我如何添加一个颜色列到下面的数据框架，使颜色='绿色'如果设置== 'Z'，和颜色='红色'否则?
Type Set
1 A Z
2 B Z
3 B X
4 C Y
当前回答
你可以使用pandas方法:
df['color'] = 'green'
df['color'] = df['color'].where(df['Set']=='Z', other='red')
# Replace values where the condition is False
or
df['color'] = 'red'
df['color'] = df['color'].mask(df['Set']=='Z', other='green')
# Replace values where the condition is True
或者，你也可以使用lambda函数的transform方法:
df['color'] = df['Set'].transform(lambda x: 'green' if x == 'Z' else 'red')
输出:
Type Set color
1 A Z green
2 B Z green
3 B X red
4 C Y red
@chai的性能比较:
import pandas as pd
import numpy as np
df = pd.DataFrame({'Type':list('ABBC')*1000000, 'Set':list('ZZXY')*1000000})
%timeit df['color1'] = 'red'; df['color1'].where(df['Set']=='Z','green')
%timeit df['color2'] = ['red' if x == 'Z' else 'green' for x in df['Set']]
%timeit df['color3'] = np.where(df['Set']=='Z', 'red', 'green')
%timeit df['color4'] = df.Set.map(lambda x: 'red' if x == 'Z' else 'green')
397 ms ± 101 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
976 ms ± 241 ms per loop
673 ms ± 139 ms per loop
796 ms ± 182 ms per loop
其他回答
如果你在处理海量数据，记忆方法是最好的:
# First create a dictionary of manually stored values
color_dict = {'Z':'red'}
# Second, build a dictionary of "other" values
color_dict_other = {x:'green' for x in df['Set'].unique() if x not in color_dict.keys()}
# Next, merge the two
color_dict.update(color_dict_other)
# Finally, map it to your column
df['color'] = df['Set'].map(color_dict)
当您有许多重复的值时，这种方法将是最快的。我的一般经验法则是记住data_size > 10**4 & n_distinct < data_size/4
在一种情况下，记忆10,000行，不同值不超过2,500。
使用.apply()方法的一行代码如下:
df['color'] = df['Set'].apply(lambda set_: 'green' if set_=='Z' else 'red')
之后，df数据帧是这样的:
>>> print(df)
Type Set color
0 A Z green
1 B Z green
2 B X red
3 C Y red
你可以使用pandas方法:
df['color'] = 'green'
df['color'] = df['color'].where(df['Set']=='Z', other='red')
# Replace values where the condition is False
or
df['color'] = 'red'
df['color'] = df['color'].mask(df['Set']=='Z', other='green')
# Replace values where the condition is True
或者，你也可以使用lambda函数的transform方法:
df['color'] = df['Set'].transform(lambda x: 'green' if x == 'Z' else 'red')
输出:
Type Set color
1 A Z green
2 B Z green
3 B X red
4 C Y red
@chai的性能比较:
import pandas as pd
import numpy as np
df = pd.DataFrame({'Type':list('ABBC')*1000000, 'Set':list('ZZXY')*1000000})
%timeit df['color1'] = 'red'; df['color1'].where(df['Set']=='Z','green')
%timeit df['color2'] = ['red' if x == 'Z' else 'green' for x in df['Set']]
%timeit df['color3'] = np.where(df['Set']=='Z', 'red', 'green')
%timeit df['color4'] = df.Set.map(lambda x: 'red' if x == 'Z' else 'green')
397 ms ± 101 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
976 ms ± 241 ms per loop
673 ms ± 139 ms per loop
796 ms ± 182 ms per loop
如果你只有两种选择:
df['color'] = np.where(df['Set']=='Z', 'green', 'red')
例如,
import pandas as pd
import numpy as np
df = pd.DataFrame({'Type':list('ABBC'), 'Set':list('ZZXY')})
df['color'] = np.where(df['Set']=='Z', 'green', 'red')
print(df)
收益率
Set Type color
0 Z A green
1 Z B green
2 X B red
3 Y C red
如果你有两个以上的条件，那么使用np.select。例如，如果你想要颜色
黄色时(df['设置']= = ' Z ') & (df(“类型”)= =“一”) 否则蓝色当(df['设置']= = ' Z ') & (df(“类型”)= = ' B ') 否则为紫色，当(df['Type'] == 'B') 否则黑,
然后使用
df = pd.DataFrame({'Type':list('ABBC'), 'Set':list('ZZXY')})
conditions = [
(df['Set'] == 'Z') & (df['Type'] == 'A'),
(df['Set'] == 'Z') & (df['Type'] == 'B'),
(df['Type'] == 'B')]
choices = ['yellow', 'blue', 'purple']
df['color'] = np.select(conditions, choices, default='black')
print(df)
的收益率
Set Type color
0 Z A yellow
1 Z B blue
2 X B purple
3 Y C black
这是另一种方法，使用字典将新值映射到列表中的键:
def map_values(row, values_dict):
return values_dict[row]
values_dict = {'A': 1, 'B': 2, 'C': 3, 'D': 4}
df = pd.DataFrame({'INDICATOR': ['A', 'B', 'C', 'D'], 'VALUE': [10, 9, 8, 7]})
df['NEW_VALUE'] = df['INDICATOR'].apply(map_values, args = (values_dict,))
它看起来像什么:
df
Out[2]:
INDICATOR VALUE NEW_VALUE
0 A 10 1
1 B 9 2
2 C 8 3
3 D 7 4
当你有很多ifelse类型语句要执行时(例如，很多唯一值要替换)，这种方法非常强大。
当然你可以这样做:
df['NEW_VALUE'] = df['INDICATOR'].map(values_dict)
但在我的机器上，这种方法比上面的apply方法慢三倍多。
你也可以使用dict.get:
df['NEW_VALUE'] = [values_dict.get(v, None) for v in df['INDICATOR']]
