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.
Answers:
Thank you for visiting the Q&A section on Magenaut. Please note that all the answers may not help you solve the issue immediately. So please treat them as advisements. If you found the post helpful (or not), leave a comment & I’ll get back to you as soon as possible.
Method 1
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
Method 2
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” column to some new list, use DataFrame.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 that 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.
Method 3
df3.set_value(1, 'B', abc) works for any dataframe. Take care of the data type of column ‘B’. For example, a list can not be inserted into a float column, at that case df['B'] = df['B'].astype(object) can help.
Method 4
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]
Method 5
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.
Method 6
As mentionned in this post pandas: how to store a list in a dataframe?; the dtypes in the dataframe may influence the results, as well as calling a dataframe or not to be assigned to.
Method 7
I’ve got a solution that’s pretty simple to implement.
Make a temporary class just to wrap the list object and later call the value from the class.
Here’s a practical example:
- Let’s say you want to insert list object into the dataframe.
df = pd.DataFrame([
{'a': 1},
{'a': 2},
{'a': 3},
])
df.loc[:, 'b'] = [
[1,2,4,2,],
[1,2,],
[4,5,6]
] # This works. Because the list has the same length as the rows of the dataframe
df.loc[:, 'c'] = [1,2,4,5,3] # This does not work.
>>> ValueError: Must have equal len keys and value when setting with an iterable
## To force pandas to have list as value in each cell, wrap the list with a temporary class.
class Fake(object):
def __init__(self, li_obj):
self.obj = li_obj
df.loc[:, 'c'] = Fake([1,2,5,3,5,7,]) # This works.
df.c = df.c.apply(lambda x: x.obj) # Now extract the value from the class. This works.
Creating a fake class to do this might look like a hassle but it can have some practical applications. For an example you can use this with apply when the return value is list.
Pandas would normally refuse to insert list into a cell but if you use this method, you can force the insert.
Method 8
I prefer .at and .loc. It is important to note, that the target column needs a dtype (object), which can handle the list.
import numpy as np
import pandas as pd
df = pd.DataFrame({
'A': [0, 1, 2, 3],
'B': np.array([np.nan]*3 + [[3, 33]], dtype=object),
})
print('df to start with:', df, 'ndtypes:', df.dtypes, sep='n')
df.at[0, 'B'] = [0, 100] # at assigns single elemnt
df.loc[1, 'B'] = [[ [1, 11] ]] # loc expects 2d input
print('df modified:', df, 'ndtypes:', df.dtypes, sep='n')
output
df to start with: A B 0 0 NaN 1 1 NaN 2 2 NaN 3 3 [3, 33] dtypes: A int64 B object dtype: object df modified: A B 0 0 [0, 100] 1 1 [[1, 11]] 2 2 NaN 3 3 [3, 33] dtypes: A int64 B object dtype: object
Method 9
first set the cell to blank. next use at to assign the abc list to the cell at 1, ‘B’
abc = ['foo', 'bar']
df =pd.DataFrame({'A':[12,23],'B':[np.nan,np.nan]})
df.loc[1,'B']=''
df.at[1,'B']=abc
print(df)
All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0