在其他列为NaN的情况下，填写相同数量的字符 [英] Fill in same amount of characters where other column is NaN

问题描述

我有以下虚拟数据框:

df = pd.DataFrame({'Col1':['a,b,c,d', 'e,f,g,h', 'i,j,k,l,m'],
                   'Col2':['aa~bb~cc~dd', np.NaN, 'ii~jj~kk~ll~mm']})

        Col1            Col2
0    a,b,c,d     aa~bb~cc~dd
1    e,f,g,h             NaN
2  i,j,k,l,m  ii~jj~kk~ll~mm

真实数据集的形状为 500000，90 .

我需要将这些值嵌套到行中，并且为此使用了新的 explode 方法，该方法工作正常.

I need to unnest these values to rows and I'm using the new explode method for this, which works fine.

问题是 NaN ，它们会在 explode 之后导致不相等的长度，因此我需要填写与填充值相同数量的定界符.在这种情况下， ~~~ 因为第1行具有三个逗号.

The problem is the NaN, these will cause unequal lengths after the explode, so I need to fill in the same amount of delimiters as the filled values. In this case ~~~ since row 1 has three comma's.

预期产量

        Col1            Col2
0    a,b,c,d     aa~bb~cc~dd
1    e,f,g,h             ~~~
2  i,j,k,l,m  ii~jj~kk~ll~mm

尝试1 :

df['Col2'].fillna(df['Col1'].str.count(',')*'~')

尝试2:

np.where(df['Col2'].isna(), df['Col1'].str.count(',')*'~', df['Col2'])

---

这可行，但我觉得有一种更简单的方法:

---

This works, but I feel like there's an easier method for this:

characters = df['Col1'].str.replace('\w', '').str.replace(',', '~')
df['Col2'] = df['Col2'].fillna(characters)

print(df)

        Col1            Col2
0    a,b,c,d     aa~bb~cc~dd
1    e,f,g,h             ~~~
2  i,j,k,l,m  ii~jj~kk~ll~mm

d1 = df.assign(Col1=df['Col1'].str.split(',')).explode('Col1')[['Col1']]
d2 = df.assign(Col2=df['Col2'].str.split('~')).explode('Col2')[['Col2']]

final = pd.concat([d1,d2], axis=1)
print(final)

  Col1 Col2
0    a   aa
0    b   bb
0    c   cc
0    d   dd
1    e     
1    f     
1    g     
1    h     
2    i   ii
2    j   jj
2    k   kk
2    l   ll
2    m   mm

---

问题:有没有更简单，更通用的方法?还是我的方法没问题?

---

Question: is there an easier and more generalized method for this? Or is my method fine as is.

推荐答案

pd.concat

delims = {'Col1': ',', 'Col2': '~'}
pd.concat({
    k: df[k].str.split(delims[k], expand=True)
    for k in df}, axis=1
).stack()

    Col1 Col2
0 0    a   aa
  1    b   bb
  2    c   cc
  3    d   dd
1 0    e  NaN
  1    f  NaN
  2    g  NaN
  3    h  NaN
2 0    i   ii
  1    j   jj
  2    k   kk
  3    l   ll
  4    m   mm

这会在 df 中的列上循环.在 delims 词典中的键上循环可能更明智.

This loops on columns in df. It may be wiser to loop on keys in the delims dictionary.

delims = {'Col1': ',', 'Col2': '~'}
pd.concat({
    k: df[k].str.split(delims[k], expand=True)
    for k in delims}, axis=1
).stack()

---

同一件事，不同的外观

delims = {'Col1': ',', 'Col2': '~'}
def f(c): return df[c].str.split(delims[c], expand=True)
pd.concat(map(f, delims), keys=delims, axis=1).stack()

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