import pandas as pd 创建一个示例 DataFrame data = {'A': [1, 2, 3], 'B': [4, 5, 6], 'C': [7, 8, 9]} df = pd.DataFrame(data)获取 DataFrame 的所有列标签 columns_list = df.columns.tolist()打印结果 print(columns_list)运行上述代码后，你将获得
"A","B","B","C","C","C","D"],"value1":[10,20,30,40,50,60,70,80]}df_1=pd.DataFrame(dict_1,columns=["time","pos","value1"])print("原数据","\n",df_1,"\n")print("\n按行输出")list_fields
创建DataFrame import pandas as pd #创建空DataFrame df = pd.DataFrame() #由List创建DataFrame l = ['l1','l2','l3'] df = pd.DataFrame(l) # 0 #0 l1 #1 l2 #2 l3 #由字典创建DataFrame dict = {'name':['Jack','Rose','Three'],'number':['112','132','177'],'level':['low',...
'San Francisco','Los Angeles']}df=pd.DataFrame(data)name_list=['Alice','Bob']df.query("Name...
DataFrame 转成内置数据结构 假设有这样一个 DataFrame: importpandasaspd df = pd.DataFrame({"name": ["Satori","Koishi","Marisa"], "score": [99,98,100], "rank": [2,3,1]}) print(df) """ name score rank 0 Satori 99 2 1 Koishi 98 3 ...
# 由列表或元组组成的列表创建DataFrame data_rows = [[1, 'a'], [2, 'b']] df_rows = pd.DataFrame(data_rows, columns=['Column1', 'Column2']) print("\nDataFrame from list of lists/tuples:\n", df_rows) # 另一个DataFrame创建DataFrame ...
df = pd.DataFrame(np.random.rand(12).reshape(3,4)*100, index = ['one','two','three'], columns = ['a','b','c','d']) print(df) print('---') data1 = df['a'] data2 = df[['b','c']] # 尝试输入 data2 = df[['b','c','e']] print...
原理就是这个是把一个list切成多个内嵌的list,然后就可以直接pd.dataframe了。 后来【隔壁😼山楂】给了个代码,如下所示: pd.DataFrame(df.groupby(['group'])['data'].agg(pd.Series).values.tolist()) 顺利地解决了粉丝的问题。 确实还真没留意到有一列可以分组!
import pandas as pd df = pd.DataFrame({ 'data': ['A1', 'D3', 'B2', 'C4', 'A1', 'A2', 'B2', 'B3', 'C3', 'C4', 'D5', 'D3'], 'new': ['A1', 'A1', 'D3', 'D3', 'B2', 'B2', 'C4', 'C4', 'A2', 'B3', 'C3', 'D5'] ...
DataFrame.apply(func[, axis, broadcast, …])应用函数 DataFrame.applymap(func)Apply a function to a DataFrame that is intended to operate elementwise, i.e. DataFrame.aggregate(func[, axis])Aggregate using callable, string, dict, or list of string/callables ...