In a column risklevels I want to replace Small with 1, Medium with 5 and High with 15. I tried:
在列冒险级别中,我想将Small替换为1,将Medium替换为5,将High替换为15.我尝试:
dfm.replace({'risk':{'Small': '1'}},{'risk':{'Medium': '5'}},{'risk':{'High': '15'}})
But only the medium were replaced. What is wrong ?
但只有媒体被替换了。哪里不对 ?
4 个解决方案
#1
22
Your replace format is off
您的替换格式已关闭
In [21]: df = pd.DataFrame({'a':['Small', 'Medium', 'High']})
In [22]: df
Out[22]:
a
0 Small
1 Medium
2 High
[3 rows x 1 columns]
In [23]: df.replace({'a' : { 'Medium' : 2, 'Small' : 1, 'High' : 3 }})
Out[23]:
a
0 1
1 2
2 3
[3 rows x 1 columns]
#2
6
You could define a dict and call map
你可以定义一个dict和调用map
In [256]:
df = pd.DataFrame({'a':['Small', 'Medium', 'High']})
df
Out[256]:
a
0 Small
1 Medium
2 High
[3 rows x 1 columns]
In [258]:
vals_to_replace = {'Small':'1', 'Medium':'5', 'High':'15'}
df['a'] = df['a'].map(vals_to_replace)
df
Out[258]:
a
0 1
1 5
2 15
[3 rows x 1 columns]
In [279]:
val1 = [1,5,15]
df['risk'].update(pd.Series(val1))
df
Out[279]:
risk
0 1
1 5
2 15
[3 rows x 1 columns]
#3
2
In [123]: import pandas as pd
In [124]: state_df = pd.DataFrame({'state':['Small', 'Medium', 'High', 'Small', 'High']})
In [125]: state_df
Out[125]:
state
0 Small
1 Medium
2 High
3 Small
4 High
In [126]: replace_values = {'Small' : 1, 'Medium' : 2, 'High' : 3 }
In [127]: state_df = state_df.replace({"state": replace_values})
In [128]: state_df
Out[128]:
state
0 1
1 2
2 3
3 1
4 3
#4
0
String replace each string (Small, Medium, High) for the new string (1,5,15)\
String替换新字符串(1,5,15)的每个字符串(小,中,高)
If dfm is the dataframe name, column is the column name.
如果dfm是数据框名称,则列是列名称。
dfm.column = dfm.column.str.replace('Small', '1')
dfm.column = dfm.column.str.replace('Medium', '5')
dfm.column = dfm.column.str.replace('High', '15')
#1
22
Your replace format is off
您的替换格式已关闭
In [21]: df = pd.DataFrame({'a':['Small', 'Medium', 'High']})
In [22]: df
Out[22]:
a
0 Small
1 Medium
2 High
[3 rows x 1 columns]
In [23]: df.replace({'a' : { 'Medium' : 2, 'Small' : 1, 'High' : 3 }})
Out[23]:
a
0 1
1 2
2 3
[3 rows x 1 columns]
#2
6
You could define a dict and call map
你可以定义一个dict和调用map
In [256]:
df = pd.DataFrame({'a':['Small', 'Medium', 'High']})
df
Out[256]:
a
0 Small
1 Medium
2 High
[3 rows x 1 columns]
In [258]:
vals_to_replace = {'Small':'1', 'Medium':'5', 'High':'15'}
df['a'] = df['a'].map(vals_to_replace)
df
Out[258]:
a
0 1
1 5
2 15
[3 rows x 1 columns]
In [279]:
val1 = [1,5,15]
df['risk'].update(pd.Series(val1))
df
Out[279]:
risk
0 1
1 5
2 15
[3 rows x 1 columns]
#3
2
In [123]: import pandas as pd
In [124]: state_df = pd.DataFrame({'state':['Small', 'Medium', 'High', 'Small', 'High']})
In [125]: state_df
Out[125]:
state
0 Small
1 Medium
2 High
3 Small
4 High
In [126]: replace_values = {'Small' : 1, 'Medium' : 2, 'High' : 3 }
In [127]: state_df = state_df.replace({"state": replace_values})
In [128]: state_df
Out[128]:
state
0 1
1 2
2 3
3 1
4 3
#4
0
String replace each string (Small, Medium, High) for the new string (1,5,15)\
String替换新字符串(1,5,15)的每个字符串(小,中,高)
If dfm is the dataframe name, column is the column name.
如果dfm是数据框名称,则列是列名称。
dfm.column = dfm.column.str.replace('Small', '1')
dfm.column = dfm.column.str.replace('Medium', '5')
dfm.column = dfm.column.str.replace('High', '15')