如何根据另一个列表中两个元素的连续出现在列表中插入字符? Python [英] How to insert a character in a list, based on the consecutive appearance of two elements from another list? Python

问题描述

我有一个带有很多字符串的DataFrame和一个根据条件在字符串中插入一个字符的函数.问题是我的代码以某种方式错误地看到了输入字符串s.

I have a DataFrame with a lot of strings and a function that inserts a character in a string, according to a condition. The problem is that my code somehow sees the input string s wrongly.

列表s由两个列表中的元素组成:(1)成员列表(2)非成员列表.我的目标是在任何成员"后跟任意两个相应的非成员"时，在s中插入数字"1". s中最多可以添加一个"1".这是代码.

The list s consists of elements from two lists: (1) member list (2) non-member list. My goal is to insert the number '1' into s when any 'member' is followed by any two consequtive 'non-members'. A maximum of one '1' should be added to s. This is the code.

import pandas as pd
members = ['AA', 'BBB', 'CC', 'DDDD']
non_members = ['EEEE', 'FF', 'GGG', 'HHHHH', 'III', 'JJ']
s = ['AA', 'EEEE', 'GGG', 'FF']

所以我要为s实现的结果是获得以下输出:

So the result I am trying to achieve for s is to get the following output:

['AA', '1', 'EEEE', 'GGG', 'FF']

这是我的代码:

df = pd.DataFrame(s, columns =['string'])
d = df['string']

def func(row):
    out = ""
    look = 2
    for i in range(len(row)-look):
        out += row[i]
        if (row[i] in members) & \
           (row[i+1] in non_members) & \
           (row[i+2] in non_members):
            out += '1' + row[i+1:]
            break
    return out

e = d.apply(func)
print(e)

这将产生以下结果:

0      
1    EE
2     G
3      

但是我希望得到的是:

['AA', '1', 'EEEE', 'GGG', 'FF']

任何建议如何解决这个问题?

Any suggestions how to solve this?

这个问题与此有关:推荐答案

您可以使用shift找出满足该条件的位置.

You can use shift to figure out where that condition is met.

mask = ((df['string'].isin(members)) 
         & (df['string'].shift(-1).isin(non_members)) 
         & (df['string'].shift(-2).isin(non_members)))

插入这两个行之间的一种方法可能是执行类似的操作.

One way to insert in the rows in between might be to do something like this.

import numpy as np
df.set_index(df.index*2, inplace=True)

indices = mask[mask==True].index.values+1

df_add = pd.DataFrame(data=np.repeat(1, len(indices)), index=indices, columns=['string'])

pd.concat([df, df_add]).sort_index()
  string
0     AA
1      1
2   EEEE
4    GGG
6     FF

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