pandas中当某列满足制定条件时，对另一列赋值的问题

Question

各位大神，想请教一个简单的问题，上网各种查都查不到，我现在想实现df数据中当点评状态列不为null的时候，是否有评价列的值都为1。完全摸不到头脑，这个应该怎么写呢？一点头绪都没有，希望大家帮忙，谢谢

我尝试用了一下这个代码，结果那是相当的不对了，应该怎么办呢，新手基础非常薄弱，恳请大家指教

import pandas as pd
import numpy as np
df = pd.read_excel('D:/谷歌下载/明细.xlsx', sheet_name='sheet1')
df['是否有房评']=df['是否有房评'].apply(str)
df.loc[df['点评状态']!='null', '是否有房评'] = "1"

Answer 1

用列表解析式进行判断：

df["has_comment_or_not"] = ["1" if status != "null" else "0" 
                            for status in df["comment_status"].apply(str)]

具体例子如下：

In [1]: import pandas as pd

In [2]: import numpy as np

In [3]: df = pd.DataFrame(
    {"has_comment_or_not":['0','0','0','1','0','0'],
    "comment_status":["null","null","120047007","null","null","122121"]}
    )

In [4]: df
Out[4]:
  has_comment_or_not comment_status
0                  0           null
1                  0           null
2                  0      120047007
3                  1           null
4                  0           null
5                  0         122121


In [5]: df["has_comment_or_not"] = ["1" if status != "null" else "0" 
for status in df["comment_status"].apply(str)]

In [6]: df
Out[6]:
  has_comment_or_not comment_status
0                  0           null
1                  0           null
2                  1      120047007
3                  0           null
4                  0           null
5                  1         122121

如果只想修改comment_status有内容，改has_comment_or_not为1，只需要对处理的结果，再进行列表解析判断即可

df["has_comment_or_not"] =[item if item == '1' else df["has_comment_or_not"].values[index] for index, item in enumerate(["1" if status != "null" else "0" for status
    ...:  in df["comment_status"].apply(str)])]

例子如下：（修改了has_comment_or_not的原始值，添加了2，方便区分）

In [44]: import pandas as pd

In [45]: import numpy as np

In [46]: df = pd.DataFrame(
    ...:     {"has_comment_or_not":['1','2','0','1','0','0'],
    ...:     "comment_status":["null","null","120047007","null","null","122121"]}
    ...:     )

In [47]: ["1" if status != "null" else "0"
    ...: for status in df["comment_status"].apply(str)]
Out[47]: ['0', '0', '1', '0', '0', '1']

In [49]: [ item if item =='1' else df["has_comment_or_not"].values[index] for index, item in enumerate(['0', '0', '1', '0', '0', '1'])]
Out[49]: ['1', '2', '1', '1', '0', '1']

In [50]: df["has_comment_or_not"] =[item if item == '1' else df["has_comment_or_not"].values[index] for index, item in enumerate(["1" if status != "null" else "0" for status
    ...:  in df["comment_status"].apply(str)])]
    
In [51]: df
Out[51]:
  has_comment_or_not comment_status
0                  1           null
1                  2           null
2                  1      120047007
3                  1           null
4                  0           null
5                  1         122121

当然也可以使用iterrows方法进行判断修改：

In [58]: row_list = []
    ...: for index, row in df.iterrows():
    ...:     if row["comment_status"] !="null":
    ...:         row["has_comment_or_not"] = "1"
    ...:     row_list.append(row)

In [60]: df = pd.DataFrame(row_list)

In [61]: df
Out[61]:
  has_comment_or_not comment_status
0                  1           null
1                  2           null
2                  1      120047007
3                  1           null
4                  0           null
5                  1         122121

Answer 2

f = lambda s: 1 if s["点评状态"]!="null" else 0

df['是否有房评'] = df.apply(f, axis=1)

Answer 3

# coding: utf-8

import pandas as pd

data = [
    {'是否有房': 0, '点评状态': 'null'},
    {'是否有房': 0, '点评状态': '111'},
    {'是否有房': 0, '点评状态': 'null'},
]

df = pd.DataFrame(data)
df['是否有房'] = df['点评状态'].apply(lambda x: 1 if x != 'null' else 0)
print df

Answer 4

貌似比较简单和高效的写法应该是：

df.loc[df['点评状态'].notnull(), '是否有房评'] = 1