在 Pandas 中组合系列

Question

我需要组合多个包含字符串值的 Pandas Series。该系列是由多个验证步骤产生的消息。我尝试将这些消息组合成 1 个Series 以将其附加到DataFrame。问题是结果是空的。

这是一个例子：

import pandas as pd

df = pd.DataFrame({'a': ['a', 'b', 'c', 'd'], 'b': ['aa', 'bb', 'cc', 'dd']})

index1 = df[df['a'] == 'b'].index
index2 = df[df['a'] == 'a'].index

series = df.iloc[index1].apply(lambda x: x['b'] + '-bbb', axis=1)
series += df.iloc[index2].apply(lambda x: x['a'] + '-aaa', axis=1)

print series
# >>> series
# 0    NaN
# 1    NaN

更新

import pandas as pd

df = pd.DataFrame({'a': ['a', 'b', 'c', 'd'], 'b': ['aa', 'bb', 'cc', 'dd']})

index1 = df[df['a'] == 'b'].index
index2 = df[df['a'] == 'a'].index

series1 = df.iloc[index1].apply(lambda x: x['b'] + '-bbb', axis=1)
series2 = df.iloc[index2].apply(lambda x: x['a'] + '-aaa', axis=1)
series3 = df.iloc[index2].apply(lambda x: x['a'] + '-ccc', axis=1)

# series3 causes a ValueError: cannot reindex from a duplicate axis
series = pd.concat([series1, series2, series3])
df['series'] = series
print df

更新2

在此示例中，索引似乎混淆了。

import pandas as pd

df = pd.DataFrame({'a': ['a', 'b', 'c', 'd'], 'b': ['aa', 'bb', 'cc', 'dd']})

index1 = df[df['a'] == 'a'].index
index2 = df[df['a'] == 'b'].index
index3 = df[df['a'] == 'c'].index

series1 = df.iloc[index1].apply(lambda x: x['a'] + '-aaa', axis=1)
series2 = df.iloc[index2].apply(lambda x: x['a'] + '-bbb', axis=1)
series3 = df.iloc[index3].apply(lambda x: x['a'] + '-ccc', axis=1)

print series1
print
print series2
print
print series3
print

df['series'] = pd.concat([series1, series2, series3], ignore_index=True)
print df
print

df['series'] = pd.concat([series2, series1, series3], ignore_index=True)
print df
print

df['series'] = pd.concat([series3, series2, series1], ignore_index=True)
print df
print

这会导致以下输出：

0    a-aaa
dtype: object

1    b-bbb
dtype: object

2    c-ccc
dtype: object

   a   b series
0  a  aa  a-aaa
1  b  bb  b-bbb
2  c  cc  c-ccc
3  d  dd    NaN

   a   b series
0  a  aa  b-bbb
1  b  bb  a-aaa
2  c  cc  c-ccc
3  d  dd    NaN

   a   b series
0  a  aa  c-ccc
1  b  bb  b-bbb
2  c  cc  a-aaa
3  d  dd    NaN

我希望row0中只有a，row1中只有b，row2中只有c，但事实并非如此......

更新 3

这是一个更好的例子，它应该展示预期的行为。正如我所说，用例是对于给定的DataFrame，一个函数评估每一行，并可能将某些行的错误消息返回为Series（包含一些索引，一些不包含；如果没有错误返回，错误序列为空）。

In [12]:

s1 = pd.Series(['b', 'd'], index=[1, 3])
s2 = pd.Series(['a', 'b'], index=[0, 1])
s3 = pd.Series(['c', 'e'], index=[2, 4])
s4 = pd.Series([], index=[])
pd.concat([s1, s2, s3, s4]).sort_index()

# I'd like to get:
#
# 0    a
# 1    b b
# 2    c
# 3    d
# 4    e
Out[12]:
0    a
1    b
1    b
2    c
3    d
4    e
dtype: object

Answer 1

concat 怎么样？

s1 = df.iloc[index1].apply(lambda x: x['b'] + '-bbb', axis=1)
s2 = df.iloc[index2].apply(lambda x: x['a'] + '-aaa', axis=1)


s = pd.concat([s1,s2])
print s

1    bb-bbb
0    a-aaa
dtype: object

Answer 2

连接时默认使用现有索引，但如果它们发生冲突，则会引发ValueError，正如您所发现的，因此您需要设置ignore_index=True：

In [33]:

series = pd.concat([series1, series2, series3], ignore_index=True)
df['series'] = series
print (df)
   a   b  series
0  a  aa  bb-bbb
1  b  bb   a-aaa
2  c  cc   a-ccc
3  d  dd     NaN

编辑

我想我现在知道您想要什么，您可以通过将系列转换为数据框然后使用索引进行合并来实现您想要的：

In [96]:

df = pd.DataFrame({'a': ['a', 'b', 'c', 'd'], 'b': ['aa', 'bb', 'cc', 'dd']})

index1 = df[df['a'] == 'b'].index
index2 = df[df['a'] == 'a'].index

series1 = df.iloc[index1].apply(lambda x: x['b'] + '-bbb', axis=1)
series2 = df.iloc[index2].apply(lambda x: x['a'] + '-aaa', axis=1)
series3 = df.iloc[index2].apply(lambda x: x['a'] + '-ccc', axis=1)
# we now don't ignore the index in order to preserve the identity of the row we want to merge back to later
series = pd.concat([series1, series2, series3])
# construct a dataframe from the series and give the column a name
df1 = pd.DataFrame({'series':series})
# perform an outer merge on both df's indices
df.merge(df1, left_index=True, right_index=True, how='outer')

Out[96]:
   a   b  series
0  a  aa   a-aaa
0  a  aa   a-ccc
1  b  bb  bb-bbb
2  c  cc     NaN
3  d  dd     NaN

Answer 3

我可能已经找到了解决方案。我希望有人可以评论它...

s1 = pd.Series(['b', 'd'], index=[1, 3])
s2 = pd.Series(['a', 'b'], index=[0, 1])
s3 = pd.Series(['c', 'e'], index=[2, 4])
s4 = pd.Series([], index=[])
pd.concat([s1, s2, s3, s4]).sort_index()


df1 = pd.DataFrame(s1)
df2 = pd.DataFrame(s2)
df3 = pd.DataFrame(s3)
df4 = pd.DataFrame(s4)

d = pd.DataFrame({0:[]})
d = pd.merge(df1, d, how='outer', left_index=True, right_index=True)
d = d.fillna('')
d = pd.DataFrame(d['0_x'] + d['0_y'])

d = pd.merge(df2, d, how='outer', left_index=True, right_index=True)
d = d.fillna('')
d = pd.DataFrame(d['0_x'] + d['0_y'])

d = pd.merge(df3, d, how='outer', left_index=True, right_index=True)
d = d.fillna('')
d = pd.DataFrame(d['0_x'] + d['0_y'])

d = pd.merge(df4, d, how='outer', left_index=True, right_index=True)
d = d.fillna('')
d = pd.DataFrame(d['0_x'] + d['0_y'])
print d

返回

    0
0   a
1  bb
2   c
3   d
4   e