为什么DataFrame.loc [[1]]比df.ix [[1]]慢1,800倍,比df.loc [1]慢3,500？

Question

亲自尝试一下:

import pandas as pd
s=pd.Series(xrange(5000000))
%timeit s.loc[[0]] # You need pandas 0.15.1 or newer for it to be that slow
1 loops, best of 3: 445 ms per loop

更新:这是大熊猫的合法错误,可能在2014年8月左右的0.15.1中引入.解决方法:在使用旧版本的pandas时等待新版本; 得到一个前沿的开发.来自github的版本; 在您的发布中手动执行单行修改pandas; 暂时使用.ix而不是.loc.

我有480万行的数据帧,以及选择使用的单个行.iloc[[ id ]](具有单个元素的列表)需要489毫秒,几乎一半的第二,慢1,800x倍的相同.ix[[ id ]],并慢于3,500x倍 .iloc[id](使ID作为一个值,而不是一个列表).公平地说,.loc[list]无论列表的长度如何都需要大约相同的时间,但我不想花费489毫秒,特别是当它.ix快一千倍时,并产生相同的结果.我的理解.ix应该是慢一点,不是吗？

我正在使用熊猫0.15.1.关于索引和选择数据的优秀教程表明,它.ix在某种程度上更为通用,并且可能比.loc和更慢.iloc.具体来说,它说

但是,当轴基于整数时,仅支持基于标签的访问而非位置访问.因此,在这种情况下,通常最好是明确的并使用.iloc或.loc.

这是一个带有基准的iPython会话:

    print 'The dataframe has %d entries, indexed by integers that are less than %d' % (len(df), max(df.index)+1)
    print 'df.index begins with ', df.index[:20]
    print 'The index is sorted:', df.index.tolist()==sorted(df.index.tolist())

    # First extract one element directly. Expected result, no issues here.
    id=5965356
    print 'Extract one element with id %d' % id
    %timeit df.loc[id]
    %timeit df.ix[id]
    print hash(str(df.loc[id])) == hash(str(df.ix[id])) # check we get the same result

    # Now extract this one element as a list.
    %timeit df.loc[[id]] # SO SLOW. 489 ms vs 270 microseconds for .ix, or 139 microseconds for .loc[id]
    %timeit df.ix[[id]] 
    print hash(str(df.loc[[id]])) == hash(str(df.ix[[id]]))  # this one should be True
    # Let's double-check that in this case .ix is the same as .loc, not .iloc, 
    # as this would explain the difference.
    try:
        print hash(str(df.iloc[[id]])) == hash(str(df.ix[[id]]))
    except:
        print 'Indeed, %d is not even a valid iloc[] value, as there are only %d rows' % (id, len(df))

    # Finally, for the sake of completeness, let's take a look at iloc
    %timeit df.iloc[3456789]    # this is still 100+ times faster than the next version
    %timeit df.iloc[[3456789]]

输出:

The dataframe has 4826616 entries, indexed by integers that are less than 6177817
df.index begins with  Int64Index([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20], dtype='int64')
The index is sorted: True
Extract one element with id 5965356
10000 loops, best of 3: 139 µs per loop
10000 loops, best of 3: 141 µs per loop
True
1 loops, best of 3: 489 ms per loop
1000 loops, best of 3: 270 µs per loop
True
Indeed, 5965356 is not even a valid iloc[] value, as there are only 4826616 rows
10000 loops, best of 3: 98.9 µs per loop
100 loops, best of 3: 12 ms per loop

Answer 1

看起来这个问题在熊猫中没有出现0.14.我用line_profiler描述了它,我想我知道发生了什么.由于pandas 0.15.1,KeyError如果给定的索引不存在,则现在会引发a.看起来当您使用.loc[list]语法时,它正在对整个轴上的索引进行详尽的搜索,即使已找到它.也就是说,首先,在找到元素的情况下没有提前终止,其次,在这种情况下搜索是强力的.

File: .../anaconda/lib/python2.7/site-packages/pandas/core/indexing.py,

  1278                                                       # require at least 1 element in the index
  1279         1          241    241.0      0.1              idx = _ensure_index(key)
  1280         1       391040 391040.0     99.9              if len(idx) and not idx.isin(ax).any():
  1281                                           
  1282                                                           raise KeyError("None of [%s] are in the [%s]" %

Answer 2

熊猫索引是疯狂的慢,我切换到numpy索引

df=pd.DataFrame(some_content)
# takes forever!!
for iPer in np.arange(-df.shape[0],0,1):
    x = df.iloc[iPer,:].values
    y = df.iloc[-1,:].values
# fast!        
vals = np.matrix(df.values)
for iPer in np.arange(-vals.shape[0],0,1):
    x = vals[iPer,:]
    y = vals[-1,:]