问题：Python Pandas将列表插入单元格
我有一个列表“ abc”和一个数据框“ df”：
abc = ['foo', 'bar']
df =
A B
0 12 NaN
1 23 NaN
我想将列表插入单元格1B中，所以我想要这个结果：
A B
0 12 NaN
1 23 ['foo', 'bar']
我能做到吗？
1）如果我使用这个：
df.ix[1,'B'] = abc
我收到以下错误消息：
ValueError: Must have equal len keys and value when setting with an iterable
因为它尝试将列表（具有两个元素）插入行/列而不插入单元格。
2）如果我使用这个：
df.ix[1,'B'] = [abc]
然后插入一个只有一个元素的列表，即“ abc”列表（ [['foo', 'bar']]）。
3）如果我使用这个：
df.ix[1,'B'] = ', '.join(abc)
然后插入一个字符串：（ foo, bar），但不插入列表。
4）如果我使用这个：
df.ix[1,'B'] = [', '.join(abc)]
然后插入一个列表，但只有一个元素（['foo, bar']），但没有两个我想要的元素（['foo', 'bar']）。
感谢帮助！
编辑
我的新数据框和旧列表：
abc = ['foo', 'bar']
df2 =
A B C
0 12 NaN 'bla'
1 23 NaN 'bla bla'
另一个数据框：
df3 =
A B C D
0 12 NaN 'bla' ['item1', 'item2']
1 23 NaN 'bla bla' [11, 12, 13]
我想将“ abc”列表插入df2.loc[1,'B']和/或df3.loc[1,'B']。
如果数据框仅包含具有整数值和/或NaN值和/或列表值的列，则将列表插入到单元格中的效果很好。如果数据框仅包含具有字符串值和/或NaN值和/或列表值的列，则将列表插入到单元格中的效果很好。但是，如果数据框具有包含整数和字符串值的列以及其他列，那么如果我使用此错误消息，则会出现错误消息：df2.loc[1,'B'] = abc或df3.loc[1,'B'] = abc。
另一个数据框：
df4 =
A B
0 'bla' NaN
1 'bla bla' NaN
这些插入可以完美地工作：df.loc[1,'B'] = abc或df4.loc[1,'B'] = abc。
I have a list ‘abc’ and a dataframe ‘df’:
abc = ['foo', 'bar']
df =
A B
0 12 NaN
1 23 NaN
I want to insert the list into cell 1B, so I want this result:
A B
0 12 NaN
1 23 ['foo', 'bar']
Ho can I do that?
1) If I use this:
df.ix[1,'B'] = abc
I get the following error message:
ValueError: Must have equal len keys and value when setting with an iterable
because it tries to insert the list (that has two elements) into a row / column but not into a cell.
2) If I use this:
df.ix[1,'B'] = [abc]
then it inserts a list that has only one element that is the ‘abc’ list ( [['foo', 'bar']] ).
3) If I use this:
df.ix[1,'B'] = ', '.join(abc)
then it inserts a string: ( foo, bar ) but not a list.
4) If I use this:
df.ix[1,'B'] = [', '.join(abc)]
then it inserts a list but it has only one element ( ['foo, bar'] ) but not two as I want ( ['foo', 'bar'] ).
Thanks for help!
EDIT
My new dataframe and the old list:
abc = ['foo', 'bar']
df2 =
A B C
0 12 NaN 'bla'
1 23 NaN 'bla bla'
Another dataframe:
df3 =
A B C D
0 12 NaN 'bla' ['item1', 'item2']
1 23 NaN 'bla bla' [11, 12, 13]
I want insert the ‘abc’ list into df2.loc[1,'B'] and/or df3.loc[1,'B'].
If the dataframe has columns only with integer values and/or NaN values and/or list values then inserting a list into a cell works perfectly. If the dataframe has columns only with string values and/or NaN values and/or list values then inserting a list into a cell works perfectly. But if the dataframe has columns with integer and string values and other columns then the error message appears if I use this: df2.loc[1,'B'] = abc or df3.loc[1,'B'] = abc.
Another dataframe:
df4 =
A B
0 'bla' NaN
1 'bla bla' NaN
These inserts work perfectly: df.loc[1,'B'] = abc or df4.loc[1,'B'] = abc.
回答 0
由于自0.21.0版set_value开始不推荐使用，因此您现在应该使用at。它可以插入一个列表的小区没有抚养ValueError的loc一样。我认为这是因为at 总是引用单个值，而loc可以引用值以及行和列。
df = pd.DataFrame(data={'A': [1, 2, 3], 'B': ['x', 'y', 'z']})
df.at[1, 'B'] = ['m', 'n']
df =
A B
0 1 x
1 2 [m, n]
2 3 z
您还需要确保要插入的列具有dtype=object。例如
>>> df = pd.DataFrame(data={'A': [1, 2, 3], 'B': [1,2,3]})
>>> df.dtypes
A int64
B int64
dtype: object
>>> df.at[1, 'B'] = [1, 2, 3]
ValueError: setting an array element with a sequence
>>> df['B'] = df['B'].astype('object')
>>> df.at[1, 'B'] = [1, 2, 3]
>>> df
A B
0 1 1
1 2 [1, 2, 3]
2 3 3
Since set_value has been deprecated since version 0.21.0, you should now use at. It can insert a list into a cell without raising a ValueError as loc does. I think this is because at always refers to a single value, while loc can refer to values as well as rows and columns.
df = pd.DataFrame(data={'A': [1, 2, 3], 'B': ['x', 'y', 'z']})
df.at[1, 'B'] = ['m', 'n']
df =
A B
0 1 x
1 2 [m, n]
2 3 z
You also need to make sure the column you are inserting into has dtype=object. For example
>>> df = pd.DataFrame(data={'A': [1, 2, 3], 'B': [1,2,3]})
>>> df.dtypes
A int64
B int64
dtype: object
>>> df.at[1, 'B'] = [1, 2, 3]
ValueError: setting an array element with a sequence
>>> df['B'] = df['B'].astype('object')
>>> df.at[1, 'B'] = [1, 2, 3]
>>> df
A B
0 1 1
1 2 [1, 2, 3]
2 3 3
回答 1
df3.set_value(1, 'B', abc)适用于任何数据框。注意列“ B”的数据类型。例如。不能将列表插入浮点列，在这种情况下df['B'] = df['B'].astype(object)可以提供帮助。
df3.set_value(1, 'B', abc) works for any dataframe. Take care of the data type of column ‘B’. Eg. a list can not be inserted into a float column, at that case df['B'] = df['B'].astype(object) can help.
回答 2
熊猫> = 0.21
set_value已不推荐使用。 现在，您可以使用DataFrame.at按标签DataFrame.iat设置和按整数位置设置。
# Setup
df = pd.DataFrame({'A': [12, 23], 'B': [['a', 'b'], ['c', 'd']]})
df
A B
0 12 [a, b]
1 23 [c, d]
df.dtypes
A int64
B object
dtype: object
如果要将“ B”第二行中的值设置为一些新列表，请使用DataFrane.at：
df.at[1, 'B'] = ['m', 'n']
df
A B
0 12 [a, b]
1 23 [m, n]
您也可以使用 DataFrame.iat
df.iat[1, df.columns.get_loc('B')] = ['m', 'n']
df
A B
0 12 [a, b]
1 23 [m, n]
如果得到了ValueError: setting an array element with a sequence怎么办？
我将尝试通过以下方式重现该内容：
df
A B
0 12 NaN
1 23 NaN
df.dtypes
A int64
B float64
dtype: object
df.at[1, 'B'] = ['m', 'n']
# ValueError: setting an array element with a sequence.
这是因为您的对象是float64dtype，而列表是objects，所以那里不匹配。在这种情况下，您要做的是先将列转换为对象。
df['B'] = df['B'].astype(object)
df.dtypes
A int64
B object
dtype: object
然后，它起作用：
df.at[1, 'B'] = ['m', 'n']
df
A B
0 12 NaN
1 23 [m, n]
可能，但是哈基
更古怪的是，我发现DataFrame.loc如果传递嵌套列表，您可以破解以实现相似的目的。
df.loc[1, 'B'] = [['m'], ['n'], ['o'], ['p']]
df
A B
0 12 [a, b]
1 23 [m, n, o, p]
您可以在这里阅读更多有关其工作原理的信息。
Pandas >= 0.21
set_value has been deprecated. You can now use DataFrame.at to set by label, and DataFrame.iat to set by integer position.
Setting Cell Values with at/iat
# Setup
df = pd.DataFrame({'A': [12, 23], 'B': [['a', 'b'], ['c', 'd']]})
df
A B
0 12 [a, b]
1 23 [c, d]
df.dtypes
A int64
B object
dtype: object
If you want to set a value in second row of the “B” to some new list, use DataFrane.at:
df.at[1, 'B'] = ['m', 'n']
df
A B
0 12 [a, b]
1 23 [m, n]
You can also set by integer position using DataFrame.iat
df.iat[1, df.columns.get_loc('B')] = ['m', 'n']
df
A B
0 12 [a, b]
1 23 [m, n]
What if I get ValueError: setting an array element with a sequence?
I’ll try to reproduce this with:
df
A B
0 12 NaN
1 23 NaN
df.dtypes
A int64
B float64
dtype: object
df.at[1, 'B'] = ['m', 'n']
# ValueError: setting an array element with a sequence.
This is because of a your object is of float64 dtype, whereas lists are objects, so there’s a mismatch there. What you would have to do in this situation is to convert the column to object first.
df['B'] = df['B'].astype(object)
df.dtypes
A int64
B object
dtype: object
Then, it works:
df.at[1, 'B'] = ['m', 'n']
df
A B
0 12 NaN
1 23 [m, n]
Possible, But Hacky
Even more wacky, I’ve found you can hack through DataFrame.loc to achieve something similar if you pass nested lists.
df.loc[1, 'B'] = [['m'], ['n'], ['o'], ['p']]
df
A B
0 12 [a, b]
1 23 [m, n, o, p]
You can read more about why this works here.
回答 3
回答 4
快速解决
只需将列表括在新列表中，就像在下面的数据框中对col2所做的那样。它起作用的原因是python获取（列表的）外部列表，并将其转换为列，就好像它包含普通标量项目一样，在我们的例子中是列表，而不是普通标量。
mydict={'col1':[1,2,3],'col2':[[1, 4], [2, 5], [3, 6]]}
data=pd.DataFrame(mydict)
data
col1 col2
0 1 [1, 4]
1 2 [2, 5]
2 3 [3, 6]
Quick work around
Simply enclose the list within a new list, as done for col2 in the data frame below. The reason it works is that python takes the outer list (of lists) and converts it into a column as if it were containing normal scalar items, which is lists in our case and not normal scalars.
mydict={'col1':[1,2,3],'col2':[[1, 4], [2, 5], [3, 6]]}
data=pd.DataFrame(mydict)
data
col1 col2
0 1 [1, 4]
1 2 [2, 5]
2 3 [3, 6]
回答 5
也得到
ValueError: Must have equal len keys and value when setting with an iterable，
在我的情况下，使用.at而不是.loc并没有任何区别，但是强制使用dataframe列的数据类型可以解决问题：
df['B'] = df['B'].astype(object)
然后，我可以将列表，numpy数组和所有类型的东西设置为数据帧中的单个单元格值。
Also getting
ValueError: Must have equal len keys and value when setting with an iterable,
using .at rather than .loc did not make any difference in my case, but enforcing the datatype of the dataframe column did the trick:
df['B'] = df['B'].astype(object)
Then I could set lists, numpy array and all sorts of things as single cell values in my dataframes.
