Question

我试图创建一个标志变量（即一个包含二进制值的新列，如1表示True，0表示False） - 我已经尝试了np.where {{3} }）和df.where无济于事。

使用df.where：

df.where(((df['MOSL_Rating'] == 'Highly Effective') & (df['MOTP_Rating'] == 'Developing')) | ((df['MOSL_Rating'] == 'Highly Effective') & (df['MOTP_Rating'] == 'Ineffective')) | ((df['MOSL_Rating'] == 'Effective') & (df['MOTP_Rating'] == 'Ineffective')) | ((df['MOSL_Rating'] == 'Ineffective') & (df['MOTP_Rating'] == 'Highly Effective')) | ((df['MOSL_Rating'] == 'Ineffective') & (df['MOTP_Rating'] == 'Effective')) | ((df['MOSL_Rating'] == 'Developing') & (df['MOTP_Rating'] == 'Highly Effective')), df['disp_rating'], 1, axis=1)

但这会返回ValueError: For argument "inplace" expected type bool, received type int.

如果我将代码从df['disp_rating'], 1, axis=1更改为df['disp_rating'], True, axis=1，则返回T ypeError: Cannot do inplace boolean setting on mixed-types with a non np.nan value

我还试过np.where，但返回ValueError: either both or neither of x and y should be given

我还阅读as per this post，看起来很相似。但是，当我使用其中提供的解决方案时，它返回： KeyError: 'disp_rating'

如果我提前创建变量（以避免Key Error），我只会收到有关其他内容的其他错误。

我认为根据一些基本条件创建一个新变量会非常简单，但我已经坚持了一段时间，尽管阅读了文档和很多SO帖子但我并没有取得任何进展

编辑：为了更清楚，我尝试根据其他2列（＆＃39; MOSL_Rating中的值）创建新列（名为＆＃39; disp_rating＆＃39;）并且在同一df内的＆＃39; MOTP_Rating＆＃39;）满足某些条件。我只有1个数据帧，因此我没有尝试比较2个数据帧。在SQL中我会使用CASE WHEN语句，在SAS中我会使用IF / THEN / ELSE语句。

我的df通常看起来像这样：

ID  Loc  MOSL_rating MOTP_Rating
12  54X  D           E   
45  86I  D           I    
98  65R  H           H

Answer 1

我找不到为什么哪里不起作用，但这是一种方法：

创建代码来创建你的df：

def make_row():
    import random
    dico = {"MOSL_Rating" : ['Highly Effective', 'Effective', 'Ineffective', 'Developing'],
            "MOTP_Rating" : ['Developing', 'Ineffective', 'Highly Effective', 'Effective', 'Highly Effective'],
           "disp_rating" : range(100)}

    row = {}
    for k in dico.keys():
        v = random.choice(dico[k])
        row[k] =v
    return row

def make_df(nb_row):
    import pandas as pd
    rows = [make_row() for i in range(nb_row)]
    return pd.DataFrame(rows)

我可以创建一个df：

df = make_df(3)

    MOSL_Rating MOTP_Rating disp_rating
0   Highly Effective    Ineffective 39
1   Highly Effective    Highly Effective    71
2   Effective   Ineffective 95

和一个借口：

df2 = make_df(3)
df2
    MOSL_Rating MOTP_Rating disp_rating
0   Effective   Highly Effective    24
1   Effective   Developing  38
2   Highly Effective    Ineffective 16

然后我创建了你的测试：

MOSL_high_efective   = df['MOSL_Rating'] == 'Highly Effective'
MOSL_efective        = df['MOSL_Rating'] == 'Effective'
MOSL_inefective      = df['MOSL_Rating'] == 'Ineffective'
MOSL_developing      = df['MOSL_Rating'] == 'Developing'

MOTP_high_efective   = df['MOTP_Rating'] == 'Highly Effective'
MOTP_efective        = df['MOTP_Rating'] == 'Effective'
MOTP_inefective      = df['MOTP_Rating'] == 'Ineffective'
MOTP_developing      = df['MOTP_Rating'] == 'Developing'

test1 = MOSL_high_efective & MOTP_developing
test2 = MOSL_high_efective & MOTP_inefective
test3 = MOSL_efective      & MOTP_inefective
test4 = MOSL_inefective    & MOTP_high_efective
test5 = MOSL_inefective    & MOTP_efective
test6 = MOSL_developing    & MOTP_high_efective

conditions  = test1 | test2 |  test3 | test4 | test5 | test6

然后用符合条件的第二个数据帧更新第一个数据帧的值：

    lines_to_be_updates = df.loc[conditions].index.values
    df.loc[lines_to_be_updates, "disp_rating"] = df2[lines_to_be_updates]["disp_rating"]

df
    MOSL_Rating MOTP_Rating disp_rating
0   Highly Effective    Ineffective 24
1   Highly Effective    Highly Effective    71
2   Effective   Ineffective 16

Answer 2

您的逻辑过于复杂，可以通过set进行简化/优化。以下是演示。

d = {frozenset({'H', 'D'}),
     frozenset({'H', 'I'}),
     frozenset({'E', 'I'})}

df['MOSL_MOTP'] = list(map(frozenset, zip(df['MOSL_Rating'], df['MOTP_Rating'])))
df['Result'] = np.where(df['MOSL_MOTP'].isin(d), 1, 0)

#    ID  Loc MOSL_Rating MOTP_Rating MOSL_MOTP  Result
# 0  12  54X           D           E    (E, D)       0
# 1  45  86I           D           I    (D, I)       0
# 2  98  65R           H           H       (H)       0
# 3  95  66R           H           D    (D, H)       1
# 4  96  67R           D           H    (D, H)       1
# 5  97  68R           E           I    (E, I)       1

使用np.where根据条件在pandas df中创建新列

2 个答案: