如何遍历数据框的列？
How to iterate through columns of the dataframe?
我想遍历数据框的所有列。这样我就可以得到该列的特定数据，使用这些数据我必须为另一个数据框计算。
我这里有 :
DP1 DP2 DP3 DP4 DP5 DP6 DP7 DP8 DP9 DP10 Total
OP1 357848.0 1124788.0 1735330.0 2218270.0 2745596.0 3319994.0 3466336.0 3606286.0 3833515.0 3901463.0 3901463.0
OP2 352118.0 1236139.0 2170033.0 3353322.0 3799067.0 4120063.0 4647867.0 4914039.0 5339085.0 NaN 5339085.0
OP3 290507.0 1292306.0 2218525.0 3235179.0 3985995.0 4132918.0 4628910.0 4909315.0 NaN NaN 4909315.0
OP4 310608.0 1418858.0 2195047.0 3757447.0 4029929.0 4381982.0 4588268.0 NaN NaN NaN 4588268.0
OP5 443160.0 1136350.0 2128333.0 2897821.0 3402672.0 3873311.0 NaN NaN NaN NaN 3873311.0
OP6 396132.0 1333217.0 2180715.0 2985752.0 3691712.0 NaN NaN NaN NaN NaN 3691712.0
OP7 440832.0 1288463.0 2419861.0 3483130.0 NaN NaN NaN NaN NaN NaN 3483130.0
OP8 359480.0 1421128.0 2864498.0 NaN NaN NaN NaN NaN NaN NaN 2864498.0
OP9 376686.0 1363294.0 NaN NaN NaN NaN NaN NaN NaN NaN 1363294.0
OP10 344014.0 NaN NaN NaN NaN NaN NaN NaN NaN NaN 344014.0
Total 3671385.0 11614543.0 17912342.0 21930921.0 21654971.0 19828268.0 17331381.0 13429640.0 9172600.0 3901463.0 34358090.0
Latest Observation 344014.0 1363294.0 2864498.0 3483130.0 3691712.0 3873311.0 4588268.0 4909315.0 5339085.0 3901463.0 NaN
从这个 table 我想计算这个公式这个公式：在 DP1 列中，Total/Last 观察，这个答案是除以 DP2 列总数。像这样我们必须计算所有列并将其保存在另一个数据框中。
我们需要这样的行：
Weighted Average 3.491 1.747 1.457 1.174 1.104 1.086 1.054 1.077 1.018
我们试过的代码：
LDFTriangledf['Weighted Average'] =CumulativePaidTriangledf.loc['Total','DP2']/(CumulativePaidTriangledf.loc['Total','DP1'] - CumulativePaidTriangledf.loc['Latest Observation','DP1'])
您可以从 .loc 中删除列名，只删除 shift(-1, axis=1) 以获得下一列的 Total。这使您可以在一次操作中将公式应用于所有列：
CumulativePaidTriangledf.shift(-1, axis=1).loc['Total'] / (CumulativePaidTriangledf.loc['Total'] - CumulativePaidTriangledf.loc['Latest Observation'])
# DP1 3.490607
# DP2 1.747333
# DP3 1.457413
# DP4 1.173852
# DP5 1.103824
# DP6 1.086269
# DP7 1.053874
# DP8 1.076555
# DP9 1.017725
# DP10 inf
# Total NaN
# dtype: float64
以下是三个组件的作用的细分：
DP1
DP2
DP3
DP4
DP5
DP6
DP7
DP8
DP9
DP10
Total
A: .shift(-1, axis=1).loc['Total'] -- We are shifting the whole Total row to the left, so every column now has the next Total value.
1.161454e+07
1.791234e+07
2.193092e+07
2.165497e+07
1.982827e+07
1.733138e+07
1.342964e+07
9.172600e+06
3.901463e+06
34358090.0
NaN
B: .loc['Total'] -- This is the normal Total row.
3.671385e+06
1.161454e+07
1.791234e+07
2.193092e+07
2.165497e+07
1.982827e+07
1.733138e+07
1.342964e+07
9.172600e+06
3901463.0
34358090.0
C: .loc['Latest Observation'] -- This is the normal Latest Observation.
3.440140e+05
1.363294e+06
2.864498e+06
3.483130e+06
3.691712e+06
3.873311e+06
4.588268e+06
4.909315e+06
5.339085e+06
3901463.0
NaN
A / (B-C) -- This is what the code above does. It takes the shifted Total row (A) and divides it by the difference of the current Total row (B) and current Latest observation row (C).
3.490607
1.747333
1.457413
1.173852
1.103824
1.086269
1.053874
1.076555
1.017725
inf
NaN
我想遍历数据框的所有列。这样我就可以得到该列的特定数据，使用这些数据我必须为另一个数据框计算。 我这里有 :
DP1 DP2 DP3 DP4 DP5 DP6 DP7 DP8 DP9 DP10 Total
OP1 357848.0 1124788.0 1735330.0 2218270.0 2745596.0 3319994.0 3466336.0 3606286.0 3833515.0 3901463.0 3901463.0
OP2 352118.0 1236139.0 2170033.0 3353322.0 3799067.0 4120063.0 4647867.0 4914039.0 5339085.0 NaN 5339085.0
OP3 290507.0 1292306.0 2218525.0 3235179.0 3985995.0 4132918.0 4628910.0 4909315.0 NaN NaN 4909315.0
OP4 310608.0 1418858.0 2195047.0 3757447.0 4029929.0 4381982.0 4588268.0 NaN NaN NaN 4588268.0
OP5 443160.0 1136350.0 2128333.0 2897821.0 3402672.0 3873311.0 NaN NaN NaN NaN 3873311.0
OP6 396132.0 1333217.0 2180715.0 2985752.0 3691712.0 NaN NaN NaN NaN NaN 3691712.0
OP7 440832.0 1288463.0 2419861.0 3483130.0 NaN NaN NaN NaN NaN NaN 3483130.0
OP8 359480.0 1421128.0 2864498.0 NaN NaN NaN NaN NaN NaN NaN 2864498.0
OP9 376686.0 1363294.0 NaN NaN NaN NaN NaN NaN NaN NaN 1363294.0
OP10 344014.0 NaN NaN NaN NaN NaN NaN NaN NaN NaN 344014.0
Total 3671385.0 11614543.0 17912342.0 21930921.0 21654971.0 19828268.0 17331381.0 13429640.0 9172600.0 3901463.0 34358090.0
Latest Observation 344014.0 1363294.0 2864498.0 3483130.0 3691712.0 3873311.0 4588268.0 4909315.0 5339085.0 3901463.0 NaN
从这个 table 我想计算这个公式这个公式：在 DP1 列中，Total/Last 观察，这个答案是除以 DP2 列总数。像这样我们必须计算所有列并将其保存在另一个数据框中。
我们需要这样的行：
Weighted Average 3.491 1.747 1.457 1.174 1.104 1.086 1.054 1.077 1.018
我们试过的代码：
LDFTriangledf['Weighted Average'] =CumulativePaidTriangledf.loc['Total','DP2']/(CumulativePaidTriangledf.loc['Total','DP1'] - CumulativePaidTriangledf.loc['Latest Observation','DP1'])
您可以从 .loc 中删除列名，只删除 shift(-1, axis=1) 以获得下一列的 Total。这使您可以在一次操作中将公式应用于所有列：
CumulativePaidTriangledf.shift(-1, axis=1).loc['Total'] / (CumulativePaidTriangledf.loc['Total'] - CumulativePaidTriangledf.loc['Latest Observation'])
# DP1 3.490607
# DP2 1.747333
# DP3 1.457413
# DP4 1.173852
# DP5 1.103824
# DP6 1.086269
# DP7 1.053874
# DP8 1.076555
# DP9 1.017725
# DP10 inf
# Total NaN
# dtype: float64
以下是三个组件的作用的细分：
| DP1 | DP2 | DP3 | DP4 | DP5 | DP6 | DP7 | DP8 | DP9 | DP10 | Total | |
|---|---|---|---|---|---|---|---|---|---|---|---|
A: .shift(-1, axis=1).loc['Total'] -- We are shifting the whole Total row to the left, so every column now has the next Total value. |
1.161454e+07 | 1.791234e+07 | 2.193092e+07 | 2.165497e+07 | 1.982827e+07 | 1.733138e+07 | 1.342964e+07 | 9.172600e+06 | 3.901463e+06 | 34358090.0 | NaN |
B: .loc['Total'] -- This is the normal Total row. |
3.671385e+06 | 1.161454e+07 | 1.791234e+07 | 2.193092e+07 | 2.165497e+07 | 1.982827e+07 | 1.733138e+07 | 1.342964e+07 | 9.172600e+06 | 3901463.0 | 34358090.0 |
C: .loc['Latest Observation'] -- This is the normal Latest Observation. |
3.440140e+05 | 1.363294e+06 | 2.864498e+06 | 3.483130e+06 | 3.691712e+06 | 3.873311e+06 | 4.588268e+06 | 4.909315e+06 | 5.339085e+06 | 3901463.0 | NaN |
A / (B-C) -- This is what the code above does. It takes the shifted Total row (A) and divides it by the difference of the current Total row (B) and current Latest observation row (C). |
3.490607 | 1.747333 | 1.457413 | 1.173852 | 1.103824 | 1.086269 | 1.053874 | 1.076555 | 1.017725 | inf | NaN |