pandas 替换多个值一列

Question

In a column risklevels I want to replace Small with 1, Medium with 5 and High with 15. I tried:

dfm.replace({'risk':{'Small': '1'}},
            {'risk':{'Medium': '5'}},
            {'risk':{'High': '15'}})

But only the medium were replaced. What is wrong?

Answer 1

Your replace format is off

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]

Answer 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

Answer 3

You could define a dict and call 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]

Answer 4

Looks like OP may have been looking for a one-liner to solve this through consecutive calls to .str.replace :

dfm.column = dfm.column.str.replace('Small', '1') \
    .str.replace('Medium', '5') \
        .str.replace('High', '15')

OP, you were close but just needed to replace your commas with .str.replace and the column call ('risk') in a dictionary format isn't necessary. Just pass the pattern-to-match and replacement-value as arguments to replace.

Answer 5

我必须打开“regex”标志才能使其工作：

 df.replace({'a' : {'Medium':2, 'Small':1, 'High':3 }}, regex=True)

Answer 6

String replace each string (Small, Medium, High) for the new string (1,5,15)\\

If dfm is the dataframe name, column is the column name.

dfm.column = dfm.column.str.replace('Small', '1')
dfm.column = dfm.column.str.replace('Medium', '5')
dfm.column = dfm.column.str.replace('High', '15')

Answer 7

Use series.replace with lists of before and after values for greater ease:

df.risklevels = df.risklevels.replace( ['Small','Medium','High'], [1,2,3] )

See here .