Question

for k, pip in enumerate(tmp):
    pip = round(pip, 4)
    if pip in levels:
        print str(pip) + ' is already in list!'
        levels[pip] = (levels[pip] + int(round(percentiles[k]))) / 2.
    else:
        levels[pip] = int(round(percentiles[k]))

点数最多为4位小数，例如1.2345。级别是OrderDict

if永远不会评估为真;始终分配（和覆盖）值。那我该如何处理钥匙？

另外，有更多的pythonic方法吗？我不能用percentiles以某种方式压缩tmp吗？

Answer 1

>>> import collections
>>> d = collections.OrderedDict()
>>> d[1] = 2
>>> d
OrderedDict([(1, 2)])
>>> d[1.234] = 4
>>> d
OrderedDict([(1, 2), (1.234, 4)])
>>> d[1.234]
4
>>> d[round(1.2343, 3)]
4

是的，OrderedDict可以将浮点数作为键。您看到的问题是由于其他原因 - 您希望在此循环之前填充levels吗？或者你应该在循环中有多个值来舍入相同的值吗？

我快速修改了你的代码来测试它，一切顺利。你能提供一些导致你看到的错误的数据吗？

>>> cache = collections.OrderedDict()
>>> for index, value in enumerate([1.234, 2.345, 1.234]):
...     rounded_value = round(value, 2)
...     if rounded_value in cache:
...             print("we already found {0}!".format(rounded_value))
...     else:
...             print("{0} is new".format(rounded_value))
...             cache[rounded_value] = index
... 
1.23 is new
2.35 is new
we already found 1.23!

Answer 2

可能你正在寻找这个：

from odict import odict
levels=odict()
tmp=[1.01234,1.13333,1.233333,1.13333]
percentiles=[3.2,1,3,3]
for k, pip in enumerate(tmp):
    pip = round(pip, 4)
    if pip in levels:
       print str(pip) + ' is already in list!'
       #import pdb;pdb.set_trace()
       levels[pip] = (levels[pip] + int(round(percentiles[k]))) / 2.
    else:
        levels[pip] = int(round(percentiles[k]))

python OrderedDict可以在键中有小数吗？

2 个答案: