如何有效地创建数据透视表? [英] How to efficiently create a pivot table?

问题描述

我确实有一个像这样的数据框:

I do have a dataframe like this:

import pandas as pd

df = pd.DataFrame({"c0": list('ABC'),
                   "c1": [" ".join(list('ab')), " ".join(list('def')), " ".join(list('s'))],
                   "c2": list('DEF')})

  c0     c1 c2
0  A    a b  D
1  B  d e f  E
2  C      s  F

我想创建一个如下所示的数据透视表:

I want to create a pivot table that looks like this:

      c2
c0 c1   
A  a   D
   b   D
B  d   E
   e   E
   f   E
C  s   F

因此，c1中的条目被拆分，然后被视为多索引中使用的单个元素.

So, the entries in c1 are split and then treated as single elements used in a multiindex.

我这样做如下:

newdf = pd.DataFrame()

for indi, rowi in df.iterrows():

    # get all single elements in string
    n_elements = rowi['c1'].split()

    # only one element so we can just add the entire row
    if len(n_elements) == 1:
        newdf = newdf.append(rowi)
    # more than one element
    else:
        for eli in n_elements:
            # that allows to add new elements using loc, without it we will have identical index values
            if not newdf.empty:
                newdf = newdf.reset_index(drop=True)
                newdf.index = -1 * newdf.index - 1

            # add entire row
            newdf = newdf.append(rowi)
            # replace the entire string by the single element
            newdf.loc[indi, 'c1'] = eli

print newdf.reset_index(drop=True)

产生

  c0 c1 c2
0  A  a  D
1  A  b  D
2  B  d  E
3  B  e  E
4  B  f  E
5  C  s  F

那我就可以打电话

pd.pivot_table(newdf, index=['c0', 'c1'], aggfunc=lambda x: ' '.join(set(str(v) for v in x)))

这给了我想要的输出(见上文).

which gives me the desired output (see above).

对于可能非常慢的大型数据帧，所以我想知道是否存在更有效的方法.

For huge dataframes that can be quite slow, so I am wondering whether there is a more efficient way of doing this.

推荐答案

选项1

import numpy as np, pandas as pd

s = df.c1.str.split()
l = s.str.len()
newdf = df.loc[df.index.repeat(l)].assign(c1=np.concatenate(s)).set_index(['c0', 'c1'])
newdf

      c2
c0 c1   
A  a   D
   b   D
B  d   E
   e   E
   f   E
C  s   F

---

选项2
应该更快

---

Option 2
Should be faster

import numpy as np, pandas as pd

s = np.core.defchararray.split(df.c1.values.astype(str), ' ')
l = [len(x) for x in s.tolist()]
r = np.arange(len(s)).repeat(l)
i = pd.MultiIndex.from_arrays([
    df.c0.values[r],
    np.concatenate(s)
], names=['c0', 'c1'])
newdf = pd.DataFrame({'c2': df.c2.values[r]}, i)
newdf

      c2
c0 c1   
A  a   D
   b   D
B  d   E
   e   E
   f   E
C  s   F

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