Python Pandas:如何插入一个缺失的行?

时间:2018-01-15 21:02:15

标签: python pandas dataframe pandas-groupby

对于以下数据框: 每组$args = array( 'post_type' => 'lp_lesson', 'numberposts' => 99999 ); 都应该有三个值cb的第二个值应该是a的第一个和第三个值的平均值。

插入"缺失"最简单的方法是什么?在aa=48之间的b=42c=4index = 0行?

index = 1

如果我使用df_x = pd.DataFrame({"a": [47, 49, 55, 54, 53, 24, 27, 30], "b": [41, 43, 51, 52, 53, 41, 42, 43], "c": [4, 4, 5, 5, 5, 4, 4, 4]}) df_x Out[14]: a b c 0 47 41 4 1 49 43 4 2 55 51 5 3 54 52 5 4 53 53 5 5 24 41 4 6 27 42 4 7 30 43 4 groupby('c').transform(my_func),我会遇到第一次调用我的函数groupby('c').apply(my_func)两次的情况。

2 个答案:

答案 0 :(得分:1)

pandas的insert方法仅适用于列。我们可以使用numpy.insert。缺点:这将创建一个新的数据集。这可以作为pd.concatpd.appendpd.merge的替代方案。

df_x = pd.DataFrame({"a": [47, 49, 55, 54, 53, 24, 27, 30], "b": [41, 43, 51, 52, 53, 41, 42, 43], "c": [4, 4, 5, 5, 5, 4, 4, 4]})

pd.DataFrame(np.insert(df_x.values, 1, values=[48, 42, 4], axis=0))


    0   1   2
0   47  41  4
1   48  42  4
2   49  43  4
3   55  51  5
4   54  52  5
5   53  53  5
6   24  41  4
7   27  42  4
8   30  43  4

np.insert(df_x.values, 1, values=[48, 42, 4], axis=0)中,1告诉函数您想要放置新值的位置/索引。

答案 1 :(得分:1)

半乳糖溶液将如下。

有没有人有想法如何使for循环更有效/分别避免它?

import pandas as pd
df_x = pd.DataFrame({"a": [47, 49, 55, 54, 53, 24, 27, 30], "b": [41, 43, 51, 52, 53, 41, 42, 43], "c": [4, 4, 5, 5, 5, 4, 4, 4]})
df_x
Out[11]: 
    a   b  c
0  47  41  4
1  49  43  4
2  55  51  5
3  54  52  5
4  53  53  5
5  24  41  4
6  27  42  4
7  30  43  4
# make new column that allows group by
df_x['cumsum']=(df_x.c != df_x.c.shift()).cumsum()
df_x
Out[14]: 
    a   b  c  cumsum
0  47  41  4       1
1  49  43  4       1
2  55  51  5       2
3  54  52  5       2
4  53  53  5       2
5  24  41  4       3
6  27  42  4       3
7  30  43  4       3
# introduce index spreaded by 10
df_x['index10'] = df_x.index * 10
print(df_x)
    a   b  c  cumsum  index10
0  47  41  4       1        0
1  49  43  4       1       10
2  55  51  5       2       20
3  54  52  5       2       30
4  53  53  5       2       40
5  24  41  4       3       50
6  27  42  4       3       60
7  30  43  4       3       70
groupby = df_x.groupby("cumsum")
# initialize dataframe for new row
df_x_append = pd.DataFrame()
for key, item in groupby:
    # sub dataframe is too small
    if item.shape[0]!=3:
        # new entry will be in the middle of existing values
        my_index = item.index10[0]+5
        # create temporary dataframe
        df_x_single = pd.DataFrame({"index10":[my_index], "a": [(item.a[0]+item.a[1])/2], "b": [(item.b[0]+item.b[1])/2],"c":[item.c[0]]})
        # append this dataframe
        df_x_append = df_x_append.append(df_x_single)
df_x=df_x.append(df_x_append)
# sort by spreaded index
df_x=df_x.sort_values(by='index10', ascending=True, na_position='first')
print(df_x)
      a     b  c  cumsum  index10
0  47.0  41.0  4     1.0        0
0  48.0  42.0  4     NaN        5
1  49.0  43.0  4     1.0       10
2  55.0  51.0  5     2.0       20
3  54.0  52.0  5     2.0       30
4  53.0  53.0  5     2.0       40
5  24.0  41.0  4     3.0       50
6  27.0  42.0  4     3.0       60
7  30.0  43.0  4     3.0       70
# set spreaded index and remove it
df_x=df_x.set_index('index10')
df_x = df_x.reset_index().drop(["index10"], axis=1)
print(df_x)
      a     b  c  cumsum
0  47.0  41.0  4     1.0
1  48.0  42.0  4     NaN
2  49.0  43.0  4     1.0
3  55.0  51.0  5     2.0
4  54.0  52.0  5     2.0
5  53.0  53.0  5     2.0
6  24.0  41.0  4     3.0
7  27.0  42.0  4     3.0
8  30.0  43.0  4     3.0