删除数据框中的集合中的“NaN”值

问题描述 投票:1回答:1

我有一个看起来像这样但有多行的数据框:

    column_1         column_2         column_3
        1         {lk, 18m, NaN}    {kjaf, NaN} 

我想从每一组中取出NaN,但迭代遍历行导致RuntimeError: Set changed size during iteration

我到目前为止使用的代码如下:

for index, row in df.iterrows():
    col2 = row['column_2']
    col3 = row['column_3']
    for x in col2:
        col2.discard('NaN')
    for y in col3:
        col3.discard('NaN')
python pandas set nan
1个回答
1
投票

如果ifs缺少值,则可以在集合理解中使用NaN

df = pd.DataFrame({'column_1': [1, 1], 
                   'column_2': [[np.nan, '18m'], ['lk', 'r']],
                   'column_3': [['kjaf'], ['ddd']]})

print (df)
   column_1    column_2 column_3
0         1  [nan, 18m]   [kjaf]
1         1     [lk, r]    [ddd]

cols = ['column_2', 'column_3']
df[cols] = df[cols].applymap(lambda x: set([i for i in x if pd.notna(i)]))
#oldier pandas versions
#df[cols] = df[cols].applymap(lambda x: set([i for i in x if pd.notnull(i)]))
print (df)

   column_1 column_2 column_3
0         1    {18m}   {kjaf}
1         1  {r, lk}    {ddd}

如果NaNs是字符串:

df = pd.DataFrame({'column_1': [1, 1], 
                   'column_2': [['NaN', '18m'], ['lk', 'r']],
                   'column_3': [['kjaf'], ['ddd']]})

print (df)
   column_1    column_2 column_3
0         1  [NaN, 18m]   [kjaf]
1         1     [lk, r]    [ddd]

cols = ['column_2', 'column_3']
df[cols] = df[cols].applymap(lambda x: set([i for i in x if i != 'NaN']))
print (df)

   column_1 column_2 column_3
0         1    {18m}   {kjaf}
1         1  {r, lk}    {ddd}
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