Question

我需要组合多个包含字符串值的Panda Series。该系列是由多个验证步骤产生的消息。我尝试将这些消息合并到1 Series中，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

UPDATE2

在这个例子中，指数似乎变得混乱。

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

这导致此输出：

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

我希望在row0中只有a，在row1中只有b，而在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，函数计算每一行并可能返回一些行的错误消息Series（包含一些索引，一些不包含;如果没有错误返回，则错误系列为空。）

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

当连接默认值是使用现有索引时，如果它们发生碰撞，那么这会引发ValueError，因为您已找到，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

修改

我想我现在知道你想要什么，你可以通过将系列转换为数据帧然后使用索引进行合并来实现你想要的：

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 2

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 3

我可能找到了解决方案。我希望有人可以发表评论......

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

结合熊猫系列

3 个答案: