条件生成新列-Pandas
[英]Conditional generation of new column - Pandas
问题描述
I am trying to create a new column based on conditional logic on pre-existing columns.
我正在尝试根据现有列的条件逻辑创建一个新列。 I understand there may be more efficient ways to achieve this but I have a few conditions that need to be included.我知道可能有更有效的方法来实现这一点，但我有一些条件需要包括在内。 This is just the first step.这只是第一步。
The overall scope is to create two new columns that are mapped from 1 and 2 .整体 scope 是创建两个从1和2映射的新列。 These are referenced to the Object column as I can have multiple rows for each time point.这些被引用到Object列，因为每个时间点我可以有多行。
Object2 and Value determine how to map the new columns. Object2和Value确定如何 map 新列。 So if Value is == X , I want to match both Object columns to return the corresponding 1 and 2 for that time point to a new column.因此，如果Value is == X ，我想匹配两个Object列，以将该时间点的相应1和2返回到新列。 The same process should occur if Value is == Y .如果Value is == Y ，则应该发生相同的过程。 If Value is == Z , I want to insert 0, 0 .如果Value is == Z ，我想插入0, 0 。 Everything else should be NaN其他一切都应该是NaN
df = pd.DataFrame({
'Time' : ['2019-08-02 09:50:10.1','2019-08-02 09:50:10.1','2019-08-02 09:50:10.2','2019-08-02 09:50:10.3','2019-08-02 09:50:10.3','2019-08-02 09:50:10.4','2019-08-02 09:50:10.5','2019-08-02 09:50:10.6','2019-08-02 09:50:10.6'],
'Object' : ['B','A','A','A','C','C','C','B','B'],
'1' : [1,3,5,7,9,11,13,15,17],
'2' : [0,1,4,6,8,10,12,14,16],
'Object2' : ['A','A',np.nan,'C','C','C','C','B','A'],
'Value' : ['X','X',np.nan,'Y','Y','Y','Y','Z',np.nan],
})
def map_12(df):
for i in df['Value']:
if i == 'X':
df['A1'] = df['1']
df['A2'] = df['2']
elif i == 'Y':
df['A1'] = df['1']
df['A2'] = df['2']
elif i == 'Z':
df['A1'] = 0
df['A2'] = 0
else:
df['A1'] = np.nan
df['A2'] = np.nan
return df
Intended Output:预期 Output：
Time Object 1 2 Object2 Value A1 A2
0 2019-08-02 09:50:10.1 A 1 0 A X 1.0 0.0 # Match A-A at this time point, so output is 1,0
1 2019-08-02 09:50:10.1 B 3 1 A X 1.0 0.0 # Still at same time point so use 1,0
2 2019-08-02 09:50:10.2 A 5 4 NaN NaN NaN NaN # No Value so NaN
3 2019-08-02 09:50:10.3 C 7 6 C Y 7.0 6.0 # Match C-C at this time point, so output is 7,6
4 2019-08-02 09:50:10.3 A 9 8 C Y 7.0 6.0 # Still at same time point so use 7,6
5 2019-08-02 09:50:10.4 C 11 10 C Y 11.0 10.0 # Match C-C at this time point, so output is 11,10
6 2019-08-02 09:50:10.5 C 13 12 C Y 13.0 12.0 # Match C-C at this time point, so output is 13,12
7 2019-08-02 09:50:10.6 B 15 14 B Z 0.0 0.0 # Z so 0,0
8 2019-08-02 09:50:10.6 B 17 16 A NaN NaN NaN # No Value so NaN
New sample df:新样本df：
df = pd.DataFrame({
'Time' : ['2019-08-02 09:50:10.1','2019-08-02 09:50:10.1','2019-08-02 09:50:10.2','2019-08-02 09:50:10.3','2019-08-02 09:50:10.3','2019-08-02 09:50:10.4','2019-08-02 09:50:10.5','2019-08-02 09:50:10.6','2019-08-02 09:50:10.6'],
'Object' : ['B','A','A','A','C','C','C','B','B'],
'1' : [1,3,5,7,9,11,13,15,17],
'2' : [0,1,4,6,8,10,12,14,16],
'Object2' : ['A','A',np.nan,'C','C','C','C','B','A'],
'Value' : ['X','X',np.nan,'Y','Y','Y','Y','Z',np.nan],
})
Intended Output:预期 Output：
Time Object 1 2 Object2 Value A1 A2
0 2019-08-02 09:50:10.1 B 1 0 A X 3.0 1.0 # Match A-A at this time point, so output is 3,1
1 2019-08-02 09:50:10.1 A 3 1 A X 3.0 1.0 # Still at same time point so use 3,1
2 2019-08-02 09:50:10.2 A 5 4 NaN NaN NaN NaN # No Value so NaN
3 2019-08-02 09:50:10.3 A 7 6 C Y 9.0 8.0 # Match C-C at this time point, so output is 9,8
4 2019-08-02 09:50:10.3 C 9 8 C Y 9.0 8.0 # Still at same time point so use 9,8
5 2019-08-02 09:50:10.4 C 11 10 C Y 11.0 10.0 # Match C-C at this time point, so output is 11,10
6 2019-08-02 09:50:10.5 C 13 12 C Y 13.0 12.0 # Match C-C at this time point, so output is 13,12
7 2019-08-02 09:50:10.6 B 15 14 B Z 0.0 0.0 # Z so 0,0
8 2019-08-02 09:50:10.6 B 17 16 A NaN NaN NaN # No Value so NaN
4 个解决方案
解决方案1
2 已采纳 2019-11-06 10:23:23
Use DataFrame.where + DataFrame.eq to create a DataFrame similar to df[['1','2']] but only with the rows where matches is True and the rest with NaN . Use DataFrame.where + DataFrame.eq to create a DataFrame similar to df[['1','2']] but only with the rows where matches is True and the rest with NaN . Then group by time points using DataFrame.groupby and fill in the missing data of each group with the existing values where Object and Object2 ( matches==True ) coincide.然后使用 DataFrame.groupby 按时间点分组，并用DataFrame.groupby和Object2 ( matches==True ) 重合的现有值Object每组的缺失数据。 Use DataFrame.where to discart values where df['Value'] is NaN .Finally use [ DataFrame.mask ] to set 0 when Z is in the column Value使用DataFrame.where舍弃df['Value']为NaN的值。最后使用 [ DataFrame.mask ] 当Z在列Value中时设置为 0
#matches
matches=df.Object.eq(df.Object2)
#Creating conditions
condition_z=df['Value']=='Z'
not_null=df['Value'].notnull()
#Creating DataFrame to fill
df12=( df[['1','2']].where(matches)
.groupby(df['Time'],sort=False)
.apply(lambda x: x.ffill().bfill()) )
#fill 0 on Value is Z and discarting NaN
df[['A1','A2']] =df12.where(not_null).mask(condition_z,0)
print(df)
Output Output
Time Object 1 2 Object2 Value A1 A2
0 2019-08-02 09:50:10.1 B 1 0 A X 3.0 1.0
1 2019-08-02 09:50:10.1 A 3 1 A X 3.0 1.0
2 2019-08-02 09:50:10.2 A 5 4 NaN NaN NaN NaN
3 2019-08-02 09:50:10.3 A 7 6 C Y 9.0 8.0
4 2019-08-02 09:50:10.3 C 9 8 C Y 9.0 8.0
5 2019-08-02 09:50:10.4 C 11 10 C Y 11.0 10.0
6 2019-08-02 09:50:10.5 C 13 12 C Y 13.0 12.0
7 2019-08-02 09:50:10.6 B 15 14 B Z 0.0 0.0
8 2019-08-02 09:50:10.6 B 17 16 A NaN NaN NaN
We can also use GroupBy.transform :我们也可以使用GroupBy.transform ：
#matches
matches=df.Object.eq(df.Object2)
#Creating conditions
condition_z=df['Value']=='Z'
not_null=df['Value'].notnull()
#Creating DataFrame to fill
df12=( df[['1','2']].where(matches)
.groupby(df['Time'],sort=False)
.transform('first') )
#fill 0 on Value is Z and discarting NaN
df[['A1','A2']] =df12.where(not_null).mask(condition_z,0)
print(df)
解决方案2
1 2019-11-06 10:32:02
If only few conditions use DataFrame.loc for assign values by condition:如果只有少数条件使用DataFrame.loc按条件赋值：
m1 = df['Value'].isin(['X','Y'])
m2 = df['Value'] == 'Z'
df[['A1','A2']] = df.loc[m1, ['1','2']]
df.loc[m2, ['A1','A2']] = 0
print(df)
Time Object 1 2 Object2 Value A1 A2
0 2019-08-02 09:50:10.1 A 1 0 A X 1.0 0.0
1 2019-08-02 09:50:10.1 B 1 1 A X 1.0 1.0
2 2019-08-02 09:50:10.2 A 5 4 NaN NaN NaN NaN
3 2019-08-02 09:50:10.3 C 7 6 C Y 7.0 6.0
4 2019-08-02 09:50:10.3 A 9 8 C Y 9.0 8.0
5 2019-08-02 09:50:10.4 C 11 10 NaN NaN NaN NaN
6 2019-08-02 09:50:10.5 C 13 12 B NaN NaN NaN
7 2019-08-02 09:50:10.6 B 15 14 B Z 0.0 0.0
8 2019-08-02 09:50:10.6 B 17 16 B NaN NaN NaN
Another solution with numpy.select and broadcasting of masks: numpy.select和广播掩码的另一种解决方案：
m1 = df['Value'].isin(['X','Y'])
m2 = df['Value'] == 'Z'
masks = [m1.values[:, None], m2.values[:, None]]
values = [df[['1','2']].values, 0]
df[['A1','A2']] = pd.DataFrame(np.select(masks,values, default=np.nan), index=df.index)
print(df)
Time Object 1 2 Object2 Value A1 A2
0 2019-08-02 09:50:10.1 A 1 0 A X 1.0 0.0
1 2019-08-02 09:50:10.1 B 1 1 A X 1.0 1.0
2 2019-08-02 09:50:10.2 A 5 4 NaN NaN NaN NaN
3 2019-08-02 09:50:10.3 C 7 6 C Y 7.0 6.0
4 2019-08-02 09:50:10.3 A 9 8 C Y 9.0 8.0
5 2019-08-02 09:50:10.4 C 11 10 NaN NaN NaN NaN
6 2019-08-02 09:50:10.5 C 13 12 B NaN NaN NaN
7 2019-08-02 09:50:10.6 B 15 14 B Z 0.0 0.0
8 2019-08-02 09:50:10.6 B 17 16 B NaN NaN NaN
解决方案3
0 2019-11-04 05:12:52
Have a look at Dataframe apply看看Dataframe 申请
df['A1'] = df.apply(lambda row: row['1'] if row['Value'] == 'X' else np.nan, axis=1)
解决方案4
0 2019-11-06 10:05:24
I had to make a few adjustments to your dataframe, as it didn't match the desired result in your question.我不得不对您的 dataframe 进行一些调整，因为它与您问题中的预期结果不符。
df = pd.DataFrame(
{
"Time": [
"2019-08-02 09:50:10.1",
"2019-08-02 09:50:10.1",
"2019-08-02 09:50:10.2",
"2019-08-02 09:50:10.3",
"2019-08-02 09:50:10.3",
"2019-08-02 09:50:10.4",
"2019-08-02 09:50:10.5",
"2019-08-02 09:50:10.6",
"2019-08-02 09:50:10.6",
],
"Object": ["A", "B", "A", "C", "A", "C", "C", "B", "B"],
"1": [1, 1, 5, 7, 9, 11, 13, 15, 17],
"2": [0, 1, 4, 6, 8, 10, 12, 14, 16],
"Object2": ["A", "A", np.nan, "C", "C", "C", "C", "B", "A"],
"Value": ["X", "X", np.nan, "Y", "Y", "Y", "Y", "Z", np.nan],
}
)
This is a vectorized solution that should perform well over large data.这是一个矢量化解决方案，应该在大数据上表现良好。
First step is to make sure the dataframe is sorted by time.第一步是确保 dataframe 按时间排序。
df = df.sort_values("Time")
Copy columns 1 and 2复制第 1 列和第 2 列
df["A1"] = df["1"]
df["A2"] = df["2"]
Going to use the index values to obtain the first row of each time group.将使用索引值来获取每个时间组的第一行。
df = df.reset_index()
I'm not that happy with the list/isin solution.我对 list/isin 解决方案不太满意。 Curious if anyone knows a less hacky way to do this?好奇是否有人知道一种不那么老套的方法来做到这一点？
li = df.groupby("Time").index.first().tolist()
print(li)
[0, 2, 3, 5, 6, 7]
print(df)
index Time Object 1 2 Object2 Value A1 A2
0 0 2019-08-02 09:50:10.1 A 1 0 A X 1 0
1 1 2019-08-02 09:50:10.1 B 1 1 A X 1 1
2 2 2019-08-02 09:50:10.2 A 5 4 NaN NaN 5 4
3 3 2019-08-02 09:50:10.3 C 7 6 C Y 7 6
4 4 2019-08-02 09:50:10.3 A 9 8 C Y 9 8
5 5 2019-08-02 09:50:10.4 C 11 10 C Y 11 10
6 6 2019-08-02 09:50:10.5 C 13 12 C Y 13 12
7 7 2019-08-02 09:50:10.6 B 15 14 B Z 15 14
8 8 2019-08-02 09:50:10.6 B 17 16 A NaN 17 16
Filter the dataframe to get all rows except the ones in the list then set them to np.NaN过滤 dataframe 以获取除列表中的行之外的所有行，然后将它们设置为 np.NaN
df.loc[~df.index.isin(li), ["A1", "A2"]] = np.NaN
Fill forward the first row values.向前填充第一行值。
df[["A1", "A2"]] = df[["A1", "A2"]].ffill(axis=0)
Set z to 0 and np.NaN to np.NaN将 z 设置为 0 并将 np.NaN 设置为 np.NaN
df.loc[df["Value"] == "Z", ["A1", "A2"]] = 0
df.loc[df["Value"].isnull(), ["A1", "A2"]] = np.NaN
Remove index column删除索引列
df = df.drop("index", axis=1)
print(df)
Time Object 1 2 Object2 Value A1 A2
0 2019-08-02 09:50:10.1 A 1 0 A X 1.0 0.0
1 2019-08-02 09:50:10.1 B 1 1 A X 1.0 0.0
2 2019-08-02 09:50:10.2 A 5 4 NaN NaN NaN NaN
3 2019-08-02 09:50:10.3 C 7 6 C Y 7.0 6.0
4 2019-08-02 09:50:10.3 A 9 8 C Y 7.0 6.0
5 2019-08-02 09:50:10.4 C 11 10 C Y 11.0 10.0
6 2019-08-02 09:50:10.5 C 13 12 C Y 13.0 12.0
7 2019-08-02 09:50:10.6 B 15 14 B Z 0.0 0.0
8 2019-08-02 09:50:10.6 B 17 16 A NaN NaN NaN
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