鑒于以下情況：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))看起來像： col_0 col_1 col_2 0 1 2 0 1 2 0 1 21 2 2 2 2 2 2 2 2 23 4 4 4 4 4 4 4 4 4我想使用以下替換：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"},}因此col_0, col_1,col_2不變，但第二層 分別0,1,2更改為o, one, two, o2, one2, two2, 和o3, one, two3 ，給出類似的內(nèi)容： col_0 col_1 col_2 o one two o2 one2 two2 o3 one3 two31 2 2 2 2 2 2 2 2 23 4 4 4 4 4 4 4 4 4
1 回答
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TA貢獻(xiàn)1862條經(jīng)驗(yàn) 獲得超6個贊
您可以使用列名創(chuàng)建元組，然后將匹配字典與第二個參數(shù)一起用作get默認(rèn)值，因此如果不匹配則不替換：
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
測試字典中是否沒有匹配的外鍵reps：
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, 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
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
測試是否沒有匹配的內(nèi)部鍵：
reps = {
"col_0": {100: "o", 1: "one", 20: "two"},
"col_1": {0: "o2", 1: "one2", 2: "two2"},
"col_2": {0: "o3", 1: "one3", 2: "two3"},
}
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
0 one 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
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