Checking a Column If It Contains a Row Value

If Column Contains String then enter value for that row

We can use grepl to return a logical index by matching the 'D' in the 'A' column, and then with ifelse, change the logical vector to 'yes' and 'no'

df$C <- ifelse(grepl("D", df$A), "yes", "no")

check if row value value is equal to column name and access the value of the column

Use DataFrame.melt for alternative for lookup:

df1 = df.melt('colfromvaluestobepicked', ignore_index=False)

df['new']=df1.loc[df1['colfromvaluestobepicked'].str.strip("'") == df1['variable'],'value']

If statement that checking value in current row and column

If you want to check each row of the column to make after this some #code:

for row in df.iterrows():
   if t==row['A']:
     #code

If you only want to know in which rows of df['A'] the value of the element is t:

df.loc[df['A']==t, 'A']

or

df['A'].astype(str).str.contains('t')

EDIT

Check if it works for you:

df.loc[df['A']=='red','Value1']=df['Value']
df.loc[df['A']!='red','Value2']=df['Value']

If a dataframe contains a value in a column, how to perform a calculation in another column? - Pandas/ Python

You could check if column "A" values are 123 or not and use mask on "C" to replace values there:

df['C'] = df['C'].mask(df['A']==123, df['B']*0.0008)

Output:

     A     B    C
0  123  1500  1.2

Check if row has particular value in R

We can change the select subset of data to a logical matrix with ==, then apply the rowSums on the matrix, and check if the row wise sum is greater than 0

library(dplyr)
...
rowSums(select(., starts_with('type')) ==1, na.rm = TRUE) > 0

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In addition to the above, it can be done using if_any

...
if_any(starts_with('type'), ~. == 1)

Check if column contains (/,-,_, *or~) and split in another column - Pandas

Use Series.str.split with list of possible separators and create new DataFrame:

import re

L = ['-','/','*','~','_','^', '.']

#some values like `^.` are escape
pat = '|'.join(re.escape(x) for x in L)
df = df['column_name'].str.split(pat, expand=True).add_prefix('num')
print (df)
  num0 num1  num2
0  152    6     3
1  163    1     6
2  145    1  None
3  163    6     3

EDIT: If need match values before value use:

L = ["\-_",'\^|\*','~','/']

for val in L:
    df[f'before {val}'] = df['column_name'].str.extract(rf'(\d+){[val]}')

#for last value not exist separator, so match $ for end of string
df['last'] = df['column_name'].str.extract(rf'(\d+)$')
print (df)
   column_name before \-_ before \^|\* before ~ before / last
0  152/2~3_4*5          3            4        2      152    5
1  152/2~3-4^5          4            4        2      152    5
2      152/6*3        NaN            6      NaN      152    3
3      163/1-6        NaN          NaN      NaN      163    6
4        145/1        NaN          NaN      NaN      145    1
5      163/6^3          6            6      NaN      163    3

Check if columns have a nan value if certain column has a specific value in Dataframe

so you have an if-elif-else situation. Then we can use np.select for it. It needs the conditions and what to do when they are satisfied:

• your if is:    "condition is 1 and a,b,c has all nan"
• your elif is: "condition is nan"
• what remains is else, as usual

conditions = [df.condition.eq(1) & df[["a", "b", "c"]].isna().all(axis=1),
              df.condition.isna()]

what_to_do = ["O", "-"]
else_case = "X"

df["check_result"] = np.select(conditions, what_to_do, default=else_case)

df

   condition    a    b    c check_result
0        1.0  NaN  NaN  3.0            X
1        NaN  4.0    2  2.0            -
2        NaN  5.0    e  1.0            -
3        NaN  6.0    2  2.0            -
4        1.0  NaN  NaN  NaN            O

So we don't write else's condition. It goes to default.

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