我需要能够
1.计算所有行中每一列的排名，
2.找到每行的最大列标签，
3.然后在每一行中移动原始df的最大排名列。
仅处理原始df中的数据时，这很简单。但是，如果需要不同的排名调用，似乎很难完成。
下面是我的Python Pandas代码来完成此任务。但这行不通。它似乎并没有像我期望的那样解释我的声明{
materials: [
{
name: 'Mars_Atmosphere_Mat',
parameters: [
{
baseColorFactor: [1,1,1,1,],
},
{
baseColor: 'Mars_Atmosphere_Mat_baseColor',
},
{
diffuseColorFactor: null,
},
{
diffuseColor: null,
},
{
normal: null,
},
{
metallicFactor: 1,
},
{
roughnessFactor: 1,
},
{
specularFactor: null,
},
{
glossinessFactor: null,
},
{
specularGlossiness: null,
},
{
specularGlossinessCalculation: null,
},
{
metallicRoughness: null,
},
{
occlusion: null,
},
{
emissiveFactor: [0,0,0,1,],
},
{
emissive: null,
},
{
opacity: null,
},
],
source: 'build/sceneform_sdk/default_materials/gltf_material.sfm',
},
{
name: 'Mars_mat',
parameters: [
{
baseColorFactor: [1,1,1,1,],
},
{
baseColor: 'Mars_mat_baseColor',
},
{
diffuseColorFactor: null,
},
{
diffuseColor: null,
},
{
normal: 'Mars_mat_normal',
},
{
metallicFactor: 0.33000000000000002,
},
{
roughnessFactor: 0.85999999999999999,
},
{
specularFactor: null,
},
{
glossinessFactor: null,
},
{
specularGlossiness: null,
},
{
specularGlossinessCalculation: null,
},
{
metallicRoughness: null,
},
{
occlusion: null,
},
{
emissiveFactor: [0,0,0,1,],
},
{
emissive: null,
},
{
opacity: null,
},
],
source: 'build/sceneform_sdk/default_materials/gltf_material.sfm',
},
],
model: {
attributes: [
'Position',
'TexCoord',
'Orientation',
],
collision: {
skin_width: 0.75,
},
file: 'sampledata/models/Mars/Mars.gltf',
name: 'Mars',
recenter: true,
scale: 0.5,
},
samplers: [
{
file: 'sampledata/models/Mars/Mars_mat_baseColor.png',
name: 'Mars_mat_baseColor',
pipeline_name: 'Mars_mat_baseColor.png',
},
{
file: 'sampledata/models/Mars/Mars_mat_normal.png',
name: 'Mars_mat_normal',
params: {
usage_type: 'Normal',
},
pipeline_name: 'Mars_mat_normal.png',
},
{
file: 'sampledata/models/Mars/Mars_Atmosphere_Mat_baseColor.png',
name: 'Mars_Atmosphere_Mat_baseColor',
pipeline_name: 'Mars_Atmosphere_Mat_baseColor.png',
},
],
version: '0.54:1',
}
。提出的建议将不胜感激。
df1['maxV'] = df1[df1['maxR']]答案 0 :(得分：0)
迭代行并将值累加到一个数组中：
maxVals = [np.nan]*3
for index, row in df1[pd.notna(df1['maxR'])].iterrows():
maxVals.append(df1.loc[index, row['maxR']])
df1['maxV'] = maxVals
替代方法：一种不太直观的方法可能是使用索引和值来对df1进行索引，这将返回更宽的数据框（＃列等于＃行的行数），对角线处有最大值：
maxVals = [np.nan]*3
newDf = df1.loc[df1['maxR'][3:].index, df1['maxR'][3:].values]
maxVals.extend(np.diag(newDf))
df1['maxV'] = maxVals