Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
1.3k views
in Technique[技术] by (71.8m points)

python - Must have equal len keys and value when setting with an iterable

I have two dataframe as flow:

leader:
    0 11
    1 8
    2 5
    3 9
    4 8
    5 6
    [6065 rows x 2 columns]

DatasetLabel:    
    Unnamed: 0      0    1  ....    7     8    9  10  11  12  
    0               A    J  ....    1     2    5 NaN NaN NaN  
    1               B    K  ....    3     4   NaN  NaN NaN NaN  

    [4095 rows x 14 columns]

The Information dataset column names 0 to 6 are DatasetLabel about data and 7 to 12 are indexes where refer to first column of leader Dataframe.

I want to create dataset where instead of the indexes in DatasetLabel Dataset I have the value of each index from the leader dataset which is leader.iloc[index,1]

How can I do it using python features?

The output should look like:

 DatasetLabel:    
        Unnamed: 0      0    1  ....    7     8    9  10  11  12  
        0               A    J  ....    8     5    6 NaN NaN NaN  
        1               B    K  ....    9     8   NaN  NaN NaN NaN  

I have came up with following, but I get error:

    for column in DatasetLabel.ix[:,8:13]:
        DatasetLabel[DatasetLabel[column].notnull ()]=leader.iloc[DatasetLabel[DatasetLabel[column].notnull ()][column].values,1]

Error:

ValueError: Must have equal len keys and value when setting with an iterable
See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

You can use apply to index into leader and exchange values with DatasetLabel, although it's not very pretty.

One issue is that Pandas won't let us index with NaN. Converting to str provides a workaround. But that creates a second issue, namely, column 9 is of type float (because NaN is float), so 5 becomes 5.0. Once it's a string, that's "5.0", which will fail to match the index values in leader. We can remove the .0, and then this solution will work - but it's a bit of a hack.

With DatasetLabel as:

   Unnamed:0  0  1  7  8    9  10  11  12
0          0  A  J  1  2  5.0 NaN NaN NaN
1          1  B  K  3  4  NaN NaN NaN NaN

And leader as:

   0   1
0  0  11
1  1   8
2  2   5
3  3   9
4  4   8
5  5   6

Then:

cols = ["7","8","9","10","11","12"]
updated = DatasetLabel[cols].apply(
    lambda x: leader.loc[x.astype(str).str.split(".").str[0], 1].values, axis=1)

updated
     7    8    9  10  11  12
0  8.0  5.0  6.0 NaN NaN NaN
1  9.0  8.0  NaN NaN NaN NaN

Now we can concat the unmodified columns (which we'll call original) with updated:

original_cols = DatasetLabel.columns[~DatasetLabel.columns.isin(cols)]
original = DatasetLabel[original_cols]
pd.concat([original, updated], axis=1)

Output:

   Unnamed:0  0  1    7    8    9  10  11  12
0          0  A  J  8.0  5.0  6.0 NaN NaN NaN
1          1  B  K  9.0  8.0  NaN NaN NaN NaN

Note: It may be clearer to use concat here, but here's another, cleaner way of merging original and updated, using assign:

DatasetLabel.assign(**updated)

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...