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如何删除文本文件中占列数少于 10% 的行？

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如何删除文本文件中占列数少于 10% 的行？

我对 Bash 还很陌生，请耐心回答我的问题（可能有点傻）。我有一个这样的文本文件（这里只是一小部分）：

                       type test    test    test    test    test    test    test    test    test    test    test    test    control control control control control control control control control control control control control control control control
Actinomyces_odontolyticus   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0.04306 0   0   0   0   0
Actinomyces_sp_HMSC035G02   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0.00575 0   0   0   0   0
Actinomyces_sp_HPA0247  0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0.01802 0   0   0   0   0
Actinomyces_sp_ICM47    0   0   0   0   0   0   0   0   0.00244 0   0   0   0   0   0   0   0   0   0   0   0   0   0.00347 0   0   0   0   0
Actinomyces_sp_S6_Spd3  0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0.01421 0   0   0   0   0
Actinomyces_sp_oral_taxon_181   0   0   0.00045 0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0.01219 0   0   0   0   0
Aeriscardovia_aeriphila 0   0   0.00786 0.00471 0   0   0   0.00118 0.00645 0.00918 0.01208 0   0.00153 0   0   0   0   0.00923 0   0.01527 0   0.00719 0.00423 0.00177 0   0.00468 0.0047  0.01937
Alloscardovia_omnicolens    0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
Bifidobacterium_adolescentis    0.06235 0.05427 0.78772 0.11693 0.03352 0.17129 0.23957 0.25833 0.16216 0.18002 2.27324 0.23361 0.38109 0   0.59227 0   0.46423 1.06198 0.20985 0   0.26431 0.7178  0   0   0.04301 0.27795 0.06356 0.54188
Bifidobacterium_angulatum   0   0   0   0.02457 0   0.03637 0   0   0   0   0   0   0   0   0.03184 0   0   0   0   0   0   0   0   0   0.00368 0   0   0
Bifidobacterium_bifidum 0   0   0   0   0   0   0   0   0   0.08402 0   0   0   0   0.06594 0   0   0   0   0   0   0   0   0   0   0   0   0

我想删除那些在至少 10% 的列（个体）中不存在的行（细菌）。这意味着，例如，如果我有 70 个个体，我想删除那些在至少 7 个个体中不存在（即 = 0）的细菌。

谁能帮我解释一些 Bash 命令？

答案1

您可以使用此命令执行此操作awk，其中file是您的初始文件，cleaned_file是结果文件：

awk '{nzeros=0; for(col=2; col<=NF; col++) {if($col == 0) {nzeros++}} {if(nzeros < 0.9 * (NF - 1)) {print $0}}}' file > cleaned_file

解释：

  1. nzeros=0：我们初始化一个变量，其中存储每行零的数量。

  2. for(col=2; col<=NF; col++) {if($col == 0) {nzeros++}}：对于每一行，我们从第二列（col=2- 第一列是细菌类型）循环到其末尾（col<=NF-NF是字段数，即总列数）。如果某列的值为 0（if($col == 0)），我们将该列的值增加nzeros1（nzeros++）。

  3. if(nzeros < 0.9 * (NF - 1)) {print $0}}：如果零的数量小于总列数减去第一个列的 90%（0.9）if(nzeros < 0.9 * (NF - 1))，则打印该行（print $0-$0表示整行awk）。

您的样本的输出是：

                       type test    test    test    test    test    test    test    test    test    test    test    test    control control control control control control control control control control control control control control control control
Aeriscardovia_aeriphila 0   0   0.00786 0.00471 0   0   0   0.00118 0.00645 0.00918 0.01208 0   0.00153 0   0   0   0   0.00923 0   0.01527 0   0.00719 0.00423 0.00177 0   0.00468 0.0047  0.01937
Bifidobacterium_adolescentis    0.06235 0.05427 0.78772 0.11693 0.03352 0.17129 0.23957 0.25833 0.16216 0.18002 2.27324 0.23361 0.38109 0   0.59227 0   0.46423 1.06198 0.20985 0   0.26431 0.7178  0   0   0.04301 0.27795 0.06356 0.54188
Bifidobacterium_angulatum   0   0   0   0.02457 0   0.03637 0   0   0   0   0   0   0   0   0.03184 0   0   0   0   0   0   0   0   0   0.00368 0   0   0

答案2

#!/bin/bash

s_DATA_FILE="remove_bacteria_sample_data.txt"

i_PEOPLE_NUMBER=$(head -n1 ${s_DATA_FILE} | awk '{print NF-1}')
# Be aware Bash does not support decimals so 10% of 28 people is 2
# I increased the example to 50% to get at least 2 results with your sample file
i_PERCENT=50
i_MAX_ZEROS=$((i_PERCENT*i_PEOPLE_NUMBER/100))
s_BACTERIA_LIST=$(awk '!(NR==1) { print $1 }' ${s_DATA_FILE})

echo "Found ${i_PEOPLE_NUMBER} People (test and control)"
echo "Max empty readings per Bacteria are ${i_PERCENT}%: ${i_MAX_ZEROS}"
echo

for s_BACTERIA in ${s_BACTERIA_LIST}
do
    # Please be aware that space after ${s_BACTERIA} is required to avoid matching names that start the same
    # Like if you add Actinomyces_sp_HPA0247 and Actinomyces_sp_HPA0247_2
    # Space makes sure Actinomyces_sp_HPA0247 will return only one row
    i_COUNT_ZEROS=$(grep "${s_BACTERIA} " ${s_DATA_FILE} | awk '{for(i=1; i<=NF; i++) if ($i==0) {i_count_zeros++}; print i_count_zeros; exit}')
    if [[ $i_COUNT_ZEROS -le $i_MAX_ZEROS ]]; then
        echo "* ${s_BACTERIA} meets the criteria with ${i_COUNT_ZEROS} people not being tested"
    else
        echo "- Not meeting the criteria ${s_BACTERIA} with ${i_COUNT_ZEROS} people not being tested"
    fi
done

这将返回：

./remove_bacteria.sh 
Found 28 People (test and control)
Max empty readings per Bacteria are 50%: 14

- Not meeting the criteria Actinomyces_odontolyticus with 27 people not being tested
- Not meeting the criteria Actinomyces_sp_HMSC035G02 with 27 people not being tested
- Not meeting the criteria Actinomyces_sp_HPA0247 with 27 people not being tested
- Not meeting the criteria Actinomyces_sp_ICM47 with 26 people not being tested
- Not meeting the criteria Actinomyces_sp_S6_Spd3 with 27 people not being tested
- Not meeting the criteria Actinomyces_sp_oral_taxon_181 with 26 people not being tested
* Aeriscardovia_aeriphila meets the criteria with 13 people not being tested
- Not meeting the criteria Alloscardovia_omnicolens with 28 people not being tested
* Bifidobacterium_adolescentis meets the criteria with 5 people not being tested
- Not meeting the criteria Bifidobacterium_angulatum with 24 people not being tested
- Not meeting the criteria Bifidobacterium_bifidum with 26 people not being tested

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