我的数据集中有很多列&我需要改变一些变量的值。我这样做
import pandas as pd
import numpy as np
df = pd.DataFrame({'one':['a' , 'b']*5, 'two':['c' , 'd']*5, 'three':['a' , 'd']*5})
选择
df1 = df[['one', 'two']]
字典
map = { 'a' : 'd', 'b' : 'c', 'c' : 'b', 'd' : 'a'}
和循环
df2=[]
for i in df1.values:
np = [ map[x] for x in i]
df2.append(np)
然后我改变了列
df['one'] = [row[0] for row in df2]
df['two'] = [row[1] for row in df2]
它有效,但它很长。如何缩短它?
答案 0 :(得分:2)
您可以使用make
迭代列:
--print-data-base
时序:
Series.map()
答案 1 :(得分:2)
仅使用'a','b'值传递col的整个地图效率不高。首先检查df col中的值。然后只为它们映射,如下所示:
dp
这也是可能的:
dp
同时注意重叠键(例如,如果你想平行改变,可以说a到b和b到c但不喜欢a-> b-> c)......
>>> cols = ['one', 'two'];
>>> map = { 'a' : 'd', 'b' : 'c', 'c' : 'b', 'd' : 'a'};
>>> for col in cols:
... colSet = set(df[col].values);
... colMap = {k:v for k,v in map.items() if k in colSet};
... df.replace(to_replace={col:colMap},inplace=True);#not efficient like rly
...
>>> df
one three two
0 d a b
1 c d a
2 d a b
3 c d a
4 d a b
5 c d a
6 d a b
7 c d a
8 d a b
9 c d a
>>>
#OR
In [12]: %%timeit
...: for col in cols:
...: colSet = set(df[col].values);
...: colMap = {k:v for k,v in map.items() if k in colSet};
...: df[col].map(colMap)
...:
...:
1 loop, best of 3: 1.93 s per loop
#OR WHEN INPLACE
In [8]: %%timeit
...: for col in cols:
...: colSet = set(df[col].values);
...: colMap = {k:v for k,v in map.items() if k in colSet};
...: df[col]=df[col].map(colMap)
...:
...:
1 loop, best of 3: 2.18 s per loop
答案 2 :(得分:1)