系统:Windows 7 语言版本:Anaconda3-4.3.0.1-Windows-x86_64 编辑器:pycharm-community-2016.3.2 pandas:0.19.2
Part 1:场景描述
["value1", "value2", "value3", "value4"]
,行索引为0-7df_1
结果如下:列操作
行操作
Part 2:代码
import pandas as pd
import numpy as np
dict_1 = {"value1": [10, 20, 30, 40, 50, 60, 70, 80],
"value2": [100, 200, 300, 400, 500, 600, 700, 800],
"value3": [50, 20, 30, 90, 50, 60, 80, 80],
"value4": [10, 30, 90, 40, 60, 60, 70, 80]}
df_1 = pd.DataFrame(dict_1, columns=["value1", "value2", "value3", "value4"])
print("\n", "df_1", "\n", df_1, "\n")
print(type(df_1))
# 对某些列进行计算
df_2 = df_1.apply(lambda x: np.square(x) if x.name in ['value1', 'value2'] else x)
print("\n", "df_2-列平方", "\n", df_2, "\n")
df_2 = df_1.apply(lambda x: x+2 if x.name in ['value1', 'value2'] else x)
print("\n", "df_2-列+2", "\n", df_2, "\n")
# 对某些行进行计算
df_3 = df_1.apply(lambda x: np.square(x) if x.name in [1, 2] else x, axis=1)
print("\n", "df_3-行平方", "\n", df_3, "\n")
df_3 = df_1.apply(lambda x: x-3 if x.name in [1, 2] else x, axis=1)
print("\n", "df_3-行-3", "\n", df_3, "\n")
代码截图
Part 3:部分代码解读
axis=1
这个参数,默认取0df_2 = df_1.apply(lambda x: np.square(x) if x.name in ['value1', 'value2'] else x)
运用了apply方法,使用lambda函数,简单来理解就是对列名为['value1', 'value2']
的每个元素进行平方,其余保持不变。