return values from dataframe

Question

I have python pandas data frame like this with 200k to 400k rows

Index value
  1     a
  2 
  3     v
  4
  5
  6    6077
  7
  8     h

and I want this dataframe value column to be filled all below rows with the specific value based on number of string values(like here in this table we have 1 number of string value). I want my dataframe to be like this.

Index value
  1     a
  2     a 
  3     v
  4     v
  5     v
  6     v
  7     v
  8     h

Answer 1

If need repeat strings with length 1 you can use Series.str.match by regex [a-zA-Z]{1} for check if strings with length 1 , replace not matched values to NaN s by Series.where and last forward filling missing values by ffill :

df['value'] = df['value'].where(df['value'].str.match('^[a-zA-Z]{1}$', na=False)).ffill()
print (df)
   Index value
0      1     a
1      2     a
2      3     v
3      4     v
4      5     v
5      6     v
6      7     v
7      8     h

Another idea:

m1 = df['value'].str.len().eq(1)
m2 = df['value'].str.isalpha()
df['value'] = df['value'].where(m1 & m2).ffill()

Answer 2

The forward fill method in fillna is exactly for this. This should work for you:

df.fillna(method='ffill')

Answer 3

try this,

import pandas as pd

df['value'].replace('\d+', pd.np.nan, regex=True).ffill()

---

0    a
1    a
2    v
3    v
4    v
5    v
6    v
7    h
Name: value, dtype: object

Answer 4

Once you have removed all numbers, do this:

df[df['value']==""] = np.NaN
df.fillna(method='ffill')

Answer 5

Assuming that any value that is not an empty string or number should be forward filled, then the regular expression r'^\\d*$' will match both an empty string or number. These values can be replaced by np.nan and then ffill can be called:

import numpy as np

df['value'].replace(r'^\d*$', np.nan, regex=True, inplace=True)
df['value'].ffill(inplace=True)