熊猫：根据其值更改列

Question

当cluster_name包含“演示”时，我想将其更改为“未知”。

这是我所管理的最好的事情

df["cluster_name"] = "unknown" if "demo" is in df["cluster_name"] else df["cluster_name"]

但是得到：

SyntaxError：语法无效

Answer 1

您可以使用：

import numpy as np
df["cluster_name"] = np.where(df["cluster_name"].str.contains("demo"), "unknown", df["cluster_name"])

请参见以下示例：

In [814]: df1
Out[814]: 
        State  Year  Incident  new     nn
0           a  1980       513    1    0.0
1  demo is in  1981       453    0    1.0
2           b  1982       424    1  100.0
3     my demo  1983       372  100    NaN

In [816]: df1.State = np.where(df1.State.str.contains('demo'), 'unknown', df1.State)

In [817]: df1
Out[817]: 
     State  Year  Incident  new     nn
0        a  1980       513    1    0.0
1  unknown  1981       453    0    1.0
2        b  1982       424    1  100.0
3  unknown  1983       372  100    NaN

Answer 2

如果不需要搜索子字符串“ demo”，则可以使用Series.replace。 “包含”是不明确的。

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或替换df['cluster_name'] = df['cluster_name'].replace('demo','unknown')

inplace

Answer 3

一种可能的解决方案是将map与lambda函数一起使用，该函数在语法上类似于您尝试执行的操作：

简单的map解决方案：

#replaces the row with 'unknown' if it is 'demo'
df['cluster_name'] = df['cluster_name'].map(lambda x : 'unknown' if x=='demo' else x)

更广义的map解决方案：

#replaces the row with 'unknown' if contains 'demo'
df['cluster_name'] = df['cluster_name'].map(lambda x : 'unknown' if 'demo' in x else x)

示例：

>>> #simple map solution
>>> df
  cluster_name
0         demo
1         demo
2            1

>>> df['cluster_name'] = df['cluster_name'].map(lambda x : 'unknown' if x=='demo' else x)
>>> df
  cluster_name
0      unknown
1      unknown
2            1

>>> #More generalized  map solution:
>>> df1
  cluster_name
0    demo is a
1         demo
2            1
>>> df1['cluster_name'] = df1['cluster_name'].map(lambda x : 'unknown' if 'demo' in x else x)
>>> df1
  cluster_name
0      unknown
1      unknown
2            1