Question

像df.A = sr这样的简单操作（将pandas.Series分配给pandas.DataFrame中的列）似乎无害，但有许多细微差别。对于像我这样开始学习pandas的人来说，它带来了许多便利和困惑。

下面给出一个简单的示例/挑战：

df:
+----+-----+
|    |   A |
|----+-----|
|  0 |   0 |
|  1 |   0 |
|  2 |   0 |
|  3 |   0 |
|  4 |   0 |
+----+-----+

l = [777, 666, 555, 444, 333]

sr:
+----+-----+
|    |   0 |
|----+-----|
|  7 | 777 |
|  6 | 666 |
|  5 | 555 |
|  4 | 444 |
|  3 | 333 |
+----+-----+

在df之后，df.A = sr是什么样？
或
df之后，df.A = l会是什么样？

根据我目前的理解，我分解了df.A = sr中的所有隐含操作，请对其进行更正/确认/扩展： 例如，我不太确定正确的术语。

# [0] a column in a DataFrame, is a Series, is a dictionary of index and values
# all cell to cell transfers are key-lookup based, individual element in an 
# index is called a "label" for a reason.

# [1] if sr didn't have some of the index labels in df.col's index, 
# the old values in those cells in df.col gets WIPED!
df.loc[ ~df.index.isin(sr.index)] = np.nan

# [2] values are transferred from sr cells into df cells with common index-labels. 
# As expected
df.loc[ df.index.isin(sr.index), 'A'] = 
    sr.loc[ [idx for idx in sr.index if idx in df.index] ]

# [3] sr's cells, whoes index-lables are not found in df.index, are ignored and 
# doesn't get to be assigned in df
sr.loc[ ~sr.index.isin(df.index)] # goes no where.

# [4] with all the wipping and ignore from above steps, 
# there is no error message or warnings.
# it can cause your mistakes to slip thru:
"""
df = pd.DataFrame(0, columns=['A'], index=np.arange(5))
df.loc[ df.index.isin( ['A', 'B']), 'A'] = sr
print(df)

df = pd.DataFrame(0, columns=['A'], index=[])
df.A = sr
print(df)
"""

SPOILER。设置和结果：

df = pd.DataFrame(0, columns=['A'], index=np.arange(5))
l = [777, 666, 555, 444, 333]
sr = pd.Series(l, index=[7, 6, 5, 4, 3])


RESULTS:
df.A = sr
df:
+----+-----+
|    |   A |
|----+-----|
|  0 | nan |
|  1 | nan |
|  2 | nan |
|  3 | 333 |
|  4 | 444 |
+----+-----+

df.A = l
df:
+----+-----+
|    |   A |
|----+-----|
|  0 | 777 |
|  1 | 666 |
|  2 | 555 |
|  3 | 444 |
|  4 | 333 |
+----+-----+

Answer 1

所以您看到的结果是由于以下原因：

sr = pd.Series(l, index=[7, 6, 5, 4, 3])

您已将l的索引值专门分配给[7、6、5、4、3]。

当您这样做：

df.A = sr

该系列保留其索引值。然后，当您定义df时：

df = pd.DataFrame(0, columns=['A'], index=np.arange(5))

您确保最高索引值为4（index=np.arange(5)）

因此您的列输出保留了sr的索引值，并将值放在A中，因此仅显示了索引值3,4。

当您这样做时：

df.A = l

您只需将l中的值分配给A列。因此所有值都将出现。如果您将sr = pd.Series(l, index=[7, 6, 5, 4, 3])更改为sr = pd.Series(l)，请设置df.A = sr。您最终将得到与df.A = l完全相同的结果。

简单的隐藏地雷：df.A = sr

1 个答案: