How i can change invalid string pattern with default string in dataframe?

pandas replace
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pandas to_datetime

i have a dataframe like below.

name   birthdate
john   21011990
steve  14021986
alice  13020198

i want to detect invalid value in birthdate column then change value.

the birthdate column use date format is "DDMMYYYY" . but in dataframe have a invalid format also as "13020198","". i want to change invalid data to 31125000 .

i want result like below

name   birthdate
john   21011990
steve  14021986
bob    31125000
alice  31125000

thank you

You can first create non-valid date mask and then update their values:

mask = df.birthdate.apply(lambda x: pd.to_datetime(x, format='%d%m%Y', errors='coerce')).isna()

df.loc[mask, 'birthdate'] = 31125000

    name    birthdate
0   john    21011990
1   steve   14021986
2   bob     31125000
3   alice   31125000

pandas.Series.str.replace — pandas 1.1.0 documentation, Replace each occurrence of pattern/regex in the Series/Index. String can be a character sequence or regular expression. replstr flagsint, default 0 (no flags). Your first thing is easy just do df['col'] = df['col'].str.replace(',15','') for the second you can filter using a regex expression something like df[df['col'].str.contains(regex)] – EdChum Mar 13 '15 at 16:31

This would be my solution to keep the format you specify:

import pandas as pd
import numpy as np

data = {'name':['J','S','B','A'],'birthdate':[21011990,14021986,'',13020198]}
df = pd.DataFrame(data)
df['birthdate'] = pd.to_datetime(df['birthdate'],format='%d%m%Y',errors='coerce').astype(str)
df['birthdate'] = df['birthdate'].str.replace('-','',regex=True).replace('NaT',31125000,regex=True).astype(int)


  name  birthdate
0    J   19900121
1    S   19860214
2    B   31125000
3    A   31125000

Of course it'd be easier if you kept the datatime format, then you could simply use:

df['birthdate'] = pd.to_datetime(df['birthdate'],format='%d%m%Y',errors='coerce').fillna(31125000)

And you'd get:

  name            birthdate
0    J  1990-01-21 00:00:00
1    S  1986-02-14 00:00:00
2    B             31125000
3    A             31125000

pandas.DataFrame.replace — pandas 1.1.0 documentation, Regular expressions, strings and lists or dicts of such objects are also allowed. inplacebool, default False. If True, in place. Note: this will modify any other views on� Input can be 0 or 1 for Integer and ‘index’ or ‘columns’ for String inplace: It is a boolean which makes the changes in data frame itself if True. limit : This is an integer value which specifies maximum number of consequetive forward/backward NaN value fills.

Create mask by to_datetime with errors='coerce' and test missing values created if no matchin format, last set new values by Series.mask:

m = pd.to_datetime(df['birthdate'], format='%d%m%Y', errors='coerce').isna()
df['birthdate'] = df['birthdate'].mask(m, 31125000)

Or @Chris A solution from comments with DataFrame.loc:

df.loc[m, 'birthdate'] = 31125000

print (df)
    name birthdate
0   john  21011990
1  steve  14021986
2    bob  31125000
3  alice  31125000

How to convert Dataframe column type from string to date time , In this article we will discuss how to convert data type of a dataframe invalid parsing raise an exception; 'coerce': In case of invalid parsing set As this function can covert the data type of a series from string to datetime. For DOB_time column we provided time only, therefore it picked the default date i.e.� Using these functions, you can construct strings with definite patterns or even at random. You can change and modify them in any desired way. String Manipulation in R Programming. Here are a few of the string manipulation functions available in R’s base packages. We are going to look at these functions in detail. The nchar function; The

Change stringsAsFactors settings for data.frame, options(stringsAsFactors = FALSE). you change the global default setting. So every data frame you create after executing that line will not auto-convert to factors� We can also search less strict for all rows where the column ‘model’ contains the string ‘ac’ (note the difference: contains vs. match). df [df ['model']. str. contains ('ac')]

R Tip: Use stringsAsFactors = FALSE, R often uses a concept of factors to re-encode strings. This can be too early and too aggressive. re-encoding of strings by using stringsAsFactors = FALSE when creating data.frame s. As is often the case: base R works okay in default mode and works very well if you judiciously change a few defaults. Late to the party, but for posterity, the stringr package (part of the popular "tidyverse" suite of packages) now provides functions with harmonised signatures for string handling:

Handling Missing Data, Reserving a specific bit pattern in all available NumPy types would lead to an a floating-point value; there is no equivalent NaN value for integers, strings, or other types. We cannot drop single values from a DataFrame ; we can only drop full rows By default, dropna() will drop all rows in which any null value is present:. You can actually use directly map on the DataFrame. So you basically check the column 1 for the String tesla. If it's tesla, use the value S for make else you the current value of column 1. Then build a tuple with all data from the row using the indexes (zero based) (Row(row(0),make,row(2))) in my example) There is probably a better way to do it.

  • Why is 13020198 invalid? It follows the DDMMYYYY format
  • Do you want to preserve the dates as datetime, or keep the format?
  • df.loc[pd.to_datetime(df['birthdate'], format='%d%m%Y', errors='coerce').isna(), 'birthdate'] = '31125000' ..?
  • Wouldn't be enough to check if the last four digits are in a range, like between 1900 and 2020?