[ad_1] You need merge with inner join: print(‘####CURRIES###’) df1 = pd.read_csv(‘C:\\O\\df1.csv’, index_col=False, usecols=[0,1,2], names=[“EW”, “WE”, “DA”], header=None) print(df1.head()) ####CURRIES### EW WE \ 0 can v can 1.90 1 Lanus U20 v Argentinos Jrs U20 2.10 2 Botafogo RJ U20 v Toluca U20 1.83 3 Atletico Mineiro U20 v Bahia U20 2.10 4 FC Porto v … Read more
[ad_1] Let’s take the following sample DataFrame, containing 2 groups of 3 adjacent rows: C1 C2 C3 C4 C5 C6 ABC NaN NaN NaN NaN PK KJ PQR NaN NaN RR SS NaN NaN MNO PO UI NaN NaN NaN NaN XXX AA NaN NaN NaN EE NaN XX1 NaN BB NaN DD NaN FF1 … Read more
[ad_1] If you only want groups of two (as opposed to groups of n), then you can hardcode n=2 and use a list comprehension to return a list of lists. This will also create a group of one at the end of the list if the length of the list is odd: some_list = [‘a’,’b’,’c’,’d’,’e’] … Read more
[ad_1] df2 = df1[df1.B.apply(lambda x:x == 1).astype(bool)] All other answers are missing the point (except for Wen’s, which is an ok alternative) 1 [ad_2] solved How do I filter an empty DataFrame and still keep the columns of that DataFrame?
[ad_1] try this: In [16]: df.ix[df.groupby(df[‘datetime’].dt.date)[‘production’].transform(‘nunique’) < 44 * 4 * 24, ‘datetime’].dt.date.unique() Out[16]: array([datetime.date(2015, 12, 7)], dtype=object) this will give you all rows for the “problematic” days: df[df.groupby(df[‘datetime’].dt.date)[‘production’].transform(‘nunique’) < 44 * 4 * 24] PS there is a good reason why people asked you for a good reproducible sample data sets – with the … Read more
[ad_1] Assuming you have a pandas dataframe, one option is to use pandas.crosstab to return another dataframe: import pandas as pd df = pd.read_csv(‘file.csv’) res = pd.crosstab(df[‘X’], df[‘Y’]) print(res) Y 0 1 X 0 3 7 1 1 3 A collections.Counter solution is also possible if a dictionary result is required: res = Counter(zip(df[‘X’].values, df[‘Y’].values)) … Read more
[ad_1] Perhaps you mean sth like this: words=” “.join(listvalue).upper().split() idx = df.value.str.upper().isin(words) In: df[idx] Out: variable value 48 Income salary 81 Shopping clothing 13 [ad_2] solved how to match string with dataframe in python [closed]
[ad_1] Given this dataframe: df.head() complete mid_c mid_h mid_l mid_o time 0 True 0.80936 0.80943 0.80936 0.80943 2018-01-31 09:54:10+00:00 1 True 0.80942 0.80942 0.80937 0.80937 2018-01-31 09:54:20+00:00 2 True 0.80946 0.80946 0.80946 0.80946 2018-01-31 09:54:25+00:00 3 True 0.80942 0.80942 0.80940 0.80940 2018-01-31 09:54:30+00:00 4 True 0.80944 0.80944 0.80944 0.80944 2018-01-31 09:54:35+00:00 Create a 50 moving … Read more
[ad_1] You were close. Try this: import pandas as pd df = pd.DataFrame({‘a’: [1, 2, 3], ‘b’: [ 4, 5, 2], ‘c’: [2, 4, 5]}) print(df) df = df[[x for _, x in sorted(zip(df.iloc[-1], df.columns), reverse=True)]] print(df) Starting DataFrame: a b c 0 1 4 2 1 2 5 4 2 3 2 5 Columns … Read more
[ad_1] You may iterate on the element of your list, and for each test the expression value-10 < x and x < value+10 that can be written in once in python value-10 < x < value+10 def search_numbers(values, x): for value in values: if value – 10 < x < value + 10: print(f”Ok: {value}-10 … Read more
[ad_1] Calculate Year on Year, Quarter on Quarter, Month on month number of Repeated, new, lost customers & theri revenue using pandas/python [ad_2] solved Calculate Year on Year, Quarter on Quarter, Month on month number of Repeated, new, lost customers & theri revenue using pandas/python
[ad_1] If you want to do all the columns in your dataframe: for col in df.columns: sns.countplot(df[col]) . Otherwise, following the pattern in your question: for i in range(1,11): column=’id_’+”{0}”.format(i).zfill(2) sns.countplot(df[column]) that will go through the numbers 1-10 and set column to the correct column name. zfill will make sure that for single digit numbers, … Read more
[ad_1] Yes, there is the split function, however before calling it on your string, you must get rid of the [ and ], or else instead of [‘1’, ‘2’, ‘3’, ‘4’], you will get [‘[1’, ‘2’, ‘3’, ‘4]’], so instead of s.split(), we do s[1:-1].split(), also this means your list is strings instead of ints … Read more
[ad_1] Something like this should work… df = pd.DataFrame({‘date’: [‘2017-01-01 01:01:01’, ‘2017-01-02 01:01:01’, ‘2017-01-03 01:01:01’, ‘2017-01-30 01:01:01’, ‘2017-01-31 01:01:01’], ‘value’: [99,98,97,95,94]}) df[‘date’] = pd.to_datetime(df[‘date’]) def get_list(row): subset = df[(row[‘date’] – df[‘date’] <= pd.to_timedelta(‘5 days’)) & (row[‘date’] – df[‘date’] >= pd.to_timedelta(‘0 days’))] return str(subset[‘value’].tolist()) df[‘list’] = df.apply(get_list, axis=1) Output: date value list 0 2017-01-01 01:01:01 99 … Read more
[ad_1] Use groupby (http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.groupby.html) Assume your input is saved in a pandas Dataframe (or equivalently save it into csv and read it using pandas.read_csv). Now you can loop over the groups with same S.No values with the following: output = {} for key, group in df.groupby(‘S.No.’): # print key # print group output[key] = {} … Read more