Question

我有这个数据框

并且正在尝试将在前两列中具有NaNs的行向左移动，因此，现在右边的值将填充此列。这是我目前正在尝试做的事情：

（注意：match数据帧是从以下链接下载的：https://www.kaggle.com/hugomathien/soccer）

#original dataframe
<class 'pandas.core.frame.DataFrame'>
Int64Index: 21374 entries, 145 to 25978
Data columns (total 47 columns):
id                  21374 non-null int64
country_id          21374 non-null int64
league_id           21374 non-null int64
season              21374 non-null object
stage               21374 non-null int64
date                21374 non-null object
match_api_id        21374 non-null int64
home_team_api_id    21374 non-null int64
away_team_api_id    21374 non-null int64
home_team_goal      21374 non-null int64
away_team_goal      21374 non-null int64
goal                13325 non-null object
shoton              13325 non-null object
shotoff             13325 non-null object
foulcommit          13325 non-null object
card                13325 non-null object
cross               13325 non-null object
corner              13325 non-null object
possession          13325 non-null object
BSA                 11856 non-null float64
Home Team           21374 non-null object
Away Team           21374 non-null object
League              21374 non-null object
Country             21374 non-null object
home_player_1       21374 non-null object
home_player_2       21374 non-null object
home_player_3       21374 non-null object
home_player_4       21374 non-null object
home_player_5       21374 non-null object
home_player_6       21374 non-null object
home_player_7       21374 non-null object
home_player_8       21374 non-null object
home_player_9       21374 non-null object
home_player_10      21374 non-null object
home_player_11      21374 non-null object
away_player_1       21374 non-null object
away_player_2       21374 non-null object
away_player_3       21374 non-null object
away_player_4       21374 non-null object
away_player_5       21374 non-null object
away_player_6       21374 non-null object
away_player_7       21374 non-null object
away_player_8       21374 non-null object
away_player_9       21374 non-null object
away_player_10      21374 non-null object
away_player_11      21374 non-null object
winner              21374 non-null object
dtypes: float64(1), int64(9), object(37)
memory usage: 7.8+ MB

创建数据框

columns = match.columns[match.columns.get_loc('home_player_1'):match.columns.get_loc('away_player_1')+1].values
columns = list(columns)

player_appearences = match.groupby(columns[0]).size().reset_index()
player_appearences.rename(columns = {0:"Count_{}".format(player_appearences.columns[0][len(player_appearences.columns[0])-1])}, inplace = True, errors='raise')
player_appearences
for i in range(1,12):
    player_appearences2 = match.groupby(columns[i]).size().reset_index()
    player_appearences2
    player_appearences2.rename(columns = {0:"Count_{}".format(player_appearences2.columns[0][len(player_appearences2.columns[0])-1])}, inplace = True, errors='raise')
    player_appearences = player_appearences.merge(right = player_appearences2,how="outer",left_on ="{}".format(player_appearences.columns[0]),right_on = "{}".format(player_appearences2.columns[0]))
    player_appearences
    #overwrite nans in first column with names in current [i] player column

#select rows where first two columns give nan values
player_appearences.loc[(player_appearences.loc[:,"home_player_1"].isna()==True) & (player_appearences.loc[:,"Count_1"].isna()==True),["home_player_1","Count_1"]] = player_appearences.loc[(player_appearences.loc[:,"home_player_1"].isna()==True) & (player_appearences.loc[:,"Count_1"].isna()==True),["home_player_2","Count_2"]]

当我然后打印player_appearences时，数据框未更改。我不确定它是什么也不做，还是正在创建原始数据帧的副本。谁能告诉我为什么没有这种方法/建议如果有更好的方法呢？

Answer 1

使用DataFrame.rename，则只需要DataFrame.stack（默认情况下为 dropna = True ）+ DataFrame.unstack：

 df = (df.rename(columns = {'home_player_2':'home_player_1',
                           'Count_2':'Count_1'}).stack().unstack()
       .reindex(columns = df.columns[:2]))
print(df)
  home_player_1 Count_1
0         Aaron       1
1          Adam       2
2         Ziggy       3
3        Zoltan       4

或将DataFrame.shift与DataFrame.where：

df.where(df.notna(),df.shift(-1,axis = 1)).iloc[:,:2]


  home_player_1  Count_1
0         Aaron      1.0
1          Adam      2.0
2         Ziggy      3.0
3        Zoltan      4.0

详细信息

print(df.where(df.notna(),df.shift(-1,axis = 1)))
  home_player_1  Count_1 home_player_2  Count_2
0         Aaron      1.0           NaN      NaN
1          Adam      2.0           NaN      NaN
2         Ziggy      3.0         Ziggy      3.0
3        Zoltan      4.0        Zoltan      4.0

Answer 2

您可以使用shift(-1, axis=1)移动列，并使用df[df.home_player_1.isna() & df.Count_1.isna()]指定要影响的行。您要移动的行应在数据框中重写。

df = pd.DataFrame([['Aaron', 1, None, None],
                   ['Adam', 2, None, None],
                   [None, None, 'Ziggy', 3],
                   [None, None, 'Zoltan', 4]],
                  columns=['home_player_1', 'Count_1', 'home_player_2', 'Count_2'])

home_player_1   Count_1     home_player_2   Count_2
Aaron           1.0         None            NaN
Adam            2.0         None            NaN
None            NaN         Ziggy           3.0
None            NaN         Zoltan          4.0

df[df.home_player_1.isna() & df.Count_1.isna()] = df[df.home_player_1.isna() & df.Count_1.isna()].shift(-1, axis=1)

home_player_1   Count_1     home_player_2   Count_2
Aaron           1.0         None            NaN
Adam            2.0         None            NaN
Ziggy           3.0         NaN             NaN
Zoltan          4.0         NaN             NaN

转移熊猫数据框的特定列的特定行

2 个答案: