that matches the column names?
strExec = "a = table[:]"
for colobj in table.description._f_walk(type="Col"):
colName = colobj._v_pathname
strExec = strExec + '\n' + colName + " = a['" + colName + "']"
exec(strExec)
The code will be executed within an analysis environment and represents a large proportion of the wait time for the end user; so, I would like confirmation that this is the best way to achieve the establishment of a namespace dynamically based upon the PyTable column names.
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根据unutbu的建议，表复制a=table[:]是时间消耗者。建议的代码确实提供了对没有前缀的表名变量的访问。众所周知，使用这种简化变量命名空间的方法是有问题的，因为它可能不适合不言而喻的代码。
As suggested by unutbu, the table copy a=table[:] is the time consumer. The code suggested does otherwise provide access to the tablename variables without a prefix. It's acknowledged that using this method of simplifying the variable namespace is questionable as it may not lend itself to self-evident code.