Question

我一直在尝试找到最有效的方法。假设我有一个DataFrame df1，看起来像：

   time_start  time_end    
0  1548102229  1548102232  
1  1548102239  1548102242 
2  1548102249  1548102252
3  1548102259  1548102262

和另一个看起来像的DataFrame df2：

   timestamp   state    
0  1548102231  A  
1  1548102241  A 
2  1548102248  B
3  1548102251  B

考虑到df2 ['timestamp']介于df1 ['time_start']和df1 ['time_end']之间的条件，是否存在将“状态”添加到df1的方法：

   time_start  time_end    state
0  1548102229  1548102232  A
1  1548102239  1548102242  A
2  1548102249  1548102252  N/A
3  1548102259  1548102262  B

Answer 1

使用IntervalIndex和get_indexer，然后我们在.loc之后分配

idx=pd.IntervalIndex.from_arrays(df1['time_start'], df1['time_end'], closed='both')
indexmatch=idx.get_indexer(df2.timestamp)
df1['New']=df2.loc[indexmatch,'state'].values
df1
   time_start    time_end  New
0  1548102229  1548102232    A
1  1548102239  1548102242    A
2  1548102249  1548102252  NaN
3  1548102259  1548102262    B

更新

idx=pd.IntervalIndex.from_arrays(df1['time_start'], df1['time_end'], closed='both')
indexmatch=idx.get_indexer(df2.timestamp)
dfcopy=df1.copy()
df1=df1.loc[indexmatch]
df1['New']=df2.loc[indexmatch,'state'].values
df1.groupby(['time_start','time_end'],as_index=False).New.sum().combine_first(dfcopy)

Answer 2

使用np.less_equal和np.greater_equal outer ufuncs

c = np.less_equal.outer(df2.timestamp, df.time_end) & \
    np.greater_equal.outer(df2.timestamp, df.time_start)

df['state'] = df2.state.values[c.argmax(1)]

然后更正所有False个结果

df.loc[~c.any(1), 'state'] = np.nan

    time_start  time_end    state
0   1548102229  1548102232  A
1   1548102239  1548102242  A
2   1548102249  1548102252  NaN
3   1548102259  1548102262  B

给定条件将值添加到pandas DataFrame

2 个答案: