Python - 如何基于分米计将列中的单元格拆分为新行

Question

相对较新并尝试使用python从CSV文件中拆分一些数据。 我试图解析这些数据，并在出现特定分隔符时将其拆分为新行。 那些分隔符是'。' ';' 和'＃'。 COL_C也没有空格。 另外，分隔符的顺序无关紧要，如果我们找到其中一个分隔符，则自动创建新行。

这是示例数据

COL_A | COL_B |COL_C
--------------------
Hello | World | Hi.Can;You#Help

我想要获得的输出将是：

COL_A | COL_B | COL_C
----------------------
Hello | World | Hi
Hello | World | Can
Hello | World | You
Hello | World | Help


例2：

COL_A | COL_B | COL_C
----------------------
Hello | World | Hi#123;move
New | line | Can.I#parse;this.data

我想要获得的输出将是：

COL_A | COL_B | COL_C
----------------------
Hello | World | Hi
Hello | World | 123
Hello | World | move
New | Line | Can
New | Line | I
New | Line | parse
New | Line | this
New | Line | data

如果这个数据集有另一行没有Hello World并且在前两列中有世界问候，我希望显示该数据，并将相应的第三列数据解析为新行。

谢谢！

Answer 1

设定

df = pd.DataFrame({'COL_A': {0: 'Hello ', 1: 'New   '},
 'COL_B': {0: ' World ', 1: ' line  '},
 'COL_C': {0: ' Hi#123;move', 1: ' Can.I#parse;this.data '}})
Out[480]: 
    COL_A    COL_B                    COL_C
0  Hello    World               Hi#123;move
1  New      line     Can.I#parse;this.data 

解

#split COL_C by given delimeter and stack them up in a series
COL_C2 = df.COL_C.str.split('\.|;|#',expand=True).stack()
#join the new series (after setting a name and index) back to the dataframe
df.join(pd.Series(index=COL_C2.index.droplevel(1), data=COL_C2.values, name='COL_C2'))

Out[475]: 
    COL_A    COL_B                    COL_C COL_C2
0  Hello    World               Hi#123;move     Hi
0  Hello    World               Hi#123;move    123
0  Hello    World               Hi#123;move   move
1  New      line     Can.I#parse;this.data     Can
1  New      line     Can.I#parse;this.data       I
1  New      line     Can.I#parse;this.data   parse
1  New      line     Can.I#parse;this.data    this
1  New      line     Can.I#parse;this.data   data 

Answer 2

融合速度与优雅

def pir(df, c):
    colc = df[c].str.split('\.|;|#')
    clst = colc.values.tolist()
    lens = [len(l) for l in clst]

    cdf = pd.DataFrame({c: np.concatenate(clst)}, df.index.repeat(lens))
    return df.drop(c, 1).join(cdf).reset_index(drop=True)

忘记优雅，给我速度！

def pir2(df, c):
    colc = df[c].str.split('\.|;|#')
    clst = colc.values.tolist()
    lens = [len(l) for l in clst]
    j = df.columns.get_loc(c)
    v = df.values
    n, m = v.shape
    r = np.arange(n).repeat(lens)
    return pd.DataFrame(
        np.column_stack([v[r, 0:j], np.concatenate(clst), v[r, j+1:]]),
        columns=df.columns
    )

---

pir(df, 'COL_C')
# pir2(df, 'COL_C')

   COL_A  COL_B  COL_C
0  Hello  World     Hi
1  Hello  World    123
2  Hello  World   move
3    New   line    Can
4    New   line      I
5    New   line  parse
6    New   line   this
7    New   line   data

---

定时

%timeit pir(df, 'COL_C')
1000 loops, best of 3: 1.42 ms per loop

%timeit pir2(df, 'COL_C')
1000 loops, best of 3: 278 µs per loop

%timeit split_list_in_cols_to_rows(df.assign(COL_C=df.COL_C.str.split(r'[.,;#]')), lst_cols='COL_C')
100 loops, best of 3: 4.16 ms per loop

%%timeit 
COL_C2 = df.COL_C.str.split('\.|;|#').apply(pd.Series).stack()
df.drop('COL_C', 1).join(pd.Series(index=COL_C2.index.droplevel(1), data=COL_C2.values, name='COL_C')).reset_index(drop=True)
100 loops, best of 3: 2.81 ms per loop

设定

from io import StringIO
import pandas as pd

txt = """COL_A | COL_B | COL_C
Hello | World | Hi#123;move
New   | line  | Can.I#parse;this.data """

df = pd.read_csv(StringIO(txt), sep='\s*\|\s*', engine='python')

Answer 3

例1

In [107]: df
Out[107]:
   COL_A  COL_B            COL_C
0  Hello  World  Hi.Can;You#Help

解：

def split_list_in_cols_to_rows(df, lst_cols, fill_value=''):
    # make sure `lst_cols` is a list
    if lst_cols and not isinstance(lst_cols, list):
        lst_cols = [lst_cols]
    # all columns except `lst_cols`
    idx_cols = df.columns.difference(lst_cols)

    # calculate lengths of lists
    lens = df[lst_cols[0]].str.len()

    return pd.DataFrame({
        col:np.repeat(df[col].values, df[lst_cols[0]].str.len())
        for col in idx_cols
    }).assign(**{col:np.concatenate(df[col].values) for col in lst_cols}) \
      .append(df.loc[lens==0, idx_cols]).fillna(fill_value) \
      .loc[:, df.columns]

In [106]: split_list_in_cols_to_rows(df.assign(COL_C=df.COL_C.str.split(r'[.,;#]')),
                                     lst_cols='COL_C')
Out[106]:
   COL_A  COL_B COL_C
0  Hello  World    Hi
1  Hello  World   Can
2  Hello  World   You
3  Hello  World  Help

例2：

In [110]: df
Out[110]:
   COL_A  COL_B                  COL_C
0  Hello  World            Hi#123;move
1    New   line  Can.I#parse;this.data

In [111]: split_list_in_cols_to_rows(df.assign(COL_C=df.COL_C.str.split(r'[.,;#]')),
     ...:                                      lst_cols='COL_C')
Out[111]:
   COL_A  COL_B  COL_C
0  Hello  World     Hi
1  Hello  World    123
2  Hello  World   move
3    New   line    Can
4    New   line      I
5    New   line  parse
6    New   line   this
7    New   line   data