Get Pandas Index as a ListSometimes you may be required to get the pandas DataFrame index as a list, we can do this by using df.index.values. Let’s pass this into a list, it will return the index as a list. # Get the index use index.values print(list(df.index.values)) # ...
使用列名直接访问列数据：print(df['Student Name'])4. 重新设置索引.set_index()可以使用.set_index...
import pandas as pd # 创建 Index idx = pd.Index([1, 2, 3], name='MyIndex') # 查看属性 print(idx.values) # 输出数据部分 print(idx.name) # 输出名称 # 数据操作 idx_new = idx.append(pd.Index([4, 5])) print(idx_new) # 输出追加后的 Index # 索引操作 print(idx.get_loc(2))...
4),index=list('abcdef'),columns=list('ABCD'))df2ABCDa-0.541813-0.1395420.0718582.259392b0...
上述方法生成的Index对象就是一个单层索引了,Index对象具有以下基本属性 代码语言:javascript 代码运行次数:0 运行 AI代码解释 >>>a=pd.Index(list('ABCD'))>>>aIndex(['A','B','C','D'],dtype='object')# 值构成的数组>>>a.array<PandasArray>['A','B','C','D']Length:4,dtype:object ...
tt2 = ss1[ss1.notnull()] # 判断序列的非空值,效果同上 print(tt2) tt3 = ss1.dropna() # 清洗空值 print(tt3) # 序列切片 import pandas as pd import numpy as np s1 = pd.Series([1, -2, 2.3, 'hq']) s2 = pd.Series([1, -2, 2.3, 'hq'], index = ['a', 'b', 'c', ...
import pandas as pd import numpy as np frame = pd.DataFrame(np.arange(16).reshape(4, 4), index = ['Ohio', 'Colorado', 'Utah', 'New York'], columns = ['one', 'two', 'three', 'four']) print(frame) print('直接选择(列):\n',frame[['one', 'two']]) #直接选择 print('直...
nodes_id_index=pd.Index(nodes_series)print(nodes_id_index.get_loc('u_3223_4017')) [Find element's index in pandas Series] [Index.get_loc] 更多请参考[Index] 皮皮blog 检索/选择 dataframe列选择 和Series一样,在DataFrame中的一列可以通过字典记法或属性来检索,返回Series: ...
in Index.get_loc(self, key) 3804 try: -> 3805 return self._engine.get_loc(casted_key) 3806 except KeyError as err: File index.pyx:167, in pandas._libs.index.IndexEngine.get_loc() File index.pyx:196, in pandas._libs.index.IndexEngine.get_loc() File pandas/_libs/hashtable_class_...
index= [0, 4, 8, 14, 20]#获取指定索引的元素ser.take(index)#> 0 a4e8i14o20u dtype: object 14. 如何垂直和水平的拼接series ser1 = pd.Series(range(5)) ser2= pd.Series(list('abcde'))#垂直拼接df = pd.concat([ser1, ser2], axis=0)#水平拼接df = pd.concat([ser1, ser2], ...