Панды: Обновить столбец по условию с помощью регулярного выражения

Question

Итак, у меня есть столбец, который является типом объекта столбца в Pandas Dataframe .

Включает в себя следующие данные:

array(['9.4', '9.8', '10', '9.5', '10.5', '9.2', '9.9', '9.1', '9.3', '9',
   '9.7', '10.1', '10.6', '9.6', '10.8', '10.3', '13.1', '10.2',
   '10.9', '10.7', '12.9', '10.4', '13', '14', '11.5', '11.4', '12.4',
   '11', '12.2', '12.8', '12.6', '12.5', '11.7', '11.3', '12.3', '12',
   '11.9', '11.8', '8.7', '13.3', '11.2', '11.6', '11.1', '13.4',
   '12.1', '8.4', '12.7', '14.9', '13.2', '13.6', '13.5',
   '100.333.333.333.333', '9.55', '8.5', '110.666.666.666.667',
   '956.666.666.666.667', '10.55', '8.8', '135.666.666.666.667',
   '11.95', '9.95', '923.333.333.333.333', '9.25', '9.05', '10.75',
   '8.6', '8.9', '13.9', '13.7', '8', '8.0', '14.2', '11.94',
   '128.933.333.333.333', '114.666.666.666.667', '10.98',
   '114.333.333.333.333', '105.333.333.333.333',
   '953.333.333.333.333', '109.333.333.333.333',
   '113.666.666.666.667', '113.333.333.333.333',
   '973.333.333.333.333', '11.05', '9.75', '11.35', '11.45', '14.05',
   '123.333.333.333.333', '12.75', '13.8', '12.15', '13.05',
   '112.666.666.666.667', '105.666.666.666.667',
   '117.333.333.333.333', '11.75', '10.65', '109.666.666.666.667',
   '101.333.333.333.333', '10.15', '104.666.666.666.667',
   '116.333.333.333.333', '12.25', '11.85', '11.65', '13.55',
   '131.333.333.333.333', '120.666.666.666.667', '11.55',
   '963.333.333.333.333', '12.05'], dtype=object)

Я хочу обновить те, которые имеют более одной точки ('.'), Возможно, 0 или что-то еще. Я не очень знаком с regex -es, но идея состоит в том, чтобы использовать regex для этой проблемы вместо '953.333.333.333.333'!

DF.replace({'column': '953.333.333.333.333'},'0')

Большое спасибо!

Answer 1

Использование numpy.where с Series.str.count и Series.gt:

DF['column'] = np.where(DF['column'].str.count('\.').gt(1), 0, DF['column'])

[out]

array(['9.4', '9.8', '10', '9.5', '10.5', '9.2', '9.9', '9.1', '9.3', '9',
       '9.7', '10.1', '10.6', '9.6', '10.8', '10.3', '13.1', '10.2',
       '10.9', '10.7', '12.9', '10.4', '13', '14', '11.5', '11.4', '12.4',
       '11', '12.2', '12.8', '12.6', '12.5', '11.7', '11.3', '12.3', '12',
       '11.9', '11.8', '8.7', '13.3', '11.2', '11.6', '11.1', '13.4',
       '12.1', '8.4', '12.7', '14.9', '13.2', '13.6', '13.5', 0, '9.55',
       '8.5', 0, 0, '10.55', '8.8', 0, '11.95', '9.95', 0, '9.25', '9.05',
       '10.75', '8.6', '8.9', '13.9', '13.7', '8', '8.0', '14.2', '11.94',
       0, 0, '10.98', 0, 0, 0, 0, 0, 0, 0, '11.05', '9.75', '11.35',
       '11.45', '14.05', 0, '12.75', '13.8', '12.15', '13.05', 0, 0, 0,
       '11.75', '10.65', 0, 0, '10.15', 0, 0, '12.25', '11.85', '11.65',
       '13.55', 0, 0, '11.55', 0, '12.05'], dtype=object)

Answer 2

Это выражение может просто захватить . в нужных элементах:

'\d+\.\d+'|'\d+'|(\.)

с помощью этой группы захвата:

(\.)

Демо

Тест

# coding=utf8
# the above tag defines encoding for this document and is for Python 2.x compatibility

import re

regex = r"'\d+\.\d+'|'\d+'|(\.)"

test_str = ("'9.4', '9.8', '10', '9.5', '10.5', '9.2', '9.9', '9.1', '9.3', '9',\n"
    "   '9.7', '10.1', '10.6', '9.6', '10.8', '10.3', '13.1', '10.2',\n"
    "   '10.9', '10.7', '12.9', '10.4', '13', '14', '11.5', '11.4', '12.4',\n"
    "   '11', '12.2', '12.8', '12.6', '12.5', '11.7', '11.3', '12.3', '12',\n"
    "   '11.9', '11.8', '8.7', '13.3', '11.2', '11.6', '11.1', '13.4',\n"
    "   '12.1', '8.4', '12.7', '14.9', '13.2', '13.6', '13.5',\n"
    "   '100.333.333.333.333', '9.55', '8.5', '110.666.666.666.667',\n"
    "   '956.666.666.666.667', '10.55', '8.8', '135.666.666.666.667',\n"
    "   '11.95', '9.95', '923.333.333.333.333', '9.25', '9.05', '10.75',\n"
    "   '8.6', '8.9', '13.9', '13.7', '8', '8.0', '14.2', '11.94',\n"
    "   '128.933.333.333.333', '114.666.666.666.667', '10.98',\n"
    "   '114.333.333.333.333', '105.333.333.333.333',\n"
    "   '953.333.333.333.333', '109.333.333.333.333',\n"
    "   '113.666.666.666.667', '113.333.333.333.333',\n"
    "   '973.333.333.333.333', '11.05', '9.75', '11.35', '11.45', '14.05',\n"
    "   '123.333.333.333.333', '12.75', '13.8', '12.15', '13.05',\n"
    "   '112.666.666.666.667', '105.666.666.666.667',\n"
    "   '117.333.333.333.333', '11.75', '10.65', '109.666.666.666.667',\n"
    "   '101.333.333.333.333', '10.15', '104.666.666.666.667',\n"
    "   '116.333.333.333.333', '12.25', '11.85', '11.65', '13.55',\n"
    "   '131.333.333.333.333', '120.666.666.666.667', '11.55',\n"
    "   '963.333.333.333.333', '12.05'")

matches = re.finditer(regex, test_str, re.MULTILINE)

for matchNum, match in enumerate(matches, start=1):

    print ("Match {matchNum} was found at {start}-{end}: {match}".format(matchNum = matchNum, start = match.start(), end = match.end(), match = match.group()))

    for groupNum in range(0, len(match.groups())):
        groupNum = groupNum + 1

        print ("Group {groupNum} found at {start}-{end}: {group}".format(groupNum = groupNum, start = match.start(groupNum), end = match.end(groupNum), group = match.group(groupNum)))

# Note: for Python 2.7 compatibility, use ur"" to prefix the regex and u"" to prefix the test string and substitution.

Answer 3

Лучше, если вы напишите функцию, а затем используете df.column.apply(function).

Функция будет выглядеть так:

def fun(val):
    if (len(val.split('.')) > 2) :
        return '0'
    else:
        return val