如果我使用下面的代码,它将保留具有NaN的列(请参阅所附图片)。我还有其他类似的专栏。是否可以保留第二个而不是第一个?
x='abcde'
result = [x[-i:]+x[:-i] for i in range(len(x))]
# result will be ['abcde', 'eabcd', 'deabc', 'cdeab', 'bcdea']
答案 0 :(得分:2)
NaN
的列如果您只需要针对此特定情况的修复程序,并且知道所需的列中没有NaN
:
data_final2 = data_final.dropna(axis=1)
data_final.columns = ['Site_nan', 'Site', 'Dimensions_nan', 'Dimensions']
data_final2 = data_final[['Site', 'Dimensions']].copy()
答案 1 :(得分:0)
groupby
列,然后选择first
值,该值将忽略Null。
df.groupby(df.columns, 1).first()
import pandas as pd
import numpy as np
df = pd.DataFrame({'0': [1,2,3], '1': [np.NaN]*3, '2': [np.NaN]*3, '3': ['1x1', '2x2', '3x3']})
df.columns= ['Size', 'Size', 'Dims', 'Dims']
# Size Size Dims Dims
#0 1 NaN NaN 1x1
#1 2 NaN NaN 2x2
#2 3 NaN NaN 3x3
df.groupby(df.columns, 1).first()
# Dims Size
#0 1x1 1
#1 2x2 2
#2 3x3 3