Pandas - 添加一列，其值根据当前行和前一行中的另一列值计算得出
[英]Pandas - add a column with value computed based on another column value in current and previous row
问题描述
Given the dataframe below,
鉴于下面的 dataframe，
colNames = ["Time","Col2","Col3","Col4","Col5","Col6","Col7","Col8","Col9","Col10","Col11","Col12","Col13"]
colVals = [['05:17:55.703', '', '', '', '', '', '21', '', '3', '89', '891', '11', ''], ['05:17:55.703', '', '', '', '', '', '21', '', '3', '217', '891', '12', ''], ['05:17:55.703', '', '', '', '', '', '21', '', '3', '217', '891', '13', ''], ['05:17:55.703', '', '', '', '', '', '21', '', '3', '217', '891', '15', ''], ['05:17:55.703', '', '', '', '', '', '21', '', '3', '217', '891', '16', ''], ['05:17:55.703', '', '', '', '', '', '21', '', '3', '217', '891', '17', ''], ['05:17:55.703', '', '', '', '', '', '21', '', '3', '217', '891', '18', ''], ['05:17:55.707', '', '', '', '', '', '18', '', '3', '185', '892', '0', ''], ['05:17:55.707', '', '', '', '', '', '21', '', '3', '185', '892', '1', ''], ['05:17:55.707', '', '', '', '', '', '17', '', '3', '73', '892', '5', ''], ['05:17:55.707', '', '', '', '', '', '17', '', '3', '185', '892', '6', ''], ['05:17:55.707', '', '', '', '', '', '21', '', '3', '73', '892', '7', ''], ['05:17:55.708', '268', '4', '28', '-67.60', '13', '', '2', '', '', '', '', '2'], ['05:17:55.711', '', '', '', '', '', '18', '', '3', '57', '892', '10', ''], ['05:17:55.711', '', '', '', '', '', '21', '', '3', '201', '892', '11', ''], ['05:17:55.711', '', '', '', '', '', '21', '', '3', '25', '892', '12', ''], ['05:17:55.723', '', '', '', '', '', '21', '', '3', '217', '893', '11', ''], ['05:17:55.723', '', '', '', '', '', '21', '', '3', '217', '893', '15', ''], ['05:17:55.723', '', '', '', '', '', '21', '', '3', '217', '893', '16', ''], ['05:17:55.726', '268', '4', '', '-67.80', '', '', '', '', '', '', '', ''], ['05:17:55.728', '', '', '28', '', '12', '31', '2', '3', '185', '894', '0', '1'], ['05:17:55.728', '', '', '', '', '', '31', '', '3', '185', '894', '1', ''], ['05:17:55.731', '', '', '', '', '', '31', '', '3', '217', '894', '10', ''], ['05:17:55.731', '', '', '', '', '', '20', '', '3', '217', '894', '11', ''], ['05:17:55.731', '', '', '', '', '', '20', '', '3', '217', '894', '12', ''], ['05:17:55.731', '', '', '', '', '', '20', '', '3', '217', '894', '13', ''], ['05:17:55.743', '', '', '', '', '', '20', '', '3', '217', '895', '11', ''], ['05:17:55.743', '', '', '', '', '', '20', '', '3', '217', '895', '15', ''], ['05:17:55.743', '', '', '', '', '', '20', '', '3', '217', '895', '16', ''], ['05:17:55.746', '268', '4', '', '-67.82', '', '', '', '', '', '', '', ''], ['05:17:55.747', '', '', '28', '', '13', '20', '2', '3', '185', '896', '1', '2'], ['05:17:55.747', '', '', '', '', '', '20', '', '3', '185', '896', '2', ''], ['05:17:55.747', '', '', '', '', '', '30', '', '3', '217', '896', '5', ''], ['05:17:55.751', '', '', '', '', '', '18', '', '3', '217', '896', '10', ''], ['05:17:55.751', '', '', '', '', '', '21', '', '3', '217', '896', '11', ''], ['05:17:55.751', '', '', '', '', '', '21', '', '3', '217', '896', '12', ''], ['05:17:55.751', '', '', '', '', '', '21', '', '3', '217', '896', '13', ''], ['05:17:55.763', '', '', '', '', '', '31', '', '3', '217', '897', '11', ''], ['05:17:55.763', '', '', '', '', '', '30', '', '3', '217', '897', '15', ''], ['05:17:55.763', '', '', '', '', '', '20', '', '3', '217', '897', '16', ''], ['05:17:55.763', '', '', '', '', '', '20', '', '3', '217', '897', '17', ''], ['05:17:55.766', '268', '4', '', '-67.13', '', '', '', '', '', '', '', ''], ['05:17:55.768', '', '', '28', '', '12', '20', '2', '3', '185', '898', '3', '2'], ['05:17:55.768', '', '', '', '', '', '16', '', '3', '217', '898', '6', ''], ['05:17:55.771', '', '', '', '', '', '18', '', '3', '217', '898', '10', ''], ['05:17:55.771', '', '', '', '', '', '20', '', '3', '217', '898', '11', ''], ['05:17:55.771', '', '', '', '', '', '20', '', '3', '217', '898', '12', ''], ['05:17:55.784', '', '', '', '', '', '20', '', '3', '217', '899', '11', ''], ['05:17:55.784', '', '', '', '', '', '20', '', '3', '41', '899', '12', ''], ['05:17:55.784', '', '', '', '', '', '20', '', '3', '25', '899', '13', ''], ['05:17:55.784', '', '', '', '', '', '20', '', '3', '217', '899', '15', ''], ['05:17:55.784', '', '', '', '', '', '20', '', '3', 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'17', ''], ['05:17:55.823', '', '', '', '', '', '21', '', '3', '89', '903', '18', ''], ['05:17:55.826', '268', '4', '', '-67.51', '', '', '', '', '', '', '', ''], ['05:17:55.828', '', '', '28', '', '12', '18', '2', '3', '185', '904', '0', '1'], ['05:17:55.828', '', '', '', '', '', '21', '', '3', '185', '904', '1', ''], ['05:17:55.828', '', '', '', '', '', '21', '', '3', '185', '904', '2', ''], ['05:17:55.828', '', '', '', '', '', '21', '', '3', '185', '904', '3', ''], ['05:17:55.828', '', '', '', '', '', '17', '', '3', '217', '904', '5', ''], ['05:17:55.828', '', '', '', '', '', '17', '', '3', '217', '904', '6', ''], ['05:17:55.831', '', '', '', '', '', '21', '', '3', '217', '904', '7', ''], ['05:17:55.831', '', '', '', '', '', '20', '', '3', '169', '904', '11', ''], ['05:17:55.831', '', '', '', '', '', '20', '', '3', '217', '904', '12', ''], ['05:17:55.831', '', '', '', '', '', '20', '', '3', '217', '904', '13', ''], ['05:17:55.846', '268', '4', '', '-67.01', '', '', '', '', '', '', 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df = pd.DataFrame(colVals, columns=colNames)
df = df.set_index('Time')
df = df.apply(pd.to_numeric, errors='coerce')
Need to add new Col14 , such that it starts from 0 and increment by one whenever value in Col11 of current row is less than the value in Col11 of previous row - if the previous row's Col11 value is NaN , it shouldn't increment Col14 .需要添加新的Col14 ，这样当当前行的Col11中的值小于前一行的Col11中的值时，它从 0 开始并递增一 - 如果前一行的Col11值为NaN ，则不应递增Col14 。
For example例如
+-------+-------+-------------------------------------------------------------------------------------------------+
| Col11 | Col14 | |
+-------+-------+-------------------------------------------------------------------------------------------------+
| 900 | 0 | start with 0 |
| NaN | 0 | |
| 900 | 0 | |
| 903 | 0 | |
| 904 | 0 | |
| 904 | 0 | |
| 8 | 1 | increment Col14 by 1 when current row's Col11 value is less than the previous row's Col11 value |
| 8 | 1 | |
| 200 | 1 | |
| 201 | 1 | |
| NaN | 1 | if Col11 is NaN, Col12 takes same value as prev row |
| 0 | 2 | increment Col14 by 1 when current row's Col11 value is less than the previous row's Col11 value |
| 1 | 2 | |
| NaN | 2 | |
| NaN | 2 | |
| 0 | 3 | increment Col14 by 1 when current row's Col11 value is less than the previous row's Col11 value |
+-------+-------+-------------------------------------------------------------------------------------------------+
Is there a way to get reference to the previous row?有没有办法参考上一行？
1 个解决方案
解决方案1
2 已采纳 2019-09-20 15:01:01
Can you try this?你能试试这个吗？
df['col14']=df.Col11.gt(df.Col11.shift(-1)).cumsum()
or或者
df['col14a']=df.Col11.gt(df.Col11.shift(-1)).shift().fillna(0).cumsum().astype(int)
The difference between the two is that the count switches at the end of lower value (in the first one) and it switches at the start of new value (2nd one)两者之间的区别在于计数在较低值的末尾（第一个）切换，它在新值的开始（第二个）切换
If you want to consider NaN same as the previous non-Nan value, you the code below.如果您想将 NaN 视为与之前的非 Nan 值相同，请使用下面的代码。
df['col14b']=df.Col11.fillna(method='ffill').gt(df.Col11.fillna(method='ffill').shift(-1)).shift().fillna(0).cumsum().astype(int)
With the smaller dummy dataset you provided, output is below使用您提供的较小的虚拟数据集，output 如下
Col11 col14 col14a col14b
0 900.0 0 0 0
1 NaN 0 0 0
2 900.0 0 0 0
3 900.0 0 0 0
4 903.0 0 0 0
5 904.0 0 0 0
6 904.0 1 0 0
7 8.0 1 1 1
8 8.0 1 1 1
9 200.0 1 1 1
10 201.0 1 1 1
11 NaN 1 1 1
12 0.0 1 1 2
13 1.0 1 1 2
14 NaN 1 1 2
15 NaN 1 1 2
16 0.0 1 1 3
Details I am explaining the last one here, the other two are similar to this.细节我这里解释最后一个，其他两个与此类似。
With df.Col11.fillna(method='ffill') we are filling the NaN with the previous valid values, so tha out put is as below (we are not changing Coll11, we are just using this while creating the new column)使用df.Col11.fillna(method='ffill')我们用以前的有效值填充 NaN，所以输出如下（我们没有更改 Coll11，我们只是在创建新列时使用它）
0 900.0
1 900.0
2 900.0
3 900.0
4 903.0
5 904.0
6 904.0
7 8.0
8 8.0
9 200.0
10 201.0
11 201.0
12 0.0
13 1.0
14 1.0
15 1.0
16 0.0
With df.Col11.fillna(method='ffill').shift(-1) we are just shifting 1 row up so that we can do the comparison between the row & it's previous row.使用df.Col11.fillna(method='ffill').shift(-1)我们只是向上移动 1 行，以便我们可以在行和它的前一行之间进行比较。
0 900.0
1 900.0
2 900.0
3 903.0
4 904.0
5 904.0
6 8.0
7 8.0
8 200.0
9 201.0
10 201.0
11 0.0
12 1.0
13 1.0
14 1.0
15 0.0
16 NaN
With df.Col11.fillna(method='ffill').gt(df.Col11.fillna(method='ffill').shift(-1)) , we do the comparison & the result is true of false, as below.使用df.Col11.fillna(method='ffill').gt(df.Col11.fillna(method='ffill').shift(-1)) ，我们进行比较&结果为真假，如下.
0 False
1 False
2 False
3 False
4 False
5 False
6 True
7 False
8 False
9 False
10 False
11 True
12 False
13 False
14 False
15 True
16 False
We use shift to shift the value one row down since you want the number to switch when the smaller number is present.我们使用shift将值向下移动一行，因为您希望在较小的数字出现时切换数字。 When we shift rows down, the very first row becomes NaN , we are replacing that with 0 (since we have to start with zero) using .fillna(0) .当我们向下移动行时，第一行变为NaN ，我们使用.fillna(0) 。 Then we use .cumsum() .然后我们使用.cumsum() 。 .cumsum() adds only True values so, we get the values we need. .cumsum()只添加True值，所以我们得到了我们需要的值。 Finally, we assign this column as integer with .astype(int) to make all the values as integers.最后，我们将此列指定为 integer 和.astype(int)以使所有值都为整数。
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