Add suffix to column names that don't already have a suffix - python

I have a data frame with columns like
Name Date Date_x Date_y A A_x A_y..
and I need to add _z to the columns (except the Name column) that don't already have _x or _y . So, I want the output to be similar to
Name Date_z Date_x Date_y A_z A_x A_y...
I've tried
df.iloc[:,~df.columns.str.contains('x|y|Name')]=df.iloc[:,~df.columns.str.contains('x|y|Name')].add_suffix("_z")
# doesn't add suffixes and replaces columns with all nans
df.columns=df.columns.map(lambda x : x+'_z' if "x" not in x or "y" not in x else x)
#many variations of this but seems to add _z to all of the column names

How about:
df.columns = [x if x=='Name' or '_' in x else x+'_z' for x in df.columns]

You can also try:
df.rename(columns = lambda x: x if x=='Name' or '_' in x else x+'_z')
stealing slightly from Quang Hoang ;)

Add '_z' where the column stub is duplicated and without a suffix.
m = (df.columns.str.split('_').str[0].duplicated(keep=False)
& ~df.columns.str.contains('_'))
df.columns = df.columns.where(~m, df.columns+'_z')

I would use index.putmask as follows:
m = (df.columns == 'Name') | df.columns.str[-2:].isin(['_x','_y'])
df.columns = df.columns.putmask(~m, df.columns+'_z')
In [739]: df.columns
Out[739]: Index(['Name', 'Date_z', 'Date_x', 'Date_y', 'A_z', 'A_x', 'A_y'], dty
pe='object')

Related

Search columns with list of string for a specific set of text and if the text is found enter new a new string of text in a new column

I want to search for names in column col_one where I have a list of names in the variable list20. When searching, if the value of col_one matches in list20, put the same name in a new column named new_col
Most of the time, the name will be at the front, such as ZEN, W, WICE, but there will be some names.
with a symbol after the name again, such as ZEN-R, ZEN-W2, ZEN13P2302A
my data
import pandas as pd
list20 = ['ZEN', 'OOP', 'WICE', 'XO', 'WP', 'K', 'WGE', 'YGG', 'W', 'YUASA', 'XPG', 'ABC', 'WHA', 'WHAUP', 'WFX', 'WINNER', 'WIIK', 'WIN', 'YONG', 'WPH', 'KCE']
data = {
"col_one": ["ZEN", "WPH", "WICE", "YONG", "K" "XO", "WIN", "WP", "WIIK", "YGG-W1", "W-W5", "WINNER", "YUASA", "WGE", "WFX", "XPG", "WHAUP", "WHA", "KCE13P2302A", "OOP-R"],
}
df = pd.DataFrame(data)
# The code you provided will give the result like the picture below. and it's not right
# or--------
df['new_col'] = df['col_one'].str.extract('('+'|'.join(list20)+')')[0]
# or--------
import re
pattern = re.compile(r"|".join(x for x in list20))
df = (df
.assign(new=lambda x: [re.findall(pattern, string)[0] for string in x.col_one])
)
# or----------
def matcher(col_one):
for i in list20:
if i in col_one:
return i
return 'na' #adjust as you see fit
df['new_col'] = df.apply(lambda x: matcher(x['col_one']), axis=1)
The result obtained from the code above and it's not right
Expected Output
Try to sort the list first:
pattern = re.compile(r"|".join(x for x in sorted(list20, reverse=True, key=len)))
(df
.assign(new=lambda x: [re.findall(pattern, string)[0] for string in x.col_one])
)
Try with str.extract
df['new'] = df['col_one'].str.extract('('+'|'.join(list20)+')')[0]
df
Out[121]:
col_one new
0 CFER CFER
1 ABCP6P45C9 ABC
2 LOU-W5 LOU
3 CFER-R CFER
4 ABC-W1 ABC
5 LOU13C2465 LOU
One way to do this, less attractive in terms of efficiency, is to use a simple function with a lambda such that:
def matcher(col_one):
for i in list20:
if i in col_one:
return i
return 'na' #adjust as you see fit
df['new_col'] = df.apply(lambda x: matcher(x['col_one']), axis=1)
df
expected results:
col_one new_col
0 CFER CFER
1 ABCP6P45C9 ABC
2 LOU-W5 LOU
3 CFER-R CFER
4 ABC-W1 ABC
5 LOU13C2465 LOU
Another approach:
pattern = re.compile(r"|".join(x for x in list20))
(df
.assign(new=lambda x: [re.findall(pattern, string)[0] for string in x.col_one])
)

Find percent difference between two columns, that share same root name, but differ in suffix

My question is somewhat similar to subtracting-two-columns-named-in-certain-pattern
I'm having trouble performing operations on columns that share the same root substring, without a loop. Basically I want to calculate a percentage change using columns that end with '_PY' with another column that shares the same name except for the suffix.
What's a possible one line solution, or one that doesn't involve a for loop?
url = r'https://www2.arccorp.com/globalassets/forms/corpstats.csv?1653338666304'
df = pd.read_csv(url)
df = df[df['TYPE'] == 'M']
PY_cols = [col for col in df.columns if col.endswith("PY")]
reg_cols = [col.split("_PY")[0] for col in PY_cols]
for k,v in zip(reg_cols,PY_cols):
df[f"{k}_YOY%"] = round((df[k] - df[v]) / df[v] * 100,2)
df
You can use:
v = (df[df.columns[df.columns.str.endswith('_PY')]]
.rename(columns=lambda x: x.rsplit('_', maxsplit=1)[0]))
k = df[v.columns]
out = pd.concat([df, k.sub(v).div(v).mul(100).round(2).add_suffix('_YOY%')], axis=1)
Gotta subset the df into the columns you need. Then zip will pull the pairs you need to do the percent calculation.
url = r'https://www2.arccorp.com/globalassets/forms/corpstats.csv?1653338666304'
df = pd.read_csv(url)
df = df[df['TYPE'] == 'M']
df_cols = [col for col in df.columns]
PY_cols = [col for col in df.columns if col.endswith("PY")]
# find the matching column, where the names match without the suffix.
PY_use = [col for col in PY_cols if col.split("_PY")[0] in df_cols]
df_use = [col.split("_PY")[0] for col in PY_use]
for k,v in zip(df_use,PY_use):
df[f"{k}_YOY%"] = round((df[k] - df[v]) / df[v] * 100,2)
You can take advantage of numpy:
py_df_array = (df[df_use].values, df[PY_use].values)
perc_dif = np.round((py_df_array[0] - py_df_array[1]) / py_df_array[1] * 100, 2)
df_perc = pd.DataFrame(perc_def, columns=[f"{col}_YOY%" for col in df_use])
df = pd.concat([df, df_perc], axis=1)

Python: strip pair-wise column names

I have a DataFrame with columns that look like this:
df=pd.DataFrame(columns=['(NYSE_close, close)','(NYSE_close, open)','(NYSE_close, volume)', '(NASDAQ_close, close)','(NASDAQ_close, open)','(NASDAQ_close, volume)'])
df:
(NYSE_close, close) (NYSE_close, open) (NYSE_close, volume) (NASDAQ_close, close) (NASDAQ_close, open) (NASDAQ_close, volume)
I want to remove everything after the underscore and append whatever comes after the comma to get the following:
df:
NYSE_close NYSE_open NYSE_volume NASDAQ_close NASDAQ_open NASDAQ_volume
I tried to strip the column name but it replaced it with nan. Any suggestions on how to do that?
Thank you in advance.
You could use re.sub to extract the appropriate parts of the column names to replace them with:
import re
df=pd.DataFrame(columns=['(NYSE_close, close)','(NYSE_close, open)','(NYSE_close, volume)', '(NASDAQ_close, close)','(NASDAQ_close, open)','(NASDAQ_close, volume)'])
df.columns = [re.sub(r'\(([^_]+_)\w+, (\w+)\)', r'\1\2', c) for c in df.columns]
Output:
Empty DataFrame
Columns: [NYSE_close, NYSE_open, NYSE_volume, NASDAQ_close, NASDAQ_open, NASDAQ_volume]
Index: []
You could:
import re
def cvt_col(x):
s = re.sub('[()_,]', ' ', x).split()
return s[0] + '_' + s[2]
df.rename(columns = cvt_col)
Empty DataFrame
Columns: [NYSE_close, NYSE_open, NYSE_volume, NASDAQ_close, NASDAQ_open, NASDAQ_volume]
Index: []
Use a list comprehension, twice:
step1 = [ent.strip('()').split(',') for ent in df]
df.columns = ["_".join([left.split('_')[0], right.strip()])
for left, right in step1]
df
Empty DataFrame
Columns: [NYSE_close, NYSE_open, NYSE_volume, NASDAQ_close, NASDAQ_open, NASDAQ_volume]
Index: []

Rename variously formatted column headers in pandas

I'm working on a small tool that does some calculations on a dataframe, let's say something like this:
df['column_c'] = df['column_a'] + df['column_b']
for this to work the dataframe need to have the columns 'column_a' and 'column_b'. I would like this code to work if the columns are named slightly different named in the import file (csv or xlsx). For example 'columnA', 'Col_a', ect).
The easiest way would be renaming the columns inside the imported file, but let's assume this is not possible. Therefore I would like to do some think like this:
if column name is in list ['columnA', 'Col_A', 'col_a', 'a'... ] rename it to 'column_a'
I was thinking about having a dictionary with possible column names, when a column name would be in this dictionary it will be renamed to 'column_a'. An additional complication would be the fact that the columns can be in arbitrary order.
How would one solve this problem?
I recommend you formulate the conversion logic and write a function accordingly:
lst = ['columnA', 'Col_A', 'col_a', 'a']
def converter(x):
return 'column_'+x[-1].lower()
res = list(map(converter, lst))
['column_a', 'column_a', 'column_a', 'column_a']
You can then use this directly in pd.DataFrame.rename:
df = df.rename(columns=converter)
Example usage:
df = pd.DataFrame(columns=['columnA', 'col_B', 'c'])
df = df.rename(columns=converter)
print(df.columns)
Index(['column_a', 'column_b', 'column_c'], dtype='object')
Simply
for index, column_name in enumerate(df.columns):
if column_name in ['columnA', 'Col_A', 'col_a' ]:
df.columns[index] = 'column_a'
with dictionary
dico = {'column_a':['columnA', 'Col_A', 'col_a' ], 'column_b':['columnB', 'Col_B', 'col_b' ]}
for index, column_name in enumerate(df.columns):
for name, ex_names in dico:
if column_name in ex_names:
df.columns[index] = name
This should solve it:
df=pd.DataFrame({'colA':[1,2], 'columnB':[3,4]})
def rename_df(col):
if col in ['columnA', 'Col_A', 'colA' ]:
return 'column_a'
if col in ['columnB', 'Col_B', 'colB' ]:
return 'column_b'
return col
df = df.rename(rename_df, axis=1)
if you have the list of other names like list_othername_A or list_othername_B, you can do:
for col_name in df.columns:
if col_name in list_othername_A:
df = df.rename(columns = {col_name : 'column_a'})
elif col_name in list_othername_B:
df = df.rename(columns = {col_name : 'column_b'})
elif ...
EDIT: using the dictionary of #djangoliv, you can do even shorter:
dico = {'column_a':['columnA', 'Col_A', 'col_a' ], 'column_b':['columnB', 'Col_B', 'col_b' ]}
#create a dict to rename, kind of reverse dico:
dict_rename = {col:key for key in dico.keys() for col in dico[key]}
# then just rename:
df = df.rename(columns = dict_rename )
Note that this method does not work if in df you have two columns 'columnA' and 'Col_A' but otherwise, it should work as rename does not care if any key in dict_rename is not in df.columns.

How Can I create three new columns for my data?

I've got some data that looks like
tweet_id worker_id option
397921751801147392 A1DZLZE63NE1ZI pro-vaccine
397921751801147392 A3UJO2A7THUZTV pro-vaccine
397921751801147392 A3G00Q5JV2BE5G pro-vaccine
558401694862942208 A1G94QON7A9K0N other
558401694862942208 ANMWPCK7TJMZ8 other
What I would like is a single line for each tweet id, and three 6 columns identifying the worker id and the option.
It the desired output is something like
tweet_id worker_id_1 option_1 worker_id_2 option_2 worker_id_3 option 3
397921751801147392 A1DZLZE63NE1ZI pro-vaccine A3UJO2A7THUZTV pro_vaccine A3G00Q5JV2BE5G pro_vaccine
How can I achieve this with pandas?
This is about reshaping data from long to wide format. You can create a grouped count column as id to spread as new column headers and then use pivot_table(), finally rename the columns by pasting the multi-level together.
df['count'] = df.groupby('tweet_id').cumcount() + 1
df1 = df.pivot_table(values = ['worker_id', 'option'], index = 'tweet_id',
columns = 'count', aggfunc='sum')
df1.columns = [x + "_" + str(y) for x, y in df1.columns]
An alternative option to pivot_table() is unstack():
df['count'] = df.groupby('tweet_id').cumcount() + 1
df1 = df.set_index(['tweet_id', 'count']).unstack(level = 1)
df1.columns = [x + "_" + str(y) for x, y in df1.columns]

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