pandas df.loc[z,x]=y 如何提高速度？

Question

我已经识别出一个 pandas 命令

timeseries.loc[z, x] = y

负责迭代中花费的大部分时间。现在我正在寻找更好的方法来加速它。该循环甚至不包含 50k 个元素（生产目标是 ~250k 或更多），但已经需要 20 秒了。

这是我的代码（忽略上半部分，它只是计时助手）

def populateTimeseriesTable(df, observable, timeseries):
    """
    Go through all rows of df and 
    put the observable into the timeseries 
    at correct row (symbol), column (tsMean).
    """

    print "len(df.index)=", len(df.index)  # show number of rows

    global bf, t
    bf = time.time()                       # set 'before' to now
    t = dict([(i,0) for i in range(5)])    # fill category timing with zeros

    def T(i):
        """
        timing helper: Add passed time to category 'i'. Then set 'before' to now.
        """
        global bf, t 
        t[i] = t[i] + (time.time()-bf)
        bf = time.time()        

    for i in df.index:             # this is the slow loop
        bf = time.time()

        sym = df["symbol"][i]
        T(0)

        tsMean = df["tsMean"][i]
        T(1)

        tsMean = tsFormatter(tsMean)
        T(2)

        o = df[observable][i]
        T(3)

        timeseries.loc[sym, tsMean] = o
        T(4)

    from pprint import pprint
    print "times needed (total = %.1f seconds) for each command:" % sum(t.values())
    pprint (t)

    return timeseries

与（不重要，不慢）

def tsFormatter(ts):
    "as human readable string, only up to whole seconds"
    return time.strftime("%Y-%m-%d %H:%M:%S", time.gmtime(ts))

。 .

--> 待优化的代码在for循环中。

（T 和 t 只是辅助函数和字典，用于计时。）

我已经为每一步计时了。绝大多数时间：

len(df.index)= 47160
times needed (total = 20.2 seconds) for each command:
{0: 1.102,
 1: 0.741,
 2: 0.243,
 3: 0.792,
 4: 17.371}

在最后一步花费了

timeseries.loc[sym, tsMean] = o

我已经下载并安装了 pypy - 但遗憾的是，它还不支持 pandas。

有什么想法可以加快填充二维数组吗？

谢谢！

---

编辑：抱歉，没有提到 -“时间序列”也是一个数据框：

timeseries = pd.DataFrame({"name": titles}, index=index)

Answer 1

更新：从 Pandas 0.20.1 开始.ix 索引器已弃用，转而使用更严格的 .iloc 和 .loc 索引器。

==================================================== ======================

@jezrael 提供了一个有趣的比较，我决定使用更多索引方法和 10M 行 DF 重复它（实际上在这种特殊情况下大小并不重要）：

设置：

In [15]: df = pd.DataFrame(np.random.rand(10**7, 5), columns=list('abcde'))

In [16]: df.info()
<class 'pandas.core.frame.DataFrame'>
RangeIndex: 10000000 entries, 0 to 9999999
Data columns (total 5 columns):
a    float64
b    float64
c    float64
d    float64
e    float64
dtypes: float64(5)
memory usage: 381.5 MB

In [17]: df.shape
Out[17]: (10000000, 5)

时间：

In [37]: %timeit df.loc[random.randint(0, 10**7), 'b']
1000 loops, best of 3: 502 µs per loop

In [38]: %timeit df.iloc[random.randint(0, 10**7), 1]
1000 loops, best of 3: 394 µs per loop

In [39]: %timeit df.at[random.randint(0, 10**7), 'b']
10000 loops, best of 3: 66.8 µs per loop

In [41]: %timeit df.iat[random.randint(0, 10**7), 1]
10000 loops, best of 3: 32.9 µs per loop

In [42]: %timeit df.ix[random.randint(0, 10**7), 'b']
10000 loops, best of 3: 64.8 µs per loop

In [43]: %timeit df.ix[random.randint(0, 10**7), 1]
1000 loops, best of 3: 503 µs per loop

结果为条形图：

计时数据为 DF：

In [88]: r
Out[88]:
       method  timing
0         loc   502.0
1        iloc   394.0
2          at    66.8
3         iat    32.9
4    ix_label    64.8
5  ix_integer   503.0

In [89]: r.to_dict()
Out[89]:
{'method': {0: 'loc',
  1: 'iloc',
  2: 'at',
  3: 'iat',
  4: 'ix_label',
  5: 'ix_integer'},
 'timing': {0: 502.0,
  1: 394.0,
  2: 66.799999999999997,
  3: 32.899999999999999,
  4: 64.799999999999997,
  5: 503.0}}

绘图

ax = sns.barplot(data=r, x='method', y='timing')
ax.tick_params(labelsize=16)
[ax.annotate(str(round(p.get_height(),2)), (p.get_x() + 0.2, p.get_height() + 5)) for p in ax.patches]
ax.set_xlabel('indexing method', size=20)
ax.set_ylabel('timing (microseconds)', size=20)

Answer 2

我一直认为

at

是最快的，但事实并非如此。

ix

更快：

import pandas as pd

df = pd.DataFrame({'A':[1,2,3],
                   'B':[4,5,6],
                   'C':[7,8,9],
                   'D':[1,3,5],
                   'E':[5,3,6],
                   'F':[7,4,3]})

print (df)
   A  B  C  D  E  F
0  1  4  7  1  5  7
1  2  5  8  3  3  4
2  3  6  9  5  6  3

print (df.at[2, 'B'])
6
print (df.ix[2, 'B'])
6
print (df.loc[2, 'B'])
6

In [77]: %timeit df.at[2, 'B']
10000 loops, best of 3: 44.6 µs per loop

In [78]: %timeit df.ix[2, 'B']
10000 loops, best of 3: 40.7 µs per loop

In [79]: %timeit df.loc[2, 'B']
1000 loops, best of 3: 681 µs per loop

编辑：

我尝试了

MaxU

df

并且导致了差异

random.randint

功能：

df = pd.DataFrame(np.random.rand(10**7, 5), columns=list('ABCDE'))


In [4]: %timeit (df.ix[2, 'B'])
The slowest run took 25.80 times longer than the fastest. This could mean that an intermediate result is being cached.
10000 loops, best of 3: 20.7 µs per loop

In [5]: %timeit (df.ix[random.randint(0, 10**7), 'B'])
The slowest run took 9.42 times longer than the fastest. This could mean that an intermediate result is being cached.
10000 loops, best of 3: 28 µs per loop

Answer 3

如果要在循环内添加行，请考虑这些性能问题；对于大约前 1000 到 2000 条记录，“my_df.loc”性能较好，但随着循环中记录数量的增加，性能逐渐变慢。

如果你打算在一个大循环中进行精简（比如 10Mu200c 记录左右），你最好混合使用“iloc”和“append”；用 iloc 填充临时数据帧，直到大小达到 1000 左右，然后将其附加到原始数据帧，并清空临时数据帧。这将使您的表现提高约 10 倍

Answer 4

更新较新的 pandas 版本

python 3.12.1 上的 pandas 2.2.2

In [1]: import random; import numpy as np; import pandas as pd

In [2]: df = pd.DataFrame(np.random.rand(10**7, 5), columns=list("abcde"))

In [3]: df.info()
<class 'pandas.core.frame.DataFrame'>
RangeIndex: 10000000 entries, 0 to 9999999
Data columns (total 5 columns):
 #   Column  Dtype  
---  ------  -----  
 0   a       float64
 1   b       float64
 2   c       float64
 3   d       float64
 4   e       float64
dtypes: float64(5)
memory usage: 381.5 MB

In [4]: %timeit df.loc[random.randint(0, 10**7 - 1), "b"]
8.17 µs ± 130 ns per loop (mean ± std. dev. of 7 runs, 100,000 loops each)

In [5]: %timeit df.iloc[random.randint(0, 10**7 - 1), 1]
18.7 µs ± 285 ns per loop (mean ± std. dev. of 7 runs, 100,000 loops each)

In [6]: %timeit df.at[random.randint(0, 10**7 - 1), "b"]
3.91 µs ± 251 ns per loop (mean ± std. dev. of 7 runs, 100,000 loops each)

In [7]: %timeit df.iat[random.randint(0, 10**7 - 1), 1]
14.9 µs ± 371 ns per loop (mean ± std. dev. of 7 runs, 100,000 loops each)

In [8]: %timeit df.values[random.randint(0, 10**7 - 1), 1]
1.7 µs ± 29.1 ns per loop (mean ± std. dev. of 7 runs, 1,000,000 loops each)

所以，直接访问

.values

numpy 数组仍然是最快的两倍以上，而且令人惊讶的是，如果不使用这种直接访问，通过标签访问比通过索引访问要快。