Question

我有两个数据帧df1和df2。

DF1：

PartNumber
0000D3447E
0000D3447E
0000D3447E12
0000D3447E
0000D3447E
0000D3447E
0000D3447E2345
0000F2892E
0000F2892E
0000F2892E
0000F2892E34
0000F2892E
0000F2892E
0000F2892E12

DF2：

PartNumber
0000D3447E39S
0000D3447E39S
0000D3447E39S
0000D3447E39S
0000D3447E39S
0000D3447E39S
0000D3447E39S2245
0000F2892EDI1
0000F2892EDI1
0000F2892EDI1
0000F2892EDI124
0000F2892EDI1
0000F2892EDI1
0000F2892EDI1
0000D1617EAD6
0000D1617EAD6
0000D1617EAD6137
0000D1617EAD6
0000D1617EAD6
0000D1617EAD612
0000D1617EAD6
0000D3447EYG1
0000D3447EYG1
0000D3447EYG1
0000D3447EYG1
0000D3447EYG1
0000D3447EYG1
0000D3447EYG1

我需要从df1获取'0000D3447E'并获取包含此字符串的df2的列数，并通过创建新列来更新df1中的计数。

给定答案df1 ['count_of_colors'] = df1 ['PartNumber']。map（df2 ['PartNumber']。str [：10] .value_counts（））
可以解决问题，但字符串不是10个字符的常量。所以，我得到错误的字符串计数，因为字符串被限制是str [：10]字符。

感谢。

Answer 1

您可以使用地图

df1['count_of_colors'] = df1['PartNumber'].map(df2['PartNumber'].str[:10].value_counts())

DF1：

    PartNumber  count_of_colors
0   0000D3447E  14
1   0000D3447E  14
2   0000D3447E  14
3   0000D3447E  14
4   0000D3447E  14
5   0000D3447E  14
6   0000D3447E  14
7   0000F2892E  7
8   0000F2892E  7
9   0000F2892E  7
10  0000F2892E  7
11  0000F2892E  7
12  0000F2892E  7
13  0000F2892E  7

编辑：使用str.extract在df2中提取完全匹配，然后使用相同的解决方案

pat = '({})'.format('|'.join(df1['PartNumber'].unique()))

df2['PartMatch'] = df2['PartNumber'].str.extract(pat, expand = False)

df1['count_of_colors'] = df1['PartNumber'].map(df2['PartMatch'].value_counts())

您获得相同的输出，并且没有硬编码的字符数

Answer 2

我想你只需要这个

df1['count_of_colors'] =df1['PartNumber'].map(df2['PartNumber'].value_counts())

Answer 3

我的代码在这里。它适用于我..我无法复制你提供的所有数据，但是样本就在这里

public class Slot {

    private static Pattern textPattern = Pattern.compile("(\\w+) ([0-9:]+)-([0-9:]+)");
    private static DateTimeFormatter dayFormatter = DateTimeFormatter.ofPattern("EEE");

    public static Slot parse(String text) {
        Matcher textMatcher = textPattern.matcher(text);
        if (textMatcher.matches()) {
            DayOfWeek day = DayOfWeek.from(dayFormatter.parse(textMatcher.group(1)));
            LocalTime startTime = LocalTime.parse(textMatcher.group(2));
            LocalTime endTime = LocalTime.parse(textMatcher.group(3));
            return new Slot(day, startTime, endTime);
        } else {
            throw new IllegalArgumentException("Unparsable slot " + text + ", expected format Sun 12:00-14:00");
        }
    }

    private final DayOfWeek day;
    private final LocalTime startTime;
    private final LocalTime endTime;

    Slot(DayOfWeek day, LocalTime startTime, LocalTime endTime) {
        if (! endTime.isAfter(startTime)) {
            throw new IllegalArgumentException("End time must be after start time");
        }
        this.day = Objects.requireNonNull(day);
        this.startTime = startTime;
        this.endTime = endTime;
    }

    public DayOfWeek getDay() {
        return day;
    }

    public LocalTime getStartTime() {
        return startTime;
    }

    public LocalTime getEndTime() {
        return endTime;
    }

    /** 
     * @param nextSlot
     * @return A new Slot object representing the vacant Slot between this Slot and nextSlot,
     *      or an empty Optional if no gap
     */
    public Optional<Slot> slotBetween(Slot nextSlot) {
        if (! nextSlot.getDay().equals(day)) {
            throw new IllegalArgumentException("Cannot compare slots on different days of week");
        }
        if (nextSlot.getStartTime().isBefore(endTime)) {
            throw new IllegalArgumentException("Error: overlap between slots " + this + " and " + nextSlot);
        }
        if (nextSlot.getStartTime().isAfter(endTime)) { // there is a gap
            return Optional.of(new Slot(day, endTime, nextSlot.getStartTime()));
        } else {
            return Optional.empty();
        }
    }

    @Override
    public String toString() {
        return dayFormatter.format(day) + ' ' + startTime + '-' + endTime;
    }

}

获取列数包含pandas中的字符串

3 个答案: