我有以下索引DataFrame命名列和行不连续的数字:
a b c d
2 0.671399 0.101208 -0.181532 0.241273
3 0.446172 -0.243316 0.051767 1.577318
5 0.614758 0.075793 -0.451460 -0.012493
我想添加一个新列,'e',到现有的数据帧,并不想改变数据帧中的任何东西(即,新列始终具有与DataFrame相同的长度)。
0 -0.335485
1 -1.166658
2 -0.385571
dtype: float64
如何将列e添加到上面的例子中?
你可以像这样通过for循环插入新列:
for label,row in your_dframe.iterrows():
your_dframe.loc[label,"new_column_length"]=len(row["any_of_column_in_your_dframe"])
示例代码如下:
import pandas as pd
data = {
"any_of_column_in_your_dframe" : ["ersingulbahar","yagiz","TS"],
"calories": [420, 380, 390],
"duration": [50, 40, 45]
}
#load data into a DataFrame object:
your_dframe = pd.DataFrame(data)
for label,row in your_dframe.iterrows():
your_dframe.loc[label,"new_column_length"]=len(row["any_of_column_in_your_dframe"])
print(your_dframe)
输出如下:
any_of_column_in_your_dframe |
calories |
duration |
new_column_length |
ersingulbahar |
420 |
50 |
13.0 |
yagiz |
380 |
40 |
5.0 |
TS |
390 |
45 |
2.0 |
你也可以这样用:
your_dframe["new_column_length"]=your_dframe["any_of_column_in_your_dframe"].apply(len)
我得到了可怕的SettingWithCopyWarning,它没有通过使用iloc语法修复。我的DataFrame是由read_sql从ODBC源创建的。根据上面low - tech的建议,以下方法对我来说是有效的:
df.insert(len(df.columns), 'e', pd.Series(np.random.randn(sLength), index=df.index))
This worked fine to insert the column at the end. I don't know if it is the most efficient, but I don't like warning messages. I think there is a better solution, but I can't find it, and I think it depends on some aspect of the index.
Note. That this only works once and will give an error message if trying to overwrite and existing column.
Note As above and from 0.16.0 assign is the best solution. See documentation http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.assign.html#pandas.DataFrame.assign
Works well for data flow type where you don't overwrite your intermediate values.
以下是我所做的…但我对熊猫和Python都很陌生,所以不能保证。
df = pd.DataFrame([[1, 2], [3, 4], [5,6]], columns=list('AB'))
newCol = [3,5,7]
newName = 'C'
values = np.insert(df.values,df.shape[1],newCol,axis=1)
header = df.columns.values.tolist()
header.append(newName)
df = pd.DataFrame(values,columns=header)