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将具有恒定值的列添加到pandas数据框[重复]

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问题：将具有恒定值的列添加到pandas数据框[重复]
回答 0
回答 1
超简单的就地分配： df['new'] = 0
对象列的注释
生成副本： df.assign(new=0)
多列分配
Super simple in-place assignment: df['new'] = 0
Note for object columns
Generating a copy: df.assign(new=0)
Multiple column assignment
回答 2
回答 3

问题：将具有恒定值的列添加到pandas数据框[重复]

给定一个DataFrame：

np.random.seed(0)
df = pd.DataFrame(np.random.randn(3, 3), columns=list('ABC'), index=[1, 2, 3])
df

          A         B         C
1  1.764052  0.400157  0.978738
2  2.240893  1.867558 -0.977278
3  0.950088 -0.151357 -0.103219

添加包含常量值（例如0）的新列的最简单方法是什么？

          A         B         C  new
1  1.764052  0.400157  0.978738    0
2  2.240893  1.867558 -0.977278    0
3  0.950088 -0.151357 -0.103219    0

这是我的解决方案，但我不知道为什么这会将NaN放入“新”列？

df['new'] = pd.Series([0 for x in range(len(df.index))])

          A         B         C  new
1  1.764052  0.400157  0.978738  0.0
2  2.240893  1.867558 -0.977278  0.0
3  0.950088 -0.151357 -0.103219  NaN
点击查看英文原文

Given a DataFrame:

np.random.seed(0)
df = pd.DataFrame(np.random.randn(3, 3), columns=list('ABC'), index=[1, 2, 3])
df

          A         B         C
1  1.764052  0.400157  0.978738
2  2.240893  1.867558 -0.977278
3  0.950088 -0.151357 -0.103219

What is the simplest way to add a new column containing a constant value eg 0?

          A         B         C  new
1  1.764052  0.400157  0.978738    0
2  2.240893  1.867558 -0.977278    0
3  0.950088 -0.151357 -0.103219    0

This is my solution, but I don’t know why this puts NaN into ‘new’ column?

df['new'] = pd.Series([0 for x in range(len(df.index))])

          A         B         C  new
1  1.764052  0.400157  0.978738  0.0
2  2.240893  1.867558 -0.977278  0.0
3  0.950088 -0.151357 -0.103219  NaN

回答 0

之所以将其NaN放入一列中，是因为df.index和Index您右侧对象的有所不同。@zach显示了分配新的零列的正确方法。通常，pandas尝试尽可能使索引对齐。一个缺点是，当指数不对准你NaN，无论他们是不是一致。尝试使用reindex和align方法来获得一些直觉，以便对齐具有部分，完全和未对齐所有对齐索引的对象。例如，以下是DataFrame.align()部分对齐索引的工作方式：

In [7]: from pandas import DataFrame

In [8]: from numpy.random import randint

In [9]: df = DataFrame({'a': randint(3, size=10)})

In [10]:

In [10]: df
Out[10]:
   a
0  0
1  2
2  0
3  1
4  0
5  0
6  0
7  0
8  0
9  0

In [11]: s = df.a[:5]

In [12]: dfa, sa = df.align(s, axis=0)

In [13]: dfa
Out[13]:
   a
0  0
1  2
2  0
3  1
4  0
5  0
6  0
7  0
8  0
9  0

In [14]: sa
Out[14]:
0     0
1     2
2     0
3     1
4     0
5   NaN
6   NaN
7   NaN
8   NaN
9   NaN
Name: a, dtype: float64
点击查看英文原文

The reason this puts NaN into a column is because df.index and the Index of your right-hand-side object are different. @zach shows the proper way to assign a new column of zeros. In general, pandas tries to do as much alignment of indices as possible. One downside is that when indices are not aligned you get NaN wherever they aren’t aligned. Play around with the reindex and align methods to gain some intuition for alignment works with objects that have partially, totally, and not-aligned-all aligned indices. For example here’s how DataFrame.align() works with partially aligned indices:

In [7]: from pandas import DataFrame

In [8]: from numpy.random import randint

In [9]: df = DataFrame({'a': randint(3, size=10)})

In [10]:

In [10]: df
Out[10]:
   a
0  0
1  2
2  0
3  1
4  0
5  0
6  0
7  0
8  0
9  0

In [11]: s = df.a[:5]

In [12]: dfa, sa = df.align(s, axis=0)

In [13]: dfa
Out[13]:
   a
0  0
1  2
2  0
3  1
4  0
5  0
6  0
7  0
8  0
9  0

In [14]: sa
Out[14]:
0     0
1     2
2     0
3     1
4     0
5   NaN
6   NaN
7   NaN
8   NaN
9   NaN
Name: a, dtype: float64

回答 1

超简单的就地分配： df['new'] = 0

对于就地修改，执行直接分配。熊猫会为每一行广播此任务。

df = pd.DataFrame('x', index=range(4), columns=list('ABC'))
df

   A  B  C
0  x  x  x
1  x  x  x
2  x  x  x
3  x  x  x


df['new'] = 'y'
# Same as,
# df.loc[:, 'new'] = 'y'
df

   A  B  C new
0  x  x  x   y
1  x  x  x   y
2  x  x  x   y
3  x  x  x   y

对象列的注释

如果要添加一列空列表，这是我的建议：

  • 考虑不这样做。object列对于性能而言是个坏消息。重新考虑数据的结构。
  • 考虑将数据存储在稀疏数据结构中。详细信息：稀疏数据结构

  • 如果必须存储一列列表，请确保不要多次复制相同的引用。 

    # Wrong
df['new'] = [[]] * len(df)
# Right
df['new'] = [[] for _ in range(len(df))]

生成副本： df.assign(new=0)

如果您需要副本，请使用DataFrame.assign：

df.assign(new='y')

   A  B  C new
0  x  x  x   y
1  x  x  x   y
2  x  x  x   y
3  x  x  x   y

而且，如果您需要分配多个具有相同值的列，这很简单，

c = ['new1', 'new2', ...]
df.assign(**dict.fromkeys(c, 'y'))

   A  B  C new1 new2
0  x  x  x    y    y
1  x  x  x    y    y
2  x  x  x    y    y
3  x  x  x    y    y

多列分配

最后，如果需要为多个列分配不同的值，则可以使用assign字典。

c = {'new1': 'w', 'new2': 'y', 'new3': 'z'}
df.assign(**c)

   A  B  C new1 new2 new3
0  x  x  x    w    y    z
1  x  x  x    w    y    z
2  x  x  x    w    y    z
3  x  x  x    w    y    z
点击查看英文原文

Super simple in-place assignment: df['new'] = 0

For in-place modification, perform direct assignment. This assignment is broadcasted by pandas for each row.

df = pd.DataFrame('x', index=range(4), columns=list('ABC'))
df

   A  B  C
0  x  x  x
1  x  x  x
2  x  x  x
3  x  x  x


df['new'] = 'y'
# Same as,
# df.loc[:, 'new'] = 'y'
df

   A  B  C new
0  x  x  x   y
1  x  x  x   y
2  x  x  x   y
3  x  x  x   y

Note for object columns

If you want to add an column of empty lists, here is my advice:

  • Consider not doing this. object columns are bad news in terms of performance. Rethink how your data is structured.
  • Consider storing your data in a sparse data structure. More information: sparse data structures

  • If you must store a column of lists, ensure not to copy the same reference multiple times. 

    # Wrong
df['new'] = [[]] * len(df)
# Right
df['new'] = [[] for _ in range(len(df))]

Generating a copy: df.assign(new=0)

If you need a copy instead, use DataFrame.assign:

df.assign(new='y')

   A  B  C new
0  x  x  x   y
1  x  x  x   y
2  x  x  x   y
3  x  x  x   y

And, if you need to assign multiple such columns with the same value, this is as simple as,

c = ['new1', 'new2', ...]
df.assign(**dict.fromkeys(c, 'y'))

   A  B  C new1 new2
0  x  x  x    y    y
1  x  x  x    y    y
2  x  x  x    y    y
3  x  x  x    y    y

Multiple column assignment

Finally, if you need to assign multiple columns with different values, you can use assign with a dictionary.

c = {'new1': 'w', 'new2': 'y', 'new3': 'z'}
df.assign(**c)

   A  B  C new1 new2 new3
0  x  x  x    w    y    z
1  x  x  x    w    y    z
2  x  x  x    w    y    z
3  x  x  x    w    y    z

回答 2

使用现代大熊猫，您可以：

df['new'] = 0
点击查看英文原文

With modern pandas you can just do:

df['new'] = 0

回答 3

这是另一种使用lambdas的班轮（创建常数值为10的列） 

df['newCol'] = df.apply(lambda x: 10, axis=1)

之前

df
    A           B           C
1   1.764052    0.400157    0.978738
2   2.240893    1.867558    -0.977278
3   0.950088    -0.151357   -0.103219

后

df
        A           B           C           newCol
    1   1.764052    0.400157    0.978738    10
    2   2.240893    1.867558    -0.977278   10
    3   0.950088    -0.151357   -0.103219   10
点击查看英文原文

Here is another one liner using lambdas (create column with constant value = 10) 

df['newCol'] = df.apply(lambda x: 10, axis=1)

before

df
    A           B           C
1   1.764052    0.400157    0.978738
2   2.240893    1.867558    -0.977278
3   0.950088    -0.151357   -0.103219

after

df
        A           B           C           newCol
    1   1.764052    0.400157    0.978738    10
    2   2.240893    1.867558    -0.977278   10
    3   0.950088    -0.151357   -0.103219   10