如何替换熊猫多索引列集中的值

问题描述

给出以下内容：

dd = {}
for i in range(3):
    for j in range(3):
        key = (f"col_{i}",j)
        dd[key] = {1: 2,3: 4}



print(pd.DataFrame.from_dict(dd))

外观如下：

  col_0       col_1       col_2      
      0  1  2     0  1  2     0  1  2
1     2  2  2     2  2  2     2  2  2
3     4  4  4     4  4  4     4  4  4

我想使用以下替换项：

reps = {
    "col_0": {0: "o",1: "one",2: "two"},"col_1": {0: "o2",1: "one2",2: "two2"},"col_2": {0: "o3",1: "one3",2: "two3"},}

使col_0，col_1，col_2保持不变，但第二级 0,1,2更改为o,one,two，o2,one2,two2和o3,two3 分别给出类似于：

  col_0             col_1            col_2      
      o  one  two   o2 one2 two2     o3  one3  two3
1     2  2    2      2  2    2        2    2     2
3     4  4    4      4  4    4        4    4     4

解决方法

您可以使用列名称创建元组，然后将匹配字典用于document.getElementById("idOfScrollingSection").onscroll(() => { let scrollValue = document.getElementById("elementId").scrollTop; //Remove class from highlighted item let oldElement = document.getElementsByClassName("current"); oldElement.classList.remove("current"); //Add highlight class,change values depending on page position if (scrollValue < 200) { let elementToBeHighlighted = document.getElementById("idOfElementToBeHighlighted"); elementToBeHighlighted.classList.add("current"); } else if .... } else { let elementToBeHighlighted = document.getElementById("idOfElementToBeHighlighted"); elementToBeHighlighted.classList.add("current"); } }) ，并将第二个参数用作默认值，因此，如果不匹配，则不替换：

get

测试L = [(a,reps[a].get(b,b)) if a in reps else (a,b) for a,b in df.columns.tolist()] df.columns = pd.MultiIndex.from_tuples(L) print (df) col_0 col_1 col_2 o one two o2 one2 two2 o3 one3 two3 1 2 2 2 2 2 2 2 2 2 3 4 4 4 4 4 4 4 4 4 字典中是否没有匹配的外键：

reps

测试是否没有匹配的内部键：

reps = {
    "col_100": {0: "o",1: "one",2: "two"},"col_1": {0: "o2",1: "one2",2: "two2"},"col_2": {0: "o3",1: "one3",2: "two3"},}

L = [(a,b in df.columns.tolist()]
df.columns = pd.MultiIndex.from_tuples(L)
print (df)
  col_0       col_1           col_2          
      0  1  2    o2 one2 two2    o3 one3 two3
1     2  2  2     2    2    2     2    2    2
3     4  4  4     4    4    4     4    4    4