仅当没有单元格为空时，才在它们之间添加带有“，”的两列值答案

【问题标题】：Add two columns values with ',' between them only if none of the cells is null仅当没有单元格为空时，才在它们之间添加带有“，”的两列值
【发布时间】：2020-09-24 10:32:35
【问题描述】：

我有以下数据框：

>>>name   breakfast  lunch   dinner
0 Zoey    apple      egg     noodels
1 Rena    pear               pasta
2 Shila             tomato  potatoes
3 Daphni coffee             soup 
4 Dufi

我想创建一个新列，其中包含每个名字在同一天吃的所有食物值。我尝试使用“+”来做到这一点，并用“，”分隔单词，如下所示：

df['food']=df['breakfast']+','+df['lunch']+','+df['dinner']

但是如果我有空值，我在中间有'，'：


>>>name   breakfast  lunch   dinner     food
0 Zoey    apple      egg     noodels    apple,egg,noodels
1 Rena    pear               pasta      pear,,pasta
2 Shila             tomato  potatoes    ,tmatoe,potatoes
3 Daphni coffee             soup       coffee,,soupp
4. Dufi                                ,,

我想在正确的地方用 ',' 把它弄干净，例如不要放，如果有空：

>>>name   breakfast  lunch   dinner     food
0 Zoey    apple      egg     noodels    apple,egg,noodels
1 Rena    pear               pasta      pear,pasta
2 Shila             tomato  potatoes    tmatoe,potatoes
3 Daphni coffee             soup       coffee,soup
4 Dufi

有没有办法做到这一点？定义如果有空单元格不要添加它/不要放在错误的地方

【问题讨论】：

标签： python pandas cell add

【解决方案1】：

在您的索引上使用 .stack 和 groupby。

假设你的空白实际上是真正的空值

由于我们不需要名称，我们可以将其添加到索引中或将其删除，我已将其添加到此处。

df['food'] = df.set_index('name',append=True).stack().groupby(level=0).agg(','.join)

如果您的空格不是空值，我们可以这样做

df.replace(' ', np.nan).set_index('name',append=True).stack()\
                       .groupby(level=0).agg(','.join)

    name breakfast     lunch   dinner               food
0    Zoey     apple       egg  noodels  apple,egg,noodels
1    Rena      pear     pasta      NaN         pear,pasta
2   Shila    tomato  potatoes      NaN    tomato,potatoes
3  Daphni    coffee      soup      NaN        coffee,soup
4    Dufi       NaN       NaN      NaN                NaN

【讨论】：

【解决方案2】：

解决方案如果没有缺失值，只加入空字符串，只加入过滤空字符串的值：

cols = ['breakfast','lunch','dinner']
df['food'] = df[cols].apply(lambda x: ','.join(y for y in x if y != ''), axis=1)
print (df)
     name breakfast   lunch    dinner               food
0    Zoey     apple     egg   noodels  apple,egg,noodels
1    Rena      pear             pasta         pear,pasta
2   Shila            tomato  potatoes    tomato,potatoes
3  Daphni    coffee              soup        coffee,soup
4   Dufi

或者使用列表理解：

cols = ['breakfast','lunch','dinner']
df['food'] = [','.join(y for y in x if y != '') for x in df[cols].to_numpy()]
print (df)
     name breakfast   lunch    dinner               food
0    Zoey     apple     egg   noodels  apple,egg,noodels
1    Rena      pear             pasta         pear,pasta
2   Shila            tomato  potatoes    tomato,potatoes
3  Daphni    coffee              soup        coffee,soup
4   Dufi

如果缺失值相似的解决方案仅使用NaN != NaN：

cols = ['breakfast','lunch','dinner']
df['food'] = [','.join(y for y in x if y == y) for x in df[cols].to_numpy()]
print (df)
     name breakfast   lunch    dinner               food
0    Zoey     apple     egg   noodels  apple,egg,noodels
1    Rena      pear     NaN     pasta         pear,pasta
2   Shila       NaN  tomato  potatoes    tomato,potatoes
3  Daphni    coffee     NaN      soup        coffee,soup
4   Dufi        NaN     NaN       NaN

【讨论】：