我创建了这个数据框:
import pandas as pd
columns = pd.MultiIndex.from_tuples([("x", "", ""), ("values", "a", "a.b"), ("values", "c", "")])
df0 = pd.DataFrame([(0,10,20),(1,100,200)], columns=columns)
df0
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我卸载df0到excel:
df0.to_excel("test.xlsx")
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并再次加载它:
df1 = pd.read_excel("test.xlsx", header=[0,1,2])
df1
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我有Unnamed :...列名。
为了df1看起来像初始df0我运行:
def rename_unnamed(df, label=""):
for i, columns in enumerate(df.columns.levels):
columns = columns.tolist()
for j, row in enumerate(columns):
if "Unnamed: " in row:
columns[j] = ""
df.columns.set_levels(columns, level=i, inplace=True)
return df
rename_unnamed(df1)
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做得好。但是有没有大熊猫的方式来做到这一点?
从 pandas 0.21.0 开始,代码应该是这样的
def rename_unnamed(df):
"""Rename unamed columns name for Pandas DataFrame
See /sf/ask/2885475561/
Parameters
----------
df : pd.DataFrame object
Input dataframe
Returns
-------
pd.DataFrame
Output dataframe
"""
for i, columns in enumerate(df.columns.levels):
columns_new = columns.tolist()
for j, row in enumerate(columns_new):
if "Unnamed: " in row:
columns_new[j] = ""
if pd.__version__ < "0.21.0": # /sf/answers/3373088351/
df.columns.set_levels(columns_new, level=i, inplace=True)
else:
df = df.rename(columns=dict(zip(columns.tolist(), columns_new)),
level=i)
return df
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您可以numpy.where通过以下方式使用条件contains:
for i, col in enumerate(df1.columns.levels):
columns = np.where(col.str.contains('Unnamed'), '', col)
df1.columns.set_levels(columns, level=i, inplace=True)
print (df1)
x values
a c
a.b
0 0 10 20
1 1 100 200
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